Aniko

Connect quadratic roots and coefficients

Lesson progressPractice problems 0/3
Difficulty
Intermediate
Estimated time
28 minutes
Techniques
Roots-and-coefficientsSum-of-rootsProduct-of-rootsKnown-rootsRepeated-root

What you’ll learn

  1. Spot questions that connect the roots to the coefficients.
  2. Get the sum of the roots from −ba-\frac{b}{a} and their product from ca\frac{c}{a}.
  3. Build a quadratic from known roots, including a root that repeats.
  4. Decide when these patterns are quicker than Desmos, and when Desmos is quicker.

Prerequisites

Why this matters on the SAT

Answer from the coefficients, not the roots

The SAT sometimes asks for the sum or the product of a quadratic’s solutions without asking for either solution. Solving the whole equation first is work you don’t need.

Solution to the example

For any quadratic written as

ax2+bx+c=0,ax^2+bx+c=0,

the product of the roots is ca\frac{c}{a}. You’ll see why in a moment. Here,

a=6andc=−7,a=6\qquad\text{and}\qquad c=-7,

so the product is

ca=−76.\frac{c}{a}=\frac{-7}{6}.

The answer is B. You never found a root, and you never needed the middle coefficient, −11-11. That one matters for the sum, and the question asked for the product.

SAT example

What is the product of the solutions to the equation

6x2−11x−7=0?6x^2-11x-7=0?
  1. A

    −116-\frac{11}{6}

  2. B

    −76-\frac{7}{6}

  3. C

    76\frac{7}{6}

  4. D

    116\frac{11}{6}

Check what the question asks for first

Before you solve, ask one question: does it want the roots, or something about the roots?

  • If it asks for the sum or product of the roots, a missing coefficient, or something about a repeated root, work from the coefficients or the factors. You get the exact answer without finding both roots, the way ca\frac{c}{a} gave the product above in one step.
  • If it asks for the roots themselves and the quadratic doesn’t factor right away, graph it and click every xx-intercept.
  • If the coefficient work leaves you with an equation that’s hard to solve by hand, let Desmos solve that last equation.

Desmos approach: click the xx-intercepts

Here’s what the graph gives you for the opening equation. Enter

y=6x^2-11x-7

and click both xx-intercepts. Desmos shows their xx-coordinates as about −0.5-0.5 and 2.3332.333. Those are the two roots. When a question asks for the roots, or when those decimals are enough to pick out exact answer choices, this is the fast way.

The opening question asked for their product, though. Reading off two decimals and multiplying them takes longer than ca=−76\frac{c}{a}=-\frac76, and with 2.3332.333 rounded, you’d only get an estimate.

Calculator loads as you approach
Click both x-intercepts to see the roots themselves. For their sum or product, the coefficients are quicker.

See how factors, roots, and coefficients connect

Why does ca\frac{c}{a} give the product? It comes straight from the factors.

Say a quadratic has roots rr and ss. Each root comes from a factor:

x−randx−s.x-r\qquad\text{and}\qquad x-s.

Start with the simplest case, a monic quadratic, one whose leading coefficient (the number in front of x2x^2) is 11. Multiply the factors out:

(x−r)(x−s)=x2−sx−rx+rs=x2−(r+s)x+rs.\begin{aligned} (x-r)(x-s) &=x^2-sx-rx+rs\\[1.4em] &=x^2-(r+s)x+rs. \end{aligned}

Look at what came out:

  • The middle coefficient is the negative of the sum of the roots.
  • The constant term is the product of the roots.

Now let the leading coefficient be any number aa instead of 11. Multiply the whole product by aa:

a(x−r)(x−s)=a[x2−(r+s)x+rs]=ax2−a(r+s)x+ars.\begin{aligned} a(x-r)(x-s) &=a\left[x^2-(r+s)x+rs\right]\\[1.4em] &=ax^2-a(r+s)x+ars. \end{aligned}

Line this up with

ax2+bx+c,ax^2+bx+c,

term by term. The coefficients have to match:

b=−a(r+s)andc=ars.b=-a(r+s) \qquad\text{and}\qquad c=ars.

Solve each one for the root quantity, and you get

r+s=−baandrs=ca.\boxed{r+s=-\frac{b}{a}} \qquad\text{and}\qquad \boxed{rs=\frac{c}{a}}.

These are often called Vieta’s formulas for a quadratic. The name matters less than the pattern: the middle coefficient gives the sum, the constant gives the product, and each one is divided by the leading coefficient first. To keep the signs straight, remember: the sum takes a minus sign, the product doesn’t.

Common mistake:

Using ba\frac{b}{a} for the sum. The minus sign comes from the factors: they expand to −(r+s)x-(r+s)x, so b=−a(r+s)b=-a(r+s). Write the whole formula, r+s=−bar+s=-\frac{b}{a}, before you put in any numbers. You can test it on (x−2)(x−3)=x2−5x+6(x-2)(x-3)=x^2-5x+6: the roots add to 55, and −−51=5-\frac{-5}{1}=5.

Check your understanding:

For −4x2+20x+9=0-4x^2+20x+9=0, what are aa, bb, and cc? Then find the sum and the product of the roots.

Common mistake:

Reading the middle and constant terms straight off as the sum and product when a≠1a\ne1. Only a monic quadratic lets you read them off, and even then the sum is the negative of the middle coefficient. In the check above, the roots don’t add to ±20\pm20 or multiply to 99. Divide by aa first, and the sum still gets its minus sign.

Example: Find a missing coefficient from known roots

Worked example

The solutions to

3x2+kx−24=03x^2+kx-24=0

are 22 and −4-4. What is the value of kk?

  1. A

    −18-18

  2. B

    −6-6

  3. C

    66

  4. D

    1818

Step 1

Add the known roots

The question gives you both roots, and kk sits in the middle coefficient, so the sum is the one to use. Call the roots r=2r=2 and s=−4s=-4. Their sum is

r+s=2+(−4)=−2.r+s=2+(-4)=-2.

Putting −4-4 in parentheses keeps its minus sign from getting lost.

Step 2

Match each coefficient to its role

In

3x2+kx−24=0,3x^2+kx-24=0,

the leading coefficient is a=3a=3, and the middle coefficient is b=kb=k.

The sum formula ties them to the roots:

r+s=−ba.r+s=-\frac{b}{a}.

Step 3

Use the sum formula

Put in the sum and the coefficients:

−2=−k3,6=k.\begin{aligned} -2&=-\frac{k}{3},\\[1.4em] 6&=k. \end{aligned}

So k=6k=6, and the answer is C.

Step 4

Check by building the quadratic

The roots 22 and −4-4 come from the factors

(x−2)(x+4).(x-2)(x+4).

The leading coefficient is 33, so multiply the whole product by 33:

3(x−2)(x+4)=3(x2+2x−8)=3x2+6x−24.\begin{aligned} 3(x-2)(x+4) &=3(x^2+2x-8)\\[1.4em] &=3x^2+6x-24. \end{aligned}

The middle coefficient is 66, as expected, and the constant, −24-24, matches the equation too.

Common mistake:

Writing (x+2)(x−4)(x+2)(x-4) for roots 22 and −4-4. It’s easy to copy each root’s sign into its factor, but a root rr gives the factor x−rx-r, so the sign flips. To check a factor, plug in its root: x−2x-2 is 00 at x=2x=2, and x+4x+4 is 00 at x=−4x=-4.

Build a quadratic from its roots

Two known roots give you one monic quadratic:

x2−(r+s)x+rs=0.\boxed{x^2-(r+s)x+rs=0}.

For example, roots −3-3 and 55 have sum 22 and product −15-15, so they give

(x+3)(x−5)=0,x2−2x−15=0.\begin{aligned} (x+3)(x-5)&=0,\\[1.4em] x^2-2x-15&=0. \end{aligned}

But they also give every scaled version, like

2x2−4x−30=0,−7x2+14x+105=0,2x^2-4x-30=0,\qquad -7x^2+14x+105=0,

and so on. Multiplying a whole equation by a nonzero number doesn’t change which xx-values make it true. So the roots alone pin down one quadratic only when you also know the leading coefficient, or some other fact that sets the scale.

Check your understanding:

A quadratic has roots 11 and 66 and leading coefficient 44. What is the quadratic in standard form?

Count a repeated root twice

A repeated root happens when both factors are the same:

a(x−r)2=0.a(x-r)^2=0.

The equation has only one solution, but the factor x−rx-r shows up twice. So the sum and product count rr twice:

r+r=2randr⋅r=r2.r+r=2r \qquad\text{and}\qquad r\cdot r=r^2.

Here’s how that plays out. Suppose

x2+px+49=0x^2+px+49=0

has a repeated positive root. Here a=1a=1, so the product is 4949, and the repeated root satisfies

r2=49.r^2=49.

That allows r=7r=7 or r=−7r=-7, and the word positive picks r=7r=7. The sum is 2r=142r=14. The sum is also −p1=−p-\frac{p}{1}=-p, so

−p=14⟹p=−14.-p=14 \qquad\Longrightarrow\qquad p=-14.

Building the quadratic confirms it:

(x−7)2=x2−14x+49.(x-7)^2=x^2-14x+49.

What if a question says only “exactly one real solution” and tells you nothing about the root? Then the discriminant may be the quicker way in. Build from the repeated root when you know the root itself, or its sign, and can tie it to the coefficients.

Common mistake:

Using rr instead of 2r2r for the sum of a repeated root. There’s one distinct solution, but two copies of the factor x−rx-r. Above, a sum of 77 would give p=−7p=-7, and (x−7)2(x-7)^2 doesn’t expand to x2−7x+49x^2-7x+49. Write r+rr+r before you use the sum formula.

Practice problems

Each of these asks about the roots rather than for them, so start with the coefficients.

Use known roots to find a coefficient

Practice problem

The solutions to

4x2+bx−21=04x^2+bx-21=0

are 32\frac32 and −72-\frac72. What is the value of bb?

Answer choices
Calculator loads as you approach
Find b from the sum of the roots. Once you have b = 8, you can check it: graph y = 4x^2 + 8x - 21 and click both x-intercepts. They should be 1.5 and -3.5.

Use a repeated positive root

Practice problem

The equation

4x2+mx+25=04x^2+mx+25=0

has a repeated positive root. What is the value of mm?

Calculator loads as you approach
Count the repeated root twice to find m. Once you have m = -20, graph y = 4x^2 - 20x + 25. It should have one x-intercept, at x = 5/2.

Use a stated gap between roots

Practice problem

For a positive constant kk, the equation

(k+1)x2−6x+k−1=0(k+1)x^2-6x+k-1=0

has two distinct positive solutions. The greater solution is 12\frac12 greater than the lesser solution. What is the value of kk?

Calculator loads as you approach
Set up an equation in k with the sum and product first. Then graph y = (6/(x+1))^2 - 4((x-1)/(x+1)) - 1/4 and click both x-intercepts. Here x stands for k.

Finish the lesson

3 practice examples left

Finish the remaining questions correctly to complete this lesson.

Quick recap

  • In ax2+bx+c=0ax^2+bx+c=0, keep each coefficient’s sign: r+s=−bar+s=-\frac{b}{a} and rs=cars=\frac{c}{a}. The sum takes the minus sign.
  • Known roots give a(x−r)(x−s)a(x-r)(x-s), and aa sets the scale.
  • A repeated root counts twice, in the sum and in the factorization.
  • For a gap between two roots, use (r−s)2=(r+s)2−4rs(r-s)^2=(r+s)^2-4rs.
  • For the roots themselves, or a hard equation left at the end, use Desmos xx-intercepts. For a sum, product, or coefficient, go straight to the coefficients.

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