Quadratics

Study Sheet

Quadratics

Parabolas, factoring, square roots, completing the square, the quadratic formula, the discriminant, vertex form, and applications

What Is a Quadratic?

Concept
Definition and Standard Form

A quadratic function is one whose highest power of the variable is 22. Its standard form is

y=ax2+bx+c,a≠0.y = ax^2 + bx + c, \qquad a \neq 0.

A quadratic equation sets a quadratic equal to zero: ax2+bx+c=0ax^2+bx+c=0. The numbers aa, bb, and cc are the coefficients:

  • aa is the coefficient of x2x^2 (it can never be 00, or the x2x^2 would vanish);
  • bb is the coefficient of xx;
  • cc is the constant term.

A quadratic y=ax2+bx+cy = ax^2 + bx + c graphs as a parabola with a vertex at x=−b2ax = -\tfrac{b}{2a}. Solving ax2+bx+c=0ax^2 + bx + c = 0 can be done by factoring, by completing the square, or by the quadratic formula, which is completing the square done once and for all. The discriminant b2−4acb^2 - 4ac tells how many real roots there are before you solve.

Why must a≠0a \ne 0? The x2x^2 term is what makes the function turn around. A linear function changes by the same amount at every step; in a quadratic the change itself changes steadily (for y=x2y = x^2 the steps are 1,3,5,7,…1, 3, 5, 7, \ldots), and that is what bends the graph into a parabola. Remove the x2x^2 term and the bending disappears.

Example
Example: Identify aa, bb, cc in 3x2−5x+2=03x^2-5x+2=0

Match term by term:

a=3,b=−5,c=2.a = 3, \qquad b = -5, \qquad c = 2.

Carry the sign! The middle coefficient is −5-5, not 55.

Side note
Where it goes wrong

Reading bb from 3x2+2−5x=03x^2 + 2 - 5x = 0 as 22 because it is the second number written. The coefficients are defined by the power they multiply: aa goes with x2x^2, bb with xx, cc with no xx. Here a=3a = 3, b=−5b = -5, c=2c = 2.

Example
Example: A missing middle term, 2x2−7=02x^2-7=0

Here there is no xx term, so b=0b=0:

a=2,b=0,c=−7.a = 2, \qquad b = 0, \qquad c = -7.
Tip

How to recognize one: A relationship is quadratic exactly when, after simplifying, the highest power of the variable is 22. y=4x−1y=4x-1 is linear (highest power 11); y=x2−9y=x^2-9 is quadratic. If you can multiply it out to reach an x2x^2 term with a≠0a\neq0, it is quadratic.

Try it
Try it: standard form first

Write (x−4)(2x+1)=3x(x - 4)(2x + 1) = 3x in standard form and give aa, bb, cc.

Show the answer

Expand: 2x2+x−8x−4=2x2−7x−42x^2 + x - 8x - 4 = 2x^2 - 7x - 4. Move 3x3x over: 2x2−10x−4=02x^2 - 10x - 4 = 0. So a=2a = 2, b=−10b = -10, c=−4c = -4.

The Parabola: The Graph of a Quadratic

Concept
Shape, Direction, and Key Points

The graph of y=ax2+bx+cy=ax^2+bx+c is a smooth U-shaped curve called a parabola.

  • Direction: if a>0a>0 the parabola opens up (holds water, has a lowest point); if a<0a<0 it opens down (has a highest point).
  • Vertex: the turning point. Its xx-coordinate is x=−b2ax=-\dfrac{b}{2a}; find the yy-coordinate by substituting that xx back in.
  • Axis of symmetry: the vertical line x=−b2ax=-\dfrac{b}{2a} through the vertex; the two halves mirror each other.
  • Minimum vs. maximum: an up-parabola's vertex is a minimum; a down-parabola's vertex is a maximum.
  • yy-intercept: where the curve crosses the yy-axis; set x=0x=0 to get the point (0,c)(0,c).
  • xx-intercepts (roots / zeros): where the curve crosses the xx-axis; set y=0y=0 and solve ax2+bx+c=0ax^2+bx+c=0. A parabola can have two, one, or no xx-intercepts.

The axis of symmetry sits halfway between the roots, and that is where −b2a-\tfrac{b}{2a} comes from: a parabola is symmetric, so its two xx-intercepts are equally far from the axis, and by Vieta's formulas their average is 12⋅(−ba)=−b2a\tfrac12 \cdot \left(-\tfrac ba\right) = -\tfrac{b}{2a}. The formula keeps working even when there are no real roots, because the symmetry does not depend on crossing the axis.

Example
Example: Describe the parabola y=x2−6x+5y=x^2-6x+5
xy(3, −4)(1, 0)(5, 0)(0, 5)x = 3
Vertex (3,−4)(3, -4) on the axis x=3x = 3, halfway between the roots 11 and 55.

Here a=1a=1, b=−6b=-6, c=5c=5.

Direction: a=1>0 so it opens up (vertex is a minimum).Axis of symmetry: x=−b2a=−−62(1)=3.Vertex y: y=(3)2−6(3)+5=9−18+5=−4.Vertex: (3,−4).y-intercept: (0, 5) since c=5.\begin{aligned} \text{Direction: } & a=1>0 \text{ so it opens \textbf{up} (vertex is a minimum).} \\ \text{Axis of symmetry: } & x = -\frac{b}{2a} = -\frac{-6}{2(1)} = 3. \\ \text{Vertex $y$: } & y = (3)^2 - 6(3) + 5 = 9 - 18 + 5 = -4. \\ \text{Vertex: } & (3,-4). \\ \text{$y$-intercept: } & (0,\,5) \text{ since } c=5. \end{aligned}

The xx-intercepts come from x2−6x+5=(x−1)(x−5)=0x^2-6x+5=(x-1)(x-5)=0, giving (1,0)(1,0) and (5,0)(5,0).

Side note
Where it goes wrong

Dropping the sign in −b2a-\tfrac{b}{2a}: for y=x2−6x+5y = x^2 - 6x + 5, writing x=−62=−3x = \tfrac{-6}{2} = -3. With b=−6b = -6, −b2a=−−62=3-\tfrac{b}{2a} = -\tfrac{-6}{2} = 3. The roots 11 and 55 confirm it: the axis is halfway between them.

Tip

Two ways to describe the same curve: the axis of symmetry and the vertex share the same xx-value, x=−b2ax=-\frac{b}{2a}. Find that number once and you have both.

Try it
Try it: describe a parabola

For y=−x2+4x+5y = -x^2 + 4x + 5: direction, axis, vertex, yy-intercept, and xx-intercepts.

Show the answer

a=−1<0a = -1 < 0, so it opens down and the vertex is a maximum. Axis: x=−42(−1)=2x = -\tfrac{4}{2(-1)} = 2. Vertex yy: −4+8+5=9-4 + 8 + 5 = 9, so (2,9)(2, 9). yy-intercept (0,5)(0, 5). Roots: −x2+4x+5=−(x−5)(x+1)=0-x^2 + 4x + 5 = -(x - 5)(x + 1) = 0, so (5,0)(5, 0) and (−1,0)(-1, 0), symmetric about x=2x = 2.

Solving by Factoring (Zero Product Property)

Concept
The Zero Product Property

If a product of two factors equals zero, then at least one factor must be zero:

if A⋅B=0, then A=0  or  B=0.\text{if } A\cdot B = 0, \text{ then } A=0 \ \text{ or } \ B=0.

To use it, first make one side 00, factor the quadratic, set each factor equal to 00, and solve.

Zero is special among products: if neither factor is 00, the product cannot be 00 (a nonzero number times a nonzero number is nonzero). So the only way (x+3)(x+4)(x + 3)(x + 4) can be 00 is for one factor to be 00. No other number has this property, which is why the right side must be 00 before factoring helps: (x+3)(x+4)=2(x + 3)(x + 4) = 2 could happen in endless ways.

Example
Example: Solve x2+7x+12=0x^2+7x+12=0
xy−6−5−4−3−2−1−1123
The solutions are where y=(x+3)(x+4)y = (x + 3)(x + 4) touches 00: x=−4x = -4 and x=−3x = -3.

Factor into two binomials (two numbers that multiply to 1212 and add to 77: that is 33 and 44):

(x+3)(x+4)=0x+3=0  or  x+4=0x=−3  or  x=−4.\begin{aligned} (x+3)(x+4) &= 0 \\ x+3=0 \ \text{ or } \ x+4 &= 0 \\ x=-3 \ \text{ or } \ x &= -4. \end{aligned}

Check x=−3x=-3:  (−3)2+7(−3)+12=9−21+12=0.\ (-3)^2+7(-3)+12 = 9-21+12 = 0. ✓

Side note
Where it goes wrong

Solving x(x−5)=6x(x - 5) = 6 as "x=6x = 6 or x−5=6x - 5 = 6." The property needs a product equal to 00. Rewrite: x2−5x−6=0x^2 - 5x - 6 = 0, so (x−6)(x+1)=0(x - 6)(x + 1) = 0, giving x=6x = 6 or x=−1x = -1. The false answer x=11x = 11 gives 11⋅6=6611 \cdot 6 = 66, not 66.

Example
Example: A common factor first, x2+2x=0x^2+2x=0

Factor out xx:

x(x+2)=0 ⇒ x=0  or  x=−2.x(x+2)=0 \ \Rightarrow \ x=0 \ \text{ or } \ x=-2.

Do not divide both sides by xx—that would lose the solution x=0x=0.

Tip

Set it equal to zero first! The Zero Product Property only works against 00. Something like (x+3)(x+4)=2(x+3)(x+4)=2 tells you nothing; you must expand, move the 22 over, and refactor.

Try it
Try it: make it zero first

Solve x2=3x+10x^2 = 3x + 10.

Show the answer

Move everything to one side: x2−3x−10=0x^2 - 3x - 10 = 0. Two numbers that multiply to −10-10 and add to −3-3: −5-5 and 22. So (x−5)(x+2)=0(x - 5)(x + 2) = 0 and x=5x = 5 or x=−2x = -2. Check x=−2x = -2: 4=−6+104 = -6 + 10.

Solving by the Square-Root Method

Concept
Take the Square Root of Both Sides

When a quadratic has no plain xx term, isolate the squared quantity and take the square root of both sides—remembering both signs:

x2=k ⇒ x=±k,(x−h)2=k ⇒ x−h=±k.x^2 = k \ \Rightarrow \ x = \pm\sqrt{k}, \qquad\qquad (x-h)^2 = k \ \Rightarrow \ x-h = \pm\sqrt{k}.

This works whenever the variable appears only inside a square.

The ±\pm is there because squaring hides the sign: 424^2 and (−4)2(-4)^2 are both 1616. So "(x−2)2=16(x - 2)^2 = 16" only tells you that x−2x - 2 is some number whose square is 1616, and there are two of those. On the graph, the horizontal line y=16y = 16 crosses the parabola y=(x−2)2y = (x - 2)^2 twice, once on each side of its axis.

Example
Example: Solve x2=49x^2=49
x=±49=±7,so x=7 or x=−7.x = \pm\sqrt{49} = \pm 7, \qquad \text{so } x=7 \text{ or } x=-7.
Side note
Where it goes wrong

Taking the root of each term: x2+9=25x^2 + 9 = 25 solved as x+3=5x + 3 = 5. Isolate the square first: x2=16x^2 = 16, so x=±4x = \pm 4. The shortcut answer x=2x = 2 gives 4+9=13≠254 + 9 = 13 \ne 25.

Example
Example: A shifted square, (x−2)2=16(x-2)^2=16
xy−4−2246481216(−2, 16)(6, 16)
The line y=16y = 16 meets y=(x−2)2y = (x - 2)^2 at x=−2x = -2 and x=6x = 6, each 44 from the axis x=2x = 2.

Take the square root of both sides:

x−2=±16=±4x=2±4x=6  or  x=−2.\begin{aligned} x-2 &= \pm\sqrt{16} = \pm 4 \\ x &= 2 \pm 4 \\ x &= 6 \ \text{ or } \ x = -2. \end{aligned}
Example
Example: When the root is irrational, x2=12x^2=12
x=±12=±4⋅3=±23.x = \pm\sqrt{12} = \pm\sqrt{4\cdot 3} = \pm 2\sqrt{3}.

Always simplify the radical by pulling out perfect-square factors.

Tip

Never forget the ±\pm. Every positive number has two square roots, one positive and one negative. Dropping the ±\pm throws away a whole solution.

Try it
Try it: isolate, then root

Solve 3(x+1)2−5=223(x + 1)^2 - 5 = 22.

Show the answer

Add 55: 3(x+1)2=273(x + 1)^2 = 27. Divide by 33: (x+1)2=9(x + 1)^2 = 9. Square root: x+1=±3x + 1 = \pm 3, so x=2x = 2 or x=−4x = -4.

Solving by Completing the Square

Concept
Building a Perfect Square
x²3x3x9x3x3
x2+6xx^2 + 6x is a square with two strips; the corner 32=93^2 = 9 completes (x+3)2(x + 3)^2.

Completing the square rewrites x2+bxx^2+bx as a perfect square by adding (b2)2\left(\dfrac{b}{2}\right)^2. The steps for x2+bx+c=0x^2+bx+c=0:

  • Move the constant to the right side.
  • Add (b2)2\left(\frac{b}{2}\right)^2 to both sides.
  • Write the left side as (x+b2)2\left(x+\frac{b}{2}\right)^2.
  • Take the square root of both sides (±\pm!) and solve for xx.

If a≠1a\neq 1, divide every term by aa first.

Why (b2)2\left(\tfrac b2\right)^2? Draw x2+6xx^2 + 6x as a square of side xx plus a 66-by-xx strip. Cut the strip in half and attach the halves to two sides of the square: the shape is almost a square of side x+3x + 3, missing only a 33-by-33 corner. Adding that corner, (62)2=9\left(\tfrac62\right)^2 = 9, completes it: x2+6x+9=(x+3)2x^2 + 6x + 9 = (x + 3)^2. Adding 99 to one side forces adding it to the other.

Example
Example: Solve x2+6x−7=0x^2+6x-7=0 by completing the square
x2+6x=7move the constant over(62)2=32=9the number to addx2+6x+9=7+9add 9 to both sides(x+3)2=16left side is a perfect squarex+3=±4square root of both sidesx=−3±4x=1  or  x=−7.\begin{aligned} x^2+6x &= 7 &&\text{move the constant over} \\ \left(\tfrac{6}{2}\right)^2 &= 3^2 = 9 &&\text{the number to add} \\ x^2+6x+9 &= 7+9 &&\text{add $9$ to both sides} \\ (x+3)^2 &= 16 &&\text{left side is a perfect square} \\ x+3 &= \pm 4 &&\text{square root of both sides} \\ x &= -3 \pm 4 \\ x &= 1 \ \text{ or } \ x = -7. \end{aligned}

Check x=1x=1:  1+6−7=0\ 1+6-7=0. ✓

Side note
Where it goes wrong

Adding (b2)2\left(\tfrac b2\right)^2 to only one side: x2+6x+9=7x^2 + 6x + 9 = 7, so (x+3)2=7(x + 3)^2 = 7. The equation is no longer balanced; x=−3+7x = -3 + \sqrt7 fails the original (≈0.652+6(0.65)−7≠0\approx 0.65^2 + 6(0.65) - 7 \ne 0). Add 99 to both sides: (x+3)2=16(x + 3)^2 = 16.

Example
Example: An odd middle term, x2−3x−5=0x^2-3x-5=0
x2−3x=5(−32)2=94x2−3x+94=5+94=294(x−32)2=294x−32=±292x=3±292.\begin{aligned} x^2-3x &= 5 \\ \left(\tfrac{-3}{2}\right)^2 &= \tfrac{9}{4} \\ x^2-3x+\tfrac{9}{4} &= 5+\tfrac{9}{4} = \tfrac{29}{4} \\ \left(x-\tfrac{3}{2}\right)^2 &= \tfrac{29}{4} \\ x-\tfrac{3}{2} &= \pm\frac{\sqrt{29}}{2} \\ x &= \frac{3\pm\sqrt{29}}{2}. \end{aligned}

An odd bb just means the number you add, (b2)2\left(\frac{b}{2}\right)^2, is a fraction. That is fine.

Tip

Whatever you do to one side, do to the other. You must add (b2)2\left(\frac{b}{2}\right)^2 to both sides to keep the equation balanced. And if a≠1a\neq1, divide through by aa before completing the square.

Try it
Try it: complete the square

Solve x2−8x+3=0x^2 - 8x + 3 = 0 by completing the square.

Show the answer

Move the constant: x2−8x=−3x^2 - 8x = -3. Half of −8-8 is −4-4, and (−4)2=16(-4)^2 = 16; add it to both sides: x2−8x+16=13x^2 - 8x + 16 = 13, so (x−4)2=13(x - 4)^2 = 13. Then x−4=±13x - 4 = \pm\sqrt{13} and x=4±13x = 4 \pm \sqrt{13}.

The Quadratic Formula

Concept
The Formula That Always Works

For any quadratic equation ax2+bx+c=0ax^2+bx+c=0,

x=−b±b2−4ac2a.x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}.

It comes from completing the square on the general equation, so it solves every quadratic—even ones that will not factor nicely.

The formula is completing the square carried out on ax2+bx+c=0ax^2 + bx + c = 0 with letters instead of numbers, so it does in one line what the last section did in six (the derivation is in Going Deeper). Read it as two pieces: −b2a-\tfrac{b}{2a} is the axis of symmetry, and ±b2−4ac2a\pm\tfrac{\sqrt{b^2 - 4ac}}{2a} is how far the two roots sit on either side of it.

Example
Example: Solve 2x2+3x−2=02x^2+3x-2=0 with the formula
xy−3−2.5−2−1.5−1−0.50.51−4−3−2−112
The roots −2-2 and 12\tfrac12 sit 54\tfrac54 on either side of the axis x=−34x = -\tfrac34.

Identify a=2a=2, b=3b=3, c=−2c=-2, then substitute carefully:

x=−b±b2−4ac2a=−3±(3)2−4(2)(−2)2(2)=−3±9+164=−3±254=−3±54.\begin{aligned} x &= \frac{-b\pm\sqrt{b^2-4ac}}{2a} \\ &= \frac{-3\pm\sqrt{(3)^2-4(2)(-2)}}{2(2)} \\ &= \frac{-3\pm\sqrt{9+16}}{4} \\ &= \frac{-3\pm\sqrt{25}}{4} = \frac{-3\pm 5}{4}. \end{aligned}

Split the ±\pm: x=−3+54=12x=\dfrac{-3+5}{4}=\dfrac{1}{2} or x=−3−54=−2x=\dfrac{-3-5}{4}=-2.

Side note
Where it goes wrong

Dividing only the square root by 2a2a: x=−3±254x = -3 \pm \tfrac{\sqrt{25}}{4}, giving −1.75-1.75 and −4.25-4.25. The whole numerator, −b±D-b \pm \sqrt{D}, is over 2a2a: −3±54\tfrac{-3 \pm 5}{4}, giving 12\tfrac12 and −2-2.

Example
Example: An irrational answer, x2−4x+1=0x^2-4x+1=0

Here a=1a=1, b=−4b=-4, c=1c=1:

x=−(−4)±(−4)2−4(1)(1)2(1)=4±16−42=4±122=4±232=2±3.\begin{aligned} x &= \frac{-(-4)\pm\sqrt{(-4)^2-4(1)(1)}}{2(1)} \\ &= \frac{4\pm\sqrt{16-4}}{2} = \frac{4\pm\sqrt{12}}{2} \\ &= \frac{4\pm 2\sqrt{3}}{2} = 2\pm\sqrt{3}. \end{aligned}

Simplify 12=23\sqrt{12}=2\sqrt{3}, then divide every term of the top by the bottom.

Tip

Use parentheses when you substitute, especially for a negative bb: −(−4)=+4-(-4)=+4. Compute the part under the root, b2−4acb^2-4ac, first—and remember b2b^2 is always positive, even when bb is negative.

Try it
Try it: use the formula

Solve 3x2−2x−4=03x^2 - 2x - 4 = 0.

Show the answer

a=3a = 3, b=−2b = -2, c=−4c = -4. D=(−2)2−4(3)(−4)=4+48=52D = (-2)^2 - 4(3)(-4) = 4 + 48 = 52, and 52=213\sqrt{52} = 2\sqrt{13}. So x=2±2136=1±133x = \tfrac{2 \pm 2\sqrt{13}}{6} = \tfrac{1 \pm \sqrt{13}}{3}, about 1.541.54 and −0.87-0.87.

The Discriminant

Concept
The Number Under the Radical

The discriminant is the expression inside the square root of the quadratic formula:

D=b2−4ac.D = b^2 - 4ac.

Its sign tells you how many real solutions the equation has, before you ever finish solving:

  • D>0D>0: two different real solutions (parabola crosses the xx-axis twice);
  • D=0D=0: one real solution, a repeated root (parabola just touches the xx-axis);
  • D<0D<0: no real solutions (parabola never reaches the xx-axis).
Reminder — The quadratic formula:  x=−b±b2−4ac2a(a≠0)x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\qquad (a\neq 0)

The discriminant is the square of the distance between the roots, scaled: the roots are −b2a±D2a-\tfrac{b}{2a} \pm \tfrac{\sqrt D}{2a}. If D>0D > 0 the two roots are a positive distance apart; if D=0D = 0 they merge at the axis (the vertex touches the xx-axis); if D<0D < 0 there is no real square root, so no real roots at all.

Example
Example: How many real solutions does x2−3x−4=0x^2-3x-4=0 have?
D=(−3)2−4(1)(−4)=9+16=25>0.D = (-3)^2 - 4(1)(-4) = 9 + 16 = 25 > 0.

Since D>0D>0, there are two real solutions. (And because 2525 is a perfect square, they are even rational.)

Side note
Where it goes wrong

Forgetting that cc carries its sign: for x2−3x−4x^2 - 3x - 4, computing D=9−16=−7D = 9 - 16 = -7 and declaring no real roots. With c=−4c = -4, −4ac=−4(1)(−4)=+16-4ac = -4(1)(-4) = +16, so D=25D = 25, and the roots 44 and −1-1 exist.

Example
Example: A negative discriminant, x2+x+5=0x^2+x+5=0
xy−2−11234−7−6−5−4−3−2−11234567
Lowest curve x2−3x−4x^2 - 3x - 4 crosses twice (D=25D = 25); dashed x2−4x+4x^2 - 4x + 4 touches once (D=0D = 0); top x2+x+5x^2 + x + 5 misses (D=−19D = -19).
D=(1)2−4(1)(5)=1−20=−19<0.D = (1)^2 - 4(1)(5) = 1 - 20 = -19 < 0.

Since D<0D<0, there are no real solutions—the parabola sits entirely above the xx-axis.

Tip

The discriminant is a preview. It tells you the number and type of solutions without solving. A perfect-square DD also signals that the quadratic factors over the rationals.

Try it
Try it: count the roots

Without solving, how many real roots does each have: 4x2−12x+9=04x^2 - 12x + 9 = 0, 2x2+5x−1=02x^2 + 5x - 1 = 0, x2−2x+7=0x^2 - 2x + 7 = 0?

Show the answer

D=144−144=0D = 144 - 144 = 0: one repeated root. D=25+8=33>0D = 25 + 8 = 33 > 0: two roots (irrational, since 3333 is not a perfect square). D=4−28=−24<0D = 4 - 28 = -24 < 0: no real roots.

Vertex Form

Concept
Reading the Vertex Directly

The vertex form of a quadratic is

y=a(x−h)2+k,y = a(x-h)^2 + k,

where the vertex is (h,k)(h,k) and x=hx=h is the axis of symmetry. As always, a>0a>0 opens up (minimum at the vertex) and a<0a<0 opens down (maximum). Watch the sign: the form uses (x−h)(x-h), so y=(x+1)2−5y=(x+1)^2-5 has h=−1h=-1.

In y=2(x−3)2+4y = 2(x - 3)^2 + 4 the square (x−3)2(x - 3)^2 is never negative and equals 00 only at x=3x = 3. So yy is always at least 44 and reaches 44 exactly at x=3x = 3: that is the vertex, read off with no computation. The form also shows the graph as y=x2y = x^2 moved: shifted right 33, stretched by 22, raised 44.

Example
Example: Read the vertex of y=2(x−3)2+4y=2(x-3)^2+4
xy−11234524681012(3, 4)y = x²
The dashed y=x2y = x^2 shifted right 33, stretched by 22, and raised 44.

Compare to y=a(x−h)2+ky=a(x-h)^2+k: a=2a=2, h=3h=3, k=4k=4.

Vertex (3,4),opens up, so (3,4) is a minimum.\text{Vertex } (3,4), \quad \text{opens up, so } (3,4) \text{ is a \textbf{minimum}}.
Side note
Where it goes wrong

Reading the vertex of y=(x+2)2−3y = (x + 2)^2 - 3 as (2,−3)(2, -3). The form subtracts hh: (x+2)2=(x−(−2))2(x + 2)^2 = (x - (-2))^2, so h=−2h = -2 and the vertex is (−2,−3)(-2, -3). Check: x=−2x = -2 makes the square 00.

Example
Example: Convert y=x2−6x+5y=x^2-6x+5 to vertex form

Complete the square on the right side:

y=(x2−6x)+5=(x2−6x+9)+5−9add and subtract (62)2=9=(x−3)2−4.\begin{aligned} y &= (x^2-6x) + 5 \\ &= (x^2-6x+9) + 5 - 9 &&\text{add and subtract } \left(\tfrac{6}{2}\right)^2=9 \\ &= (x-3)^2 - 4. \end{aligned}

So the vertex is (3,−4)(3,-4)—matching what x=−b2ax=-\frac{b}{2a} gave us earlier.

Tip

Mind the signs. In y=a(x−h)2+ky=a(x-h)^2+k, the vertex is (h,k)(h,k) with the sign of hh flipped from what you see: y=(x+2)2−3y=(x+2)^2-3 has vertex (−2,−3)(-2,-3).

Try it
Try it: to vertex form

Write y=x2+10x+21y = x^2 + 10x + 21 in vertex form. Give the vertex and the minimum value.

Show the answer

Half of 1010 is 55, and 52=255^2 = 25. y=(x2+10x+25)+21−25=(x+5)2−4y = (x^2 + 10x + 25) + 21 - 25 = (x + 5)^2 - 4. The vertex is (−5,−4)(-5, -4), and since a=1>0a = 1 > 0 the minimum value is −4-4.

Applications

Concept
Turning Words into a Quadratic

Many real situations are modeled by quadratics:

  • Projectile height: an object thrown or launched has height h=−16t2+v0t+h0h=-16t^2+v_0 t + h_0 (feet, seconds), where v0v_0 is the starting speed and h0h_0 the starting height. Set h=0h=0 to find when it lands; use t=−b2at=-\frac{b}{2a} for the time of greatest height.
  • Area: length ×\times width often produces an x2x^2 term.

Solve, then keep only answers that make sense (a length or a time cannot be negative).

Height under gravity is quadratic because speed changes at a steady rate: every second the ball rises 3232 ft/s more slowly, so its height gains shrink by a constant amount, which is exactly the "change of the change is constant" signature of a quadratic. The top of the flight is where the rise has slowed to zero, the vertex, halfway between launch and landing.

Example
Example: A ball's height is h=−16t2+64th=-16t^2+64t
th123416324864(2, 64)(4, 0)
Up for 22 seconds to 6464 ft, down for 22 more.

When does it hit the ground? Set h=0h=0 and factor:

−16t2+64t=0−16t(t−4)=0t=0  or  t=4.\begin{aligned} -16t^2+64t &= 0 \\ -16t(t-4) &= 0 \\ t=0 \ \text{ or } \ t &= 4. \end{aligned}

t=0t=0 is the launch; the ball returns to the ground at t=4t=4 seconds.

When is it highest? At t=−b2a=−642(−16)=2t=-\frac{b}{2a}=-\frac{64}{2(-16)}=2 seconds, where h=−16(2)2+64(2)=64h=-16(2)^2+64(2)=64 feet.

Side note
Where it goes wrong

Keeping both roots of w(w+3)=54w(w + 3) = 54 and reporting "width −9-9 or 66." A width cannot be negative. The algebra is fine; it is the model that throws out −9-9.

Example
Example: A rectangle whose length is 33 more than its width has area 5454

Let the width be ww. Then w(w+3)=54w(w+3)=54:

w2+3w−54=0(w+9)(w−6)=0w=−9  or  w=6.\begin{aligned} w^2+3w-54 &= 0 \\ (w+9)(w-6) &= 0 \\ w=-9 \ \text{ or } \ w &= 6. \end{aligned}

A width cannot be negative, so w=6w=6: the rectangle is 66 by 99.

Tip

Check that your answer fits reality. A quadratic often gives two solutions, but a negative time or length is not physically possible—discard it and state only the meaningful answer with its units.

Try it
Try it: a launch

A stone is thrown upward from a 4848-ft cliff: h=−16t2+32t+48h = -16t^2 + 32t + 48. When does it hit the ground, and what is its greatest height?

Show the answer

Ground: −16t2+32t+48=0-16t^2 + 32t + 48 = 0; divide by −16-16: t2−2t−3=0t^2 - 2t - 3 = 0, so (t−3)(t+1)=0(t - 3)(t + 1) = 0 and t=3t = 3 s (discard −1-1). Top: t=−322(−16)=1t = -\tfrac{32}{2(-16)} = 1, h=−16+32+48=64h = -16 + 32 + 48 = 64 ft.

Going Deeper: Advanced Quadratics

Concept
Where the Quadratic Formula Comes From

The quadratic formula is not magic—it is completing the square done once and for all on the general equation ax2+bx+c=0ax^2+bx+c=0. Because the derivation never uses specific numbers, the result works for every quadratic. The key move is the same as before: turn x2+baxx^2+\frac{b}{a}x into a perfect square by adding (b2a)2\left(\frac{b}{2a}\right)^2 to both sides. Master this derivation and you never have to memorize the formula blindly again.

Example
Worked Example: Derive x=−b±b2−4ac2ax=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}

Start from the general equation and complete the square, tracking every step:

ax2+bx+c=0startx2+bax+ca=0divide every term by ax2+bax=−camove the constant overx2+bax+(b2a)2=(b2a)2−caadd (b2a)2 to both sides(x+b2a)2=b24a2−ca=b2−4ac4a2perfect square; common denominatorx+b2a=±b2−4ac2asquare root of both sides (±!)x=−b±b2−4ac2asubtract b2a, combine over 2a.\begin{aligned} ax^2+bx+c &= 0 &&\text{start} \\ x^2+\tfrac{b}{a}x+\tfrac{c}{a} &= 0 &&\text{divide every term by } a \\ x^2+\tfrac{b}{a}x &= -\tfrac{c}{a} &&\text{move the constant over} \\ x^2+\tfrac{b}{a}x+\left(\tfrac{b}{2a}\right)^2 &= \left(\tfrac{b}{2a}\right)^2-\tfrac{c}{a} &&\text{add } \left(\tfrac{b}{2a}\right)^2 \text{ to both sides} \\ \left(x+\tfrac{b}{2a}\right)^2 &= \frac{b^2}{4a^2}-\frac{c}{a} = \frac{b^2-4ac}{4a^2} &&\text{perfect square; common denominator} \\ x+\tfrac{b}{2a} &= \pm\frac{\sqrt{b^2-4ac}}{2a} &&\text{square root of both sides ($\pm$!)} \\ x &= \frac{-b\pm\sqrt{b^2-4ac}}{2a} &&\text{subtract } \tfrac{b}{2a}, \text{ combine over } 2a. \end{aligned}

The discriminant b2−4acb^2-4ac appears naturally as the numerator under the root—it was there in the algebra all along.

Concept
The Discriminant and the Nature of the Roots

For a quadratic with rational coefficients, the discriminant D=b2−4acD=b^2-4ac tells you not just how many roots, but what kind:

  • D>0D>0 and a perfect square: two rational roots (the quadratic factors over the rationals);
  • D>0D>0 but not a perfect square: two irrational roots, a conjugate pair −b2a±D2a\frac{-b}{2a}\pm\frac{\sqrt{D}}{2a};
  • D=0D=0: one repeated rational root (a “double root”); the parabola is tangent to the xx-axis;
  • D<0D<0: no real roots—two complex conjugate roots (see below).
Concept
Sum and Product of the Roots (Vieta's Formulas)

If r1r_1 and r2r_2 are the two roots of ax2+bx+c=0ax^2+bx+c=0, then

r1+r2=−ba,r1 r2=ca.r_1+r_2 = -\frac{b}{a}, \qquad\qquad r_1\, r_2 = \frac{c}{a}.

You can see why by expanding a(x−r1)(x−r2)=ax2−a(r1+r2)x+a r1r2a(x-r_1)(x-r_2)=ax^2-a(r_1+r_2)x+a\,r_1 r_2 and matching coefficients. For a monic quadratic (a=1a=1), this reads especially cleanly:

x2−(sum) x+(product)=0.x^2 - (\text{sum})\,x + (\text{product}) = 0.

These give a fast check on any answer, and let you build a quadratic straight from conditions on its roots.

Example
Worked Example: Build a quadratic from conditions

(a) Find a quadratic whose roots are 33 and −5-5. Use the sum and product:

sum=3+(−5)=−2,product=(3)(−5)=−15.\text{sum} = 3+(-5) = -2, \qquad \text{product} = (3)(-5) = -15.

So x2−(sum)x+(product)=x2−(−2)x+(−15)=x2+2x−15=0x^2-(\text{sum})x+(\text{product}) = x^2-(-2)x+(-15) = x^2+2x-15=0.

(b) Find a quadratic whose roots sum to 66 and multiply to 1010. Directly,

x2−6x+10=0.x^2-6x+10=0.

Check with the discriminant: D=(−6)2−4(1)(10)=36−40=−4<0D=(-6)^2-4(1)(10)=36-40=-4<0, so those two roots are a complex conjugate pair—consistent, since no two real numbers add to 66 and multiply to 1010.

Concept
Vertex Form for Maximum/Minimum Optimization

Rewriting y=ax2+bx+cy=ax^2+bx+c as y=a(x−h)2+ky=a(x-h)^2+k turns optimization into reading off the vertex: the extreme value of yy is exactly kk, reached at x=hx=h.

  • If a>0a>0, the vertex is the lowest point, so kk is the minimum value.
  • If a<0a<0, the vertex is the highest point, so kk is the maximum value.

This is the standard way to answer “what is the largest area / smallest cost / greatest height?”

Example
Worked Example: Maximize a rectangular area
wA5101520406080100(10, 100)
Area A=20w−w2A = 20w - w^2 peaks at w=10w = 10.

A farmer has 4040 m of fencing for a rectangular pen. What dimensions give the largest area? Let the width be ww; then the length is 20−w20-w (since 2w+2ℓ=402w+2\ell=40). The area is

A(w)=w(20−w)=20w−w2=−(w2−20w)=−(w2−20w+100)+100complete the square, add/subtract 100=−(w−10)2+100.\begin{aligned} A(w) &= w(20-w) = 20w-w^2 \\ &= -\left(w^2-20w\right) \\ &= -\left(w^2-20w+100\right)+100 &&\text{complete the square, add/subtract } 100 \\ &= -(w-10)^2 + 100. \end{aligned}

Since a=−1<0a=-1<0, the vertex (10,100)(10,100) is a maximum: the area is greatest at w=10w=10 m, giving a 10×1010\times 10 square with area 100\mathbf{100} m2. (A square always maximizes area for a fixed perimeter.)

Concept
A First Look at Complex Roots

When D=b2−4ac<0D=b^2-4ac<0, the quadratic formula asks for the square root of a negative number, which no real number provides. We introduce the imaginary unit

i=−1,soi2=−1.i = \sqrt{-1}, \qquad\text{so}\qquad i^2 = -1.

Then −16=16 −1=4i\sqrt{-16}=\sqrt{16}\,\sqrt{-1}=4i, and the formula still works. The two roots come out as a complex conjugate pair p±qip\pm qi: same real part p=−b2ap=-\frac{b}{2a}, opposite imaginary parts. Graphically, the parabola simply never crosses the xx-axis.

Example
Worked Example: Complex roots of x2+2x+5=0x^2+2x+5=0
xy−4−3−2−11−112345678(−1, 4)
Vertex (−1,4)(-1, 4) above the axis: no real crossings.

Here a=1a=1, b=2b=2, c=5c=5, and D=(2)2−4(1)(5)=4−20=−16<0D=(2)^2-4(1)(5)=4-20=-16<0:

x=−2±−162=−2±16 −12=−2±4i2=−1±2i.\begin{aligned} x &= \frac{-2\pm\sqrt{-16}}{2} \\ &= \frac{-2\pm\sqrt{16}\,\sqrt{-1}}{2} = \frac{-2\pm 4i}{2} \\ &= -1\pm 2i. \end{aligned}

The roots are −1+2i-1+2i and −1−2i-1-2i. Vieta check: their sum is −2=−ba-2=-\frac{b}{a} ✓ and their product is (−1)2−(2i)2=1−4i2=1+4=5=ca(-1)^2-(2i)^2=1-4i^2=1+4=5=\frac{c}{a} ✓.

Reminder — Vieta's formulas:  r1+r2=−ba,r1r2=car_1+r_2=-\frac{b}{a},\qquad r_1 r_2=\frac{c}{a}
Concept
The Line-Tangent-to-Parabola Condition

To find where a line y=mx+dy=mx+d meets a parabola y=ax2+bx+cy=ax^2+bx+c, set them equal and collect everything on one side to get a single quadratic in xx. The discriminant of that quadratic decides the geometry:

  • D>0D>0: the line cuts the parabola at two points (a secant);
  • D=0D=0: the line just touches it at one point—it is tangent;
  • D<0D<0: the line misses the parabola entirely.

So “tangent” is precisely the condition D=0D=0: one repeated intersection.

Example
Worked Example: For which kk is y=x+ky=x+k tangent to y=x2y=x^2?
xy−11−112
At k=−14k = -\tfrac14 the line touches y=x2y = x^2 once, at (12,14)\left(\tfrac12, \tfrac14\right).

Set the two expressions equal and move everything to one side:

x2=x+k⟹x2−x−k=0.x^2 = x+k \quad\Longrightarrow\quad x^2-x-k=0.

This is a quadratic in xx with a=1a=1, b=−1b=-1, c=−kc=-k. Tangency means one repeated root, i.e. D=0D=0:

D=(−1)2−4(1)(−k)=1+4k=0⟹k=−14.D=(-1)^2-4(1)(-k) = 1+4k = 0 \quad\Longrightarrow\quad k=-\tfrac{1}{4}.

At k=−14k=-\frac14 the equation becomes x2−x+14=(x−12)2=0x^2-x+\frac14=\left(x-\frac12\right)^2=0, so the single point of contact is x=12x=\frac12, y=14y=\frac14: the point (12,14)\left(\frac12,\frac14\right).

Tip

The discriminant ties it all together. The same expression b2−4acb^2-4ac counts real roots, classifies them as rational / irrational / complex, and—when applied to the “set them equal” equation—tells whether a line cuts, touches, or misses a parabola. And Vieta's formulas r1+r2=−bar_1+r_2=-\frac{b}{a}, r1r2=car_1 r_2=\frac{c}{a} give a fast check on any pair of roots, real or complex.

Why It Works

Completing the square: ax2+bx+c=0ax^2 + bx + c = 0 becomes x2+bax=−cax^2 + \tfrac bax = -\tfrac ca; add (b2a)2\left(\tfrac{b}{2a}\right)^2 to both sides so the left is a perfect square: (x+b2a)2=b2−4ac4a2\left(x + \tfrac{b}{2a}\right)^2 = \tfrac{b^2 - 4ac}{4a^2}. Take square roots and solve: x=−b±b2−4ac2ax = \tfrac{-b\pm\sqrt{b^2-4ac}}{2a}. The discriminant sits under the root, which is exactly why its sign decides whether there are two, one, or no real solutions.

Tip
Reading a proof

Do not read a proof the way you read a rule. Cover the last line, ask what the previous line forces, and only then uncover it. If a step feels like magic, that is the step to work by hand with a concrete number.

Formulas, Proofs & Tips

Tip
The quadratic formula
x=−b±b2−4ac2a(a≠0)x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\qquad (a\neq 0)

What it means. Solves every quadratic ax2+bx+c=0ax^2+bx+c=0, factorable or not.

Example. x2−5x+6=0x^2-5x+6=0: x=5±25−242x=\dfrac{5\pm\sqrt{25-24}}{2}, so x=3x=3 or x=2x=2.

Why it works. Complete the square. Divide by aa: x2+bax=−cax^2+\tfrac{b}{a}x=-\tfrac{c}{a}. Add (b2a)2\left(\tfrac{b}{2a}\right)^2 to both sides so the left is a perfect square: (x+b2a)2=b2−4ac4a2\left(x+\tfrac{b}{2a}\right)^2=\tfrac{b^2-4ac}{4a^2}. Take square roots and subtract b2a\tfrac{b}{2a}.

Tip. Write the equation as ax2+bx+c=0ax^2+bx+c=0 first — every term must be on one side, or your a,b,ca,b,c will be wrong.

Tip
The discriminant
Δ=b2−4ac\Delta = b^{2}-4ac

What it means. The part under the root. Δ>0\Delta>0: two real roots. Δ=0\Delta=0: one repeated root. Δ<0\Delta<0: two complex roots.

Example. x2−4x+4x^2-4x+4 has Δ=16−16=0\Delta=16-16=0, so one repeated real root.

Why it works. In the quadratic formula the roots differ by ±Δ2a\tfrac{\pm\sqrt{\Delta}}{2a}. A positive Δ\Delta gives two distinct real shifts, zero gives none, and a negative Δ\Delta forces the square root to be imaginary.

Tip. If Δ\Delta is a perfect square the quadratic factors over the integers — worth checking before reaching for the formula.

Tip
Vertex of a parabola
x=−b2a,y=f ⁣(−b2a)x=-\frac{b}{2a}, \qquad y=f\!\left(-\frac{b}{2a}\right)

What it means. The turning point — the minimum if a>0a>0, the maximum if a<0a<0.

Example. y=x2−6x+5y=x^2-6x+5: vertex at x=−−62=3x=-\tfrac{-6}{2}=3, y=−4y=-4, so (3,−4)(3,-4).

Why it works. The two roots of the quadratic formula sit symmetrically either side of −b2a-\tfrac{b}{2a}, so that value is the axis of symmetry. A parabola turns exactly on its axis of symmetry.

Tip. This also gives the max/min value of a quadratic without calculus — useful for word problems about greatest area or lowest cost.