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Sums and Products of Roots: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 Totals from the coefficients

For $4x^2-6x-9=0$ , give the sum and the product of its roots.
1. Hint 1
  Both totals are already inside the coefficients, so no solving is needed here.
2. Hint 2
  Divide by the leading coefficient: the sum is $-\frac{b}{a}$ and the product is $\frac{c}{a}$ .
Answer

Sum $\tfrac32$ ; product $-\tfrac94$ .

Full solution

Read the coefficients from the standard form: $a=4$ , $b=-6$ , and $c=-9$ .
The roots add to $-\frac{b}{a}$ , which here is
$r+s=-\frac{-6}{4}.$
That simplifies to $\tfrac32$ .

The roots multiply to $\frac{c}{a}$ , so
$rs=\frac{-9}{4}.$
That is $-\tfrac94$ .

Answer

Sum $\tfrac32$ ; product $-\tfrac94$ .

Key idea

The sum $-\frac{b}{a}$ and the product $\frac{c}{a}$ are read straight off the coefficients, with no solving.
Problem 1: Mark complete
Problem 2 One root already known

The equation $5x^2+4=12x$ has $2$ as one of its two roots. Write it in standard form, find the other root from the coefficients, and confirm that root against the product of the roots.
1. Hint 1
  The two root totals can only be read once every term sits on one side.
2. Hint 2
  With the equation in standard form, the roots add to $-\frac{b}{a}$ ; subtract the known root from that total.
3. Hint 3
  The roots multiply to $\frac{c}{a}$ , and the known root is not zero, so the product gives an independent check.
Answer

Standard form $5x^2-12x+4=0$ ; the other root is $\tfrac25$ ; the product check gives $2\cdot\tfrac25=\tfrac45$ , matching $\tfrac{c}{a}=\tfrac45$ .

Full solution

Move every term to one side to reach standard form:
$5x^2-12x+4=0.$
Now $a=5$ , $b=-12$ , and $c=4$ .

The roots add to $-\frac{b}{a}$ , a total of $\tfrac{12}{5}$ .
Subtracting the known root leaves
$s=\frac{12}{5}-2.$
That gives
$s=\tfrac25.$

Check with the product.
The two roots multiply to
$2\cdot\frac{2}{5}=\frac{4}{5},$
which is $\frac{c}{a}$ , so the missing root is confirmed.

Answer

Standard form $5x^2-12x+4=0$ ; the other root is $\tfrac25$ ; the product check gives $2\cdot\tfrac25=\tfrac45$ , matching $\tfrac{c}{a}=\tfrac45$ .

Key idea

Standard form comes first; then the sum recovers a missing root, and the product checks it whenever the known root is not zero.
Problem 2: Mark complete
Problem 3 A pair from its totals

Two numbers have sum $-3$ and product $-18$ . Write a monic quadratic equation whose roots are those two numbers, then find the numbers.
1. Hint 1
  A monic quadratic is built from these two totals alone, with nothing else needed.
2. Hint 2
  Put the negative of the sum on $x$ , and put the product in the constant place.
3. Hint 3
  Factoring the equation you built recovers the two numbers themselves.
Answer

$x^2+3x-18=0$ ; the numbers are $-6$ and $3$ .

Full solution

The monic template is $x^2-(r+s)x+rs$ , so the sum $-3$ enters with its sign flipped and the product $-18$ keeps its own sign:
$x^2+3x-18=0.$

The two numbers are the roots of that equation.
Since $6$ and $-3$ have product $-18$ and sum $3$ , the left side factors as
$(x+6)(x-3)=0,$
so the numbers are $-6$ and $3$ .
Their sum is $-3$ and their product is $-18$ , as required.

Answer

$x^2+3x-18=0$ ; the numbers are $-6$ and $3$ .

Key idea

A pair of real numbers is pinned down by its sum and product through the monic quadratic $x^2-(r+s)x+rs$ .
Problem 3: Mark complete
Problem 4 A shifted pair

A quadratic has roots $-1$ and $6$ . Create the monic quadratic whose roots are each $3$ greater than those roots. Give the new equation in standard form and verify the roots.
1. Hint 1
  Find the two new root values before building the equation.
2. Hint 2
  Use the negative of their sum for the linear coefficient and their product for the constant.
Answer

$x^2-11x+18=0$ ; new roots: $2,9$ .

Full solution

The shifted roots are $2$ and $9$ .
Their sum is $11$ and their product is $18$ , so the equation is
$x^2-11x+18=0.$
At $x=2$ , the left side is $4-22+18=0$ .
At $x=9$ , it is
$81-99+18=0.$

Answer

$x^2-11x+18=0$ ; new roots: $2,9$ .

Key idea

Transform the roots first, then use their sum and product to build the new monic quadratic.
Problem 4: Mark complete
Problem 5 One root twice the other

The equation $x^2+bx+32=0$ has two positive roots, and one root is twice the other. Find both roots and the value of $b$ .
1. Hint 1
  Call the smaller root a single letter, then write the larger root in terms of it.
2. Hint 2
  For a monic quadratic the two roots multiply to the constant term, which gives one equation in that letter.
3. Hint 3
  Once both roots are known, the linear coefficient is the negative of their sum.
Answer

The roots are $4$ and $8$ , and $b=-12$ .

Full solution

Let the smaller root be $r$ , so the other root is $2r$ .
For a monic quadratic the roots multiply to the constant term, so
$2r^2=32.$

Dividing by $2$ gives $r^2=16$ , so $r=4$ or $r=-4$ .
The roots are positive, so $r=4$ , and the two roots are $4$ and $8$ .

The roots add to $-b$ , and their sum is $12$ , so
$b=-12.$
As a check, $x^2-12x+32$ factors as $(x-4)(x-8)$ .

Answer

The roots are $4$ and $8$ , and $b=-12$ .

Key idea

A condition linking the two roots turns the product relation into a single equation in one unknown.
Problem 5: Mark complete
Problem 6 Clearing the fractions

The two real roots of a quadratic equation add to $\tfrac74$ and multiply to $-\tfrac32$ . Write the monic equation with those roots. Then write an equation with the same roots whose coefficients are all integers, using the smallest positive integer leading coefficient that makes every coefficient an integer.
1. Hint 1
  The monic template takes the two totals exactly as they are, fractions included.
2. Hint 2
  Multiplying every term of an equation by a nonzero number leaves its roots unchanged.
3. Hint 3
  Choose the multiplier so that both $\tfrac74$ and $-\tfrac32$ become whole numbers.
Answer

Monic: $x^2-\tfrac74x-\tfrac32=0$ ; integer form: $4x^2-7x-6=0$ .

Full solution

The monic template is $x^2-(r+s)x+rs$ , so the sum enters with its sign flipped and the product keeps its own sign:
$x^2-\frac74x-\frac32=0.$

Multiplying every term by the same nonzero number does not change the roots.
Any multiplier that turns $\tfrac74$ into a whole number must be a multiple of $4$ , so the smallest positive integer that clears both denominators is $4$ , giving
$4x^2-7x-6=0.$

Check the integer form.
Its roots add to $-\frac{b}{a}=\tfrac74$ , and they multiply to $\frac{c}{a}=-\tfrac64$ , which is $-\tfrac32$ .

Answer

Monic: $x^2-\tfrac74x-\tfrac32=0$ ; integer form: $4x^2-7x-6=0$ .

Key idea

Build the monic equation first, then multiply through by the smallest number that clears the denominators, since scaling does not move the roots.
Problem 6: Mark complete
Problem 7 Factoring numbers and roots

The trinomial $x^2+12x+20$ factors as $(x+10)(x+2)$ . Give the two roots of $x^2+12x+20=0$ , then compare the sum and the product of those roots with the sum and the product of the factoring numbers $10$ and $2$ , and say for each total whether the two values agree.
1. Hint 1
  The factoring numbers are the entries printed inside the factors, while a root is a value that makes one factor zero.
2. Hint 2
  Solve $x+10=0$ and $x+2=0$ to get the two roots before comparing anything.
3. Hint 3
  Add and multiply each pair in turn, then set the four results beside the coefficients $b=12$ and $c=20$ .
Answer

Roots $-10$ and $-2$ . The factoring numbers have sum $12$ and product $20$ ; the roots have sum $-12$ and product $20$ , so only the sum differs.

Full solution

A root is a value that makes one factor zero.
Solving $x+10=0$ gives $x=-10$ , and solving $x+2=0$ gives $x=-2$ , so the roots are $-10$ and $-2$ .

The factoring numbers are $10$ and $2$ .
They add to $12$ and multiply to $20$ , which are the coefficients $b$ and $c$ of $x^2+12x+20$ .

Now do the same with the roots.
They add to
$(-10)+(-2)=-12.$
They multiply to
$(-10)(-2)=20.$

So both pairs multiply to $20$ , while the sums are $12$ for the factoring numbers and $-12$ for the roots.
That is what the two formulas say, since the roots of a monic trinomial add to $-b$ and multiply to $c$ , while the factoring numbers add to $b$ and multiply to $c$ .

Answer

Roots $-10$ and $-2$ . The factoring numbers have sum $12$ and product $20$ ; the roots have sum $-12$ and product $20$ , so only the sum differs.

Key idea

The roots of a monic trinomial are the negatives of its factoring numbers, so the two pairs share the product and have opposite sums.
Problem 7: Mark complete
Problem 8 A pair comparison

A monic quadratic has real roots $r,s$ , and another monic quadratic has roots $-r,-s$ . Jo says the constant coefficients are equal and the linear coefficients are opposite. Is this correct, including when the first quadratic's linear coefficient is zero? Explain.
1. Hint 1
  Compare the sums and products before and after negating both roots.
2. Hint 2
  The linear coefficient is the negative root sum; the constant coefficient is the root product.
3. Hint 3
  Check separately what happens when the two roots add to zero.
Answer

Yes, including when the first quadratic's linear coefficient is zero.

Full solution

The original coefficients are $-(r+s)$ and $rs$ .
The new root sum is $-(r+s)$ , while the new product is
$(-r)(-s)=rs.$
Thus the new linear coefficient is $r+s$ , the opposite of the original, and the constant is unchanged.
If the first quadratic's linear coefficient is zero, then $r+s$ is zero, and the opposite of zero is zero, so both linear coefficients are zero and are still opposites.

Answer

Yes, including when the first quadratic's linear coefficient is zero.

Key idea

Negating both roots reverses their sum and preserves their product.
Problem 8: Mark complete
Problem 9 The coefficient record

A monic quadratic has two distinct real roots. Their product is $0$ , and their sum is negative. Determine the sign of the linear coefficient and the value of the constant coefficient, and give one equation that meets all these conditions. Justify your conclusions.
1. Hint 1
  The linear coefficient is the negative of the root sum, and the constant coefficient is the root product.
2. Hint 2
  For an example, choose one zero root and one negative root.
Answer

Positive linear coefficient; constant coefficient $0$ . For example, $x^2+2x=0$ .

Full solution

Write the monic quadratic as
$x^2+bx+c=0.$
Since the root sum is $-b$ and is negative, $b$ is positive.
The root product equals $c$ , so
$c=0.$

For an example, choose roots $0$ and $-2$ .
Their sum is $-2$ and their product is $0$ , so they give
$x^2+2x=0.$
Its factorization $x(x+2)=0$ verifies two distinct real roots with all the required properties.

Answer

Positive linear coefficient; constant coefficient $0$ . For example, $x^2+2x=0$ .

Key idea

Signs and values of the root totals constrain the coefficients of a monic quadratic.
Problem 9: Mark complete
Problem 10 Two shared totals

Can two different monic quadratics with real roots have the same root sum? Can they have both the same root sum and the same root product? Give an example for any yes answer and justify any no answer.
1. Hint 1
  Write down what each of the two totals tells you about a monic quadratic.
2. Hint 2
  Try to build two different monic quadratics with the same sum, then with the same sum and product.
Answer

Same sum: yes, for example $x^2-10x+16$ and $x^2-10x+24$ . Same sum and product: no.

Full solution

The quadratics $x^2-10x+16$ and $x^2-10x+24$ have real roots $2,8$ and $4,6$ , respectively.
Both root sums are $10$ , but their products differ.
If the common sum is $S$ and common product is $P$ , each monic quadratic must be
$x^2-Sx+P.$
Therefore the same two totals force every coefficient to agree, so the quadratics cannot be different.

Answer

Same sum: yes, for example $x^2-10x+16$ and $x^2-10x+24$ . Same sum and product: no.

Key idea

A common root sum fixes the linear coefficient, while a common root sum and product fix the whole monic quadratic.
Problem 10: Mark complete

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