Roots, Sums, and Products

A monic quadratic with roots $r$ and $s$ factors as $(x - r) (x - s)$ , 9.2's factorization written with the roots rather than the search numbers. Expanding gives $x^{2} - (r + s) x + r s$ . Against $x^{2} + b x + c$ , $b = - (r + s)$ and $c = r s$ , so the roots add to $- b$ and multiply to $c$ . In $x^{2} - 10 x + 21 = (x - 3) (x - 7)$ the roots add to $10$ and multiply to $21$ .

Problem

The roots of

x^{2} - 17 x + 52

are

r

and

s

. Find

r + s

without factoring the quadratic and without finding either root.

Show a hint

Multiply $(x - r) (x - s)$ out. The coefficient of $x$ in the expansion is $- (r + s)$ .
Match that against the coefficient of $x$ in $x^{2} - 17 x + 52$ , which is $- 17$ with its sign, and solve for $r + s$ .

Show the full solution

Expanding,

(x - r) (x - s) = x^{2} - (r + s) x + r s

, and matching the middle coefficients gives

- (r + s) = - 17

, so

r + s = 17

. Watch the sign, the coefficient of

x

- 17

, so the sum is its opposite and not

- 17

itself.

Problem

The roots of

x^{2} + 6 x - 91

are

r

and

s

. Match the quadratic against

(x - r) (x - s)

expanded and read off the product

r s

. Enter it as an integer.

Show a hint

Match $x^{2} + 6 x - 91$ against the expanded form $(x - r) (x - s) = x^{2} - (r + s) x + r s$ . The product of the roots sits in the constant slot.
For a second route, factor the quadratic into two whole-number factors, read a root off each factor, and multiply the two roots.

Show the full solution

Matching

x^{2} + 6 x - 91

with

(x - r) (x - s) = x^{2} - (r + s) x + r s

puts the product in the constant slot, so

r s = - 91

. In

x^{2} - (r + s) x + r s

the sign flip is on the sum term only, so the constant is the product with no change. Factoring gives

(x + 13) (x - 7)

, so the roots are

- 13

and

7

, and

- 13 \cdot 7 = - 91

Problem

The numbers

6

and

17

multiply to

102

and add to

23

, so

x^{2} + 23 x + 102 = (x + 6) (x + 17)

. What is the sum of the two roots?

Show a hint

The two numbers given are the constants inside the factors. A root is a value of $x$ that makes a factor zero, which is a different thing.
$x + 6$ is zero at $x = - 6$ , so the roots are $- 6$ and $- 17$ . Add those, or read $- b$ straight off the quadratic.

Show the full solution

The sum of the roots is

- b

, so

r + s = - 23

. The factors are

x + 6

and

x + 17

, and

x + 6

is zero at

- 6

, not at

6

, so the roots are

- 6

and

- 17

, which add to

- 23

. Adding the two search numbers instead gives

23

, the right size with the wrong sign.

Problem

The quadratic

8 x^{2} + 18 x - 5

factors as

(2 x + 5) (4 x - 1)

. Find its two roots and add them. Enter the sum as a fraction in lowest terms.

Show a hint

Each factor is zero at one root. Set $2 x + 5 = 0$ and $4 x - 1 = 0$ , and solve each one.
The roots are $- \frac{5}{2}$ and $\frac{1}{4}$ . Put both over the denominator $4$ before adding.

Show the full solution

From

(2 x + 5) (4 x - 1)

the roots are

- \frac{5}{2}

and

\frac{1}{4}

, so the sum is

- \frac{10}{4} + \frac{1}{4} = - \frac{9}{4}

. That is

- \frac{18}{8}

reduced, the middle coefficient over the leading one with the sign flipped, which is not a coincidence.

The map from a quadratic's coefficients to its two root facts. From

a x^{2} + b x + c

the roots add to

- \frac{b}{a}

and multiply to

\frac{c}{a}

. The leading coefficient sits under both. The minus sits on the sum alone, in the same gold as

b

's underline above it, since that is the sign readers drop. No root appears in the picture, because neither fact needs one.

Both facts are read off the coefficients in place. Put the quadratic in standard form with one side zero, and $a$ , $b$ and $c$ are already written down. For $6 x^{2} - 13 x - 3$ the sum of the roots is $- \frac{- 13}{6} = \frac{13}{6}$ and the product is $- \frac{1}{2}$ . Nothing was factored, and this one has no integer factorization to find.

Problem

A monic quadratic has two roots that add to

12

and multiply to

- 85

. Write it in the form

x^{2} + b x + c

and enter

b

, the coefficient of

x

Show a hint

A monic quadratic with roots $r$ and $s$ is $(x - r) (x - s)$ , which expands to $x^{2} - (r + s) x + r s$ . Line that up against $x^{2} + b x + c$ .
Matching the $x$ terms gives $- b = r + s$ , and the roots add to $12$ . Solve that for $b$ .

Show the full solution

The roots add to

- b

, so

- b = 12

and

b = - 12

. The quadratic is

x^{2} - 12 x - 85

, which is

(x - 17) (x + 5)

with roots

17

and

- 5

. Writing

12

straight in as the coefficient of

x

gives

x^{2} + 12 x - 85

, whose roots add to

- 12

instead.

Problem

One root of

5 x^{2} - 16 x + 3 = 0

3

. Find the other root from the sum of the roots rather than by factoring, and enter it as a fraction in lowest terms.

Show a hint

The two roots add to $- \frac{b}{a}$ , and one of them is already known, so the other is one subtraction away.
The sum is $\frac{16}{5}$ . Subtract $3$ from it. Dividing the product $\frac{3}{5}$ by $3$ is the second route.

Show the full solution

The roots add to

- \frac{- 16}{5} = \frac{16}{5}

, so the other root is

\frac{16}{5} - 3 = \frac{1}{5}

. The product gives the same value,

\frac{3}{5} \div 3 = \frac{1}{5}

Here is what the two facts are for. Some expressions in $r$ and $s$ can be rewritten so that only $r + s$ and $r s$ appear in them, and any expression like that can be evaluated from $a$ , $b$ and $c$ alone. Whether the roots are integers, fractions, or numbers no method in this chapter produces makes no difference to the work.

Problem

The quadratic

7 x^{2} - 5 x - 9

has no factorization with integer coefficients, so nothing in this chapter produces its roots. Call them

r

and

s

anyway and find

\frac{1}{r} + \frac{1}{s}

as a fraction in lowest terms.

Show a hint

Combine the two fractions over the common denominator $r s$ . The result is a sum divided by a product, and both of those come from the coefficients.
The sum is $\frac{5}{7}$ and the product is $- \frac{9}{7}$ . Divide the first by the second, and the sevens cancel.

Show the full solution

Over the common denominator

r s

\frac{1}{r} + \frac{1}{s} = \frac{r + s}{r s}

. Here

r + s = \frac{5}{7}

and

r s = - \frac{9}{7}

, so the value is

\frac{5 / 7}{- 9 / 7} = - \frac{5}{9}

. The sevens cancel in the division, so this expression is

- \frac{b}{c}

for any quadratic with

c

not zero, whatever

a

is.

Problem

The roots

r

and

s

x^{2} - 5 x - 8 = 0

are not integers, so factoring will not produce them. Find

r^{2} + s^{2}

from the sum and the product of the roots alone.

Show a hint

Neither root can be found here, but $r + s$ and $r s$ come straight from the coefficients. Build $r^{2} + s^{2}$ out of those two numbers.
Squaring $r + s$ gives $r^{2} + 2 r s + s^{2}$ , which is $2 r s$ too big. Here $r s = - 8$ , so watch the sign when you subtract $2 r s$ .

Show the full solution

Since

(r + s)^{2} = r^{2} + 2 r s + s^{2}

, subtracting

2 r s

gives

r^{2} + s^{2} = (r + s)^{2} - 2 r s

. Here

r + s = 5

and

r s = - 8

, so the value is

25 - 2 (- 8) = 25 + 16 = 41

. The product is negative, so

- 2 r s

is positive here. Reading it as

25 - 16 = 9

is the standard slip.

The coefficients are the shorter route even when the roots are findable. $20 x^{2} + x - 12$ factors as $(4 x - 3) (5 x + 4)$ , so one way to reach $\frac{1}{r} + \frac{1}{s}$ is to add $\frac{4}{3}$ and $- \frac{5}{4}$ over a common denominator. From the coefficients it is $- \frac{b}{c}$ , a single division.

Problem

The quadratic

2 x^{2} - 15 x + c

has two roots, and one of them is four times the other. Neither root is given. Find the value of

c

Show a hint

Call one root $r$ , so the other is $4 r$ . Their sum is $- \frac{b}{a}$ , so the stated ratio gives one equation in one unknown.
The sum of the roots is $\frac{15}{2}$ , so $5 r = \frac{15}{2}$ . Once both roots are known, use the product, which is $\frac{c}{2}$ rather than $c$ .

Show the full solution

Call the roots

r

and

4 r

. They add to

\frac{15}{2}

, so

5 r = \frac{15}{2}

and

r = \frac{3}{2}

, making the roots

\frac{3}{2}

and

6

. Their product is

9

, and the product of the roots is

\frac{c}{2}

, so

c = 18

. The slip here is reading that product as

c

rather than

\frac{c}{2}

, which gives

9

Problem

The two roots of

x^{2} - 24 x + c

differ by

8

. Call them

r

and

s

, and find the value of

c

without finding either root.

Show a hint

Both $(r + s)^{2}$ and $(r - s)^{2}$ expand to $r^{2} + s^{2}$ plus or minus $2 r s$ , so $(r - s)^{2} = (r + s)^{2} - 4 r s$ .
For $x^{2} - 24 x + c$ the sum of the roots is $24$ and the product is $c$ , and the squared difference is $64$ . Put all three into that identity.

Show the full solution

Here

r + s = 24

and

r s = c

, so

(r - s)^{2} = (r + s)^{2} - 4 r s

becomes

64 = 576 - 4 c

, and

c = 128

. The difference

r - s

is not symmetric and cannot be read off the coefficients, but

(r - s)^{2}

is, which is what makes this work. A sum of

24

and a difference of

8

also give the roots outright,

16

and

8

Any question that involves the roots only through $r + s$ and $r s$ can be answered from the coefficients, and there is exactly one monic quadratic with a stated sum and product. Neither fact separates $r$ from $s$ , and neither one produces a root. The last two problems use most of the lesson at once.

Problem

The quadratic

4 x^{2} + 10 x - 21

has no integer factorization, so no method in this chapter produces either root. Its roots are

r

and

s

. Find

r^{2} s + r s^{2}

as a fraction in lowest terms.

Show a hint

Both terms have a factor of $r s$ . Take it out and see what is left.
$r^{2} s + r s^{2} = r s (r + s)$ , with $r s = - \frac{21}{4}$ and $r + s = - \frac{5}{2}$ . Two negatives multiply to a positive.

Show the full solution

Both terms have a factor of

r s

, so

r^{2} s + r s^{2} = r s (r + s)

. Here

r + s = - \frac{10}{4} = - \frac{5}{2}

and

r s = - \frac{21}{4}

, so the value is

(- \frac{21}{4}) (- \frac{5}{2}) = \frac{105}{8}

. Both factors are negative, so the value is positive. A sign dropped anywhere in the setup shows up as

- \frac{105}{8}

Problem

Let

r

and

s

be the roots of

4 x^{2} - 12 x - 9

, which does not factor over the integers. Find the coefficient of

x

in the monic quadratic whose roots are

r^{2}

and

s^{2}

, as a fraction in lowest terms.

Show a hint

In a monic quadratic the coefficient of $x$ is the negative of the sum of its roots, and the roots here are $r^{2}$ and $s^{2}$ , so find $r^{2} + s^{2}$ first.
From the coefficients, $r + s = 3$ and $r s = - \frac{9}{4}$ . Use $r^{2} + s^{2} = (r + s)^{2} - 2 r s$ , then flip the sign of the result.

Show the full solution

From the coefficients,

r + s = \frac{12}{4} = 3

and

r s = - \frac{9}{4}

, so

r^{2} + s^{2} = (r + s)^{2} - 2 r s = 9 - 2 (- \frac{9}{4}) = \frac{27}{2}

. The monic quadratic with roots

r^{2}

and

s^{2}

x^{2} - (r^{2} + s^{2}) x + r^{2} s^{2}

, so its coefficient of

x

- \frac{27}{2}

. Neither root was needed anywhere in that. The sign flip is the step to watch, since

\frac{27}{2}

is the sum of the new roots and the coefficient is its negative.

Every answer in this lesson's second half came off the coefficients, and four of those quadratics have no root chapter 9 can produce. Lesson 9.5, Factoring in Action, turns these tools on equations not written as $a x^{2} + b x + c = 0$ . An equation can be a quadratic in some repeated piece, and naming that piece with one letter makes it one you can solve.

Practice these ideas

Practice

Find the sum of the two roots of

x^{2} - 15 x + 26 = 0

. Read the sum off the coefficients, and keep the sign of the middle coefficient in mind.

Show the solution

The coefficient of

x

b = - 15

, so the sum of the roots is

- b = - (- 15) = 15

. Factoring gives the same total, since

x^{2} - 15 x + 26 = (x - 2) (x - 13)

and

2 + 13 = 15

Practice

The quadratic

x^{2} - 33 - 8 x

is not written in standard form. Put it in standard form, then find the product of its two roots. The leading coefficient is

1

, so the product can be read straight off once it is.

Show the solution

The quadratic is monic, so the product of the roots is its constant term,

r s = - 33

. Check by factoring,

x^{2} - 8 x - 33 = (x - 11) (x + 3)

, whose roots

11

and

- 3

multiply to

- 33

Practice

The quadratic

x^{2} + 19 x + 34

factors as

(x + 2) (x + 17)

. Find the sum of its two roots. Read each root off its own factor first, and mind the sign as you do it.

Show the solution

The factors

x + 2

and

x + 17

are zero at

- 2

and

- 17

, so the two roots add to

- 19

. Both roots are negative, so the sum is the negative of the middle coefficient

19

, never

19

itself.

Practice

Find the product of the two roots of

10 x^{2} + 11 x - 6 = 0

. Give your answer as a fraction in lowest terms. The leading coefficient is not

1

here, so the constant term alone is not the product.

Show the solution

The product of the roots is

\frac{c}{a} = \frac{- 6}{10} = - \frac{3}{5}

. As a check,

10 x^{2} + 11 x - 6 = (5 x - 2) (2 x + 3)

has roots

\frac{2}{5}

and

- \frac{3}{2}

, and

\frac{2}{5} \cdot (- \frac{3}{2}) = - \frac{3}{5}

Practice

Find the sum of the two roots of

8 x^{2} + 30 x - 7

as a fraction in lowest terms. This quadratic has no integer factorization, so the sum has to come from the coefficients.

Show the solution

The sum of the roots is

- \frac{b}{a} = - \frac{30}{8} = - \frac{15}{4}

. The roots themselves are irrational, so no factor search would have found them, and the sum is still exact.

Practice

A monic quadratic has two roots that add to

- 6

and multiply to

- 40

. Enter the coefficient of

x

in that quadratic. The two roots themselves are not needed.

Show the solution

The roots add to

- b

, so

- b = - 6

and

b = 6

. The quadratic is

x^{2} + 6 x - 40 = (x + 10) (x - 4)

, with roots

- 10

and

4

, which add to

- 6

and multiply to

- 40

Practice

One root of the quadratic

x^{2} - 2 x - 63

9

. Find the other root from the sum of the roots rather than by factoring, then check it against the product.

Show the solution

The roots add to

- (- 2) = 2

, so the other root is

2 - 9 = - 7

. Checking against the product,

9 \cdot (- 7) = - 63

Practice

One root of

6 x^{2} - 19 x + 14

2

. The leading coefficient is

6

, not

1

, so the two roots add to

- \frac{b}{a}

rather than to

- b

. Find the other root as a fraction in lowest terms.

Show the solution

With

a = 6

and

b = - 19

, the roots add to

- \frac{b}{a} = \frac{19}{6}

, so the other root is

\frac{19}{6} - 2 = \frac{7}{6}

. The product route gives the same value, since the roots multiply to

\frac{c}{a} = \frac{7}{3}

and

\frac{7}{3} \div 2 = \frac{7}{6}

Practice

The quadratic

5 x^{2} - 14 x + 4

has no integer factorization. Its roots are

r

and

s

. Find

\frac{1}{r} + \frac{1}{s}

as a fraction in lowest terms.

Show the solution

Over a common denominator,

\frac{1}{r} + \frac{1}{s} = \frac{r + s}{r s}

. Here

r + s = \frac{14}{5}

and

r s = \frac{4}{5}

, so the value is

\frac{14}{4} = \frac{7}{2}

. The fifths divide out, so for any

a x^{2} + b x + c

this expression equals

- \frac{b}{c}

, with

b = - 14

and

c = 4

here.

Practice

Let

r

and

s

be the roots of the quadratic

x^{2} - 9 x + 11

, neither of which is an integer. Find

r^{2} + s^{2}

without finding either root.

Show the solution

Since the quadratic is monic,

r + s = 9

and

r s = 11

, so

r^{2} + s^{2} = (r + s)^{2} - 2 r s = 81 - 22 = 59

. Expanding

(r + s)^{2}

gives

r^{2} + 2 r s + s^{2}

, which exceeds

r^{2} + s^{2}

by exactly

2 r s

, so one subtraction of

2 r s

is the whole correction.

Practice

Let

r

and

s

be the roots of the quadratic

x^{2} - 6 x - 2

, which does not factor over the integers. Find

(r - s)^{2}

Show the solution

(r - s)^{2} = (r + s)^{2} - 4 r s = 36 - 4 (- 2) = 36 + 8 = 44

. The product

r s

is negative, so

- 4 r s

is positive here, and the common slip is

36 - 8 = 28

Practice

Is there a pair of real numbers

r

and

s

whose sum is

4

and whose product is

9

? Decide from the square of their difference, which is fixed by the sum and the product alone, then answer yes or no.

Show the solution

For such a pair,

(r - s)^{2} = (r + s)^{2} - 4 r s = 16 - 36 = - 20

, and no real number has a negative square, so the answer is

no

. The quadratic with that sum and product is

x^{2} - 4 x + 9

, and chapter 11 takes it up.

Practice

Both roots of

x^{2} + b x + 125

are negative, and one root is five times the other. Start from the product of the roots rather than the sum. Find

b

Show the solution

With roots

r

and

5 r

, the product is

5 r^{2} = 125

, so

r = - 5

since both roots are negative, and the roots are

- 5

and

- 25

. They add to

- 30

, and the sum of the roots is

- b

, so

b = 30

. Taking

r = 5

gives two positive roots and

b = - 30

instead.

Practice

Both roots of

25 x^{2} - 40 x + c

are the same number. Find

c

. The leading coefficient is not

1

, so use the sum and the product in their general form.

Show the solution

The roots sum to

- \frac{b}{a} = \frac{40}{25} = \frac{8}{5}

, and equal roots are each half of that, so each is

\frac{4}{5}

. Their product

\frac{16}{25}

equals

\frac{c}{25}

, so

c = 16

. Checking,

25 x^{2} - 40 x + 16 = (5 x - 4)^{2}

Practice

The quadratic

x^{2} - 14 x + c

has two roots that differ by

4

. The difference itself cannot be read off the coefficients, but its square can. Find

c

Show the solution

Here

r + s = 14

and

r s = c

, so

(r - s)^{2} = (r + s)^{2} - 4 r s

becomes

16 = 196 - 4 c

and

c = 45

. As a check, the roots must be

9

and

5

, since those sum to

14

and differ by

4

, and their product is

45

Practice

Let

r

and

s

be the roots of

2 x^{2} - 5 x - 6

, which does not factor over the integers. Find

\frac{r}{s} + \frac{s}{r}

as a fraction in lowest terms.

Show the solution

Over a common denominator,

\frac{r}{s} + \frac{s}{r} = \frac{r^{2} + s^{2}}{r s}

. From the coefficients,

r + s = \frac{5}{2}

and

r s = - 3

, so

r^{2} + s^{2} = (r + s)^{2} - 2 r s = \frac{25}{4} + 6 = \frac{49}{4}

, and the value is

\frac{49}{4} \div (- 3) = - \frac{49}{12}

. Because

r s

is negative, the

- 2 r s

term is

+ 6

and not

- 6

, which is the usual slip here.

Practice

Let

r

and

s

be the roots of

x^{2} - 7 x + 3

, which does not factor over the integers. Build the monic quadratic whose roots are

\frac{1}{r}

and

\frac{1}{s}

, and enter its coefficient of

x

as a fraction in lowest terms.

Show the solution

From the coefficients,

r + s = 7

and

r s = 3

, so the new roots add to

\frac{1}{r} + \frac{1}{s} = \frac{r + s}{r s} = \frac{7}{3}

, and a monic quadratic's coefficient of

x

is the negative of the sum of its roots, so it is

- \frac{7}{3}

. The built quadratic is

x^{2} - \frac{7}{3} x + \frac{1}{3}