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Sum and Product 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 A rearranged root equation

The equation $3x(x-2)=x+2$ has real roots. Find their sum and product without finding the roots individually.
1. Hint 1
  The formulas use the coefficients after the equation is written with zero on one side.
2. Hint 2
  Divide the linear and constant coefficients by the leading coefficient, applying the minus sign only to the sum.
Answer

Sum $\frac73$ ; product $-\frac23$ .

Full solution

Expand and move every term to one side: $3x(x-2)=3x^2-6x$ , so $3x^2-6x=x+2$ becomes
$3x^2-7x-2=0$
Thus $a=3$ , $b=-7$ , $c=-2$ , giving sum $-b/a=7/3$ and product $c/a=-2/3$ .
The discriminant is $49+24=73>0$ , consistent with the stated real roots.

Answer

Sum $\frac73$ ; product $-\frac23$ .

Key idea

Collect the equation before reading root totals from its coefficients.
Problem 1: Mark complete
Problem 2 A shifted product

The real roots of $2x^2-5x-3=0$ are $r_1,r_2$ . Find $(r_1+1)(r_2+1)$ without solving for either root.
1. Hint 1
  Expand the requested product into the root sum and product.
2. Hint 2
  Read those two quantities from the coefficients.
Answer

$2$ .

Full solution

The roots have sum $5/2$ and product $-3/2$ .
Expanding $(r_1+1)(r_2+1)=r_1r_2+r_1+r_2+1$ and substituting gives
$-\tfrac32+\tfrac52+1=2$

The roots happen to be $3$ and $-1/2$ , whose shifted product is $4(1/2)=2$ , confirming the result.

Answer

$2$ .

Key idea

A product of equally shifted roots can be evaluated from their sum and product.
Problem 2: Mark complete
Problem 3 A root at zero

One root of $4x^2+7x=0$ is zero. Find the other root using the sum of the roots.
1. Hint 1
  The root sum remains useful when a product-based division would divide by zero.
2. Hint 2
  Subtract the known root from the total.
Answer

$-\frac74$ .

Full solution

The root sum is $-7/4$ .
Subtracting the known zero gives
$r_2=-\frac74$
A direct check comes from $4x^2+7x=x(4x+7)$ , whose other zero is $-7/4$ .

Answer

$-\frac74$ .

Key idea

Use the sum when recovering a partner from a zero root.
Problem 3: Mark complete
Problem 4 A family from two values

Find all quadratic polynomials whose roots are $-2$ and $3$ and whose coefficients are real. Give the family in standard form and state the restriction on its parameter.
1. Hint 1
  The two roots determine factors but leave a common multiplier free.
2. Hint 2
  Expand their product and retain every nonzero multiplier.
Answer

$a x^2-a x-6a$ , for every real $a\ne0$ .

Full solution

Every such quadratic has the form $a(x+2)(x-3)$ with $a\ne0$ .
Expanding gives
$f(x)=ax^2-ax-6a$
The nonzero condition preserves its degree and its zero set.
Conversely, each member vanishes exactly at the two prescribed roots.

Answer

$a x^2-a x-6a$ , for every real $a\ne0$ .

Key idea

Prescribed roots determine a quadratic up to a nonzero scalar.
Problem 4: Mark complete
Problem 5 A sum of shifted squares

The real roots of $x^2-10x+7=0$ are $r_1,r_2$ . Find $(r_1-5)^2+(r_2-5)^2$ without finding either root.
1. Hint 1
  Expand the target and combine terms symmetric in the two roots.
2. Hint 2
  Use the sum to remove the linear terms and the product to find the sum of squares.
Answer

$36$ .

Full solution

The sum is $10$ and the product is $7$ .
Expand each shifted square.
$\begin{aligned} (r_1-5)^2&=r_1^2-10r_1+25 \\ (r_2-5)^2&=r_2^2-10r_2+25 \end{aligned}$
Adding these, the target equals
$(r_1^2+r_2^2)-10(r_1+r_2)+50$

First find $r_1^2+r_2^2$ from the sum and product.
$\begin{aligned} r_1^2+r_2^2&=10^2-2(7) \\ &=86 \end{aligned}$
Substituting into the expanded target gives
$86-10(10)+50=36$
The axis is $x=5$ , and the two squared distances from it are equal.

Answer

$36$ .

Key idea

Symmetric combinations of shifted roots reduce to their original sum and product.
Problem 5: Mark complete
Problem 6 Doubling both roots

Let $r_1,r_2$ be the roots of $x^2+3x-2=0$ . Find the monic quadratic whose roots are $2r_1$ and $2r_2$ , in standard form.
1. Hint 1
  Write the new roots $2r_1,2r_2$ in terms of the old ones and substitute into the definitions of the sum and the product.
2. Hint 2
  Insert the new sum and product into a monic quadratic.
Answer

$x^2+6x-8$ .

Full solution

The original sum is $-3$ and product is $-2$ .
Doubling both roots gives a sum of $-6$ and product $4(-2)=-8$ .
The required monic polynomial is
$x^2+6x-8$
As a check, it equals $4f(x/2)$ for $f(x)=x^2+3x-2$ , so each original root doubled makes it zero.

Answer

$x^2+6x-8$ .

Key idea

Scaling roots multiplies their sum by the scale and their product by its square.
Problem 6: Mark complete
Problem 7 The spacing of two roots

The real roots of $3x^2-6x-2=0$ are $r_1,r_2$ . Find their distance $|r_1-r_2|$ exactly.
1. Hint 1
  First find the squared difference from the sum and product.
2. Hint 2
  Distance uses the nonnegative square root of that squared difference.
Answer

$\frac{2\sqrt{15}}3$ .

Full solution

The sum is $2$ and product is $-2/3$ .
$\begin{aligned} (r_1-r_2)^2&=2^2-4(-\tfrac23) \\ (r_1-r_2)^2&=\frac{20}{3} \end{aligned}$

The nonnegative square root is $2\sqrt{15}/3$ .
Equivalently, $\Delta=60$ and the root distance is $\sqrt{60}/3$ , the same value.

Answer

$\frac{2\sqrt{15}}3$ .

Key idea

A root distance is the nonnegative square root of the discriminant divided by the leading coefficient’s magnitude.
Problem 7: Mark complete
Problem 8 A coefficient report

For $f(x)=5x^2+10x-3$ , a student expands $5(x-r_1)(x-r_2)$ and reports $r_1+r_2=-10$ and $r_1r_2=-3$ . Is that report correct? Show the coefficient comparisons that support your answer.
1. Hint 1
  The leading multiplier affects the linear and constant coefficients as well as the squared one.
2. Hint 2
  Compare the expanded product with each matching coefficient before isolating the root totals.
Answer

No; $r_1+r_2=-2$ and $r_1r_2=-\frac35$ .

Full solution

The discriminant is $10^2-4(5)(-3)=160>0$ , so the two roots are real.
Expanding the factored form makes the linear coefficient $-5(r_1+r_2)$ and the constant $5r_1r_2$ .
Match these to the given rule.

$-5(r_1+r_2)=10$ $5r_1r_2=-3$
Dividing the first equation by $-5$ and the second by $5$ gives the sum $-2$ and product $-3/5$ .
The report omitted this division.

Answer

No; $r_1+r_2=-2$ and $r_1r_2=-\frac35$ .

Key idea

Coefficient matching must include the quadratic’s leading multiplier in both root totals.
Problem 8: Mark complete
Problem 9 A reported separation

A quadratic with leading coefficient $-3$ has two real roots 4 units apart. A student reports its discriminant as $-48$ because the leading coefficient is negative. Is this correct? Find the discriminant.
1. Hint 1
  The discriminant relates to the square of the leading coefficient and the square of the gap.
2. Hint 2
  Track which signs survive squaring.
Answer

No; the discriminant is $144$ .

Full solution

The squared-root-gap identity is
$\Delta=a^2(r_1-r_2)^2.$
Here
$\begin{aligned} \Delta&=(-3)^2(4)^2 \\ \Delta&=144 \end{aligned}$

A negative leading coefficient reverses the parabola but does not make a squared distance negative.

Answer

No; the discriminant is $144$ .

Key idea

The discriminant measures squared root separation with a squared leading coefficient.
Problem 9: Mark complete
Problem 10 A negative product target

For every real $s$ and every real $p<0$ , a designer expects two distinct real numbers whose sum is $s$ and whose product is $p$ . Is that expectation valid? Explain using a quadratic that would have those roots.
1. Hint 1
  Build the monic quadratic whose coefficient pattern would encode the desired sum and product.
2. Hint 2
  Check the condition for its two roots to be real and distinct.
Answer

Yes; $x^2-sx+p=0$ has two distinct real roots with the required sum and product.

Full solution

The desired quadratic is
$x^2-sx+p=0$
Its discriminant is $s^2-4p$ .
Since $s^2\ge0$ and $p<0$ ,
$s^2-4p>0$
Therefore it has two distinct real roots.
The coefficient formulas give their sum as $s$ and their product as $p$ , so the construction meets both requirements.
The general existence condition is $s^2-4p\ge0$ , and these hypotheses satisfy it strictly.

Answer

Yes; $x^2-sx+p=0$ has two distinct real roots with the required sum and product.

Key idea

A negative product makes the real-pair existence condition hold for every real target sum.
Problem 10: Mark complete

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