# Relations between Roots and Coefficients

### Formulae

**Sum and product of roots**

Sum of roots = + = - b/a = -
(coefficient of x)/(coefficient of x²)

Product of roots = = c/a = (constant
term)/(coefficient of x²)

### Symmetric functions of roots

From + = -b/a and
= c/a, values of other functions of
and can be calculated:

( - )² = ( +
)² -4 ,

² + ² = ( +
)² -2 ,

³ + ³ = ( +
)³ -3
(
+ ),

^{4} + ^{4} = [(
+ )² -2
]² -2 (
)²,

² - ² = ( +
)( - ) etc.

### Formation of a quadratic equation with given roots

x² -S x +P = 0, where S = sum of roots and P = product of roots.

## Illustrative Examples

### Example

If , are roots of the equation x² -4 x +2 = 0, find the values of

- ² + ²
- ² - ²
- ³ + ³
- 1/ + 1/

### Solution

As , are roots of the equation x²
-4 x +2 = 0,

+ =
-(-4)/1 = 4, = 2/1 = 2

- ² +² = ( + )² -2 = (4)² -2 (2) = 12
- ² -² = ( +
)( -).

Now ( -)² = ( +)² -4 = (4)² -4 (2) = 8

- = ± 8 = ±2 2

² -² = ( +)( -) = (4)(±2) = ±8 - ³ +³ =
( +)³
-3
( +)

= (4)³ -3 (2)(4) = 64 -24

= 40 - 1/ + 1/ = ( + )/ = 4/2 =2

### Example

If , are roots of the quadratic equation ax² +bx +c = 0, form an equation whose roots are

- - , -
- 1/, 1/

### Solution

Since , are roots of a x 2 +b x
+c = 0,

+ = -b/a,
= c/a

- Here, S = (-)
+(-) = -(
+ ) = -(-b)/a = b/a

P = (- )(-) = = c/a

Hence the required equation is x² -S x +P = 0

i.e. x² - (b/a) x + c/a = 0 i.e. a x² -b x +c = 0 - Here, S = 1/ + 1/ = (
+)/ = - (b /
a)/(c / a) = - b/c

P = (1/).(1/) = 1/( ) = 1/(c/a) = a/c

Hence the required equation is x² -S x +P = 0

i.e. x² -(- b/c) x + a/c = 0 i.e. cx² +bx +a = 0

### Example

Find p, q if p and q are roots of the equation x² +px +q = 0.

### Solution

Since p, q are roots of x² +px +q = 0, p +q = -p and pq = q.

Now pq = q => pq -q = 0 => q(p -1) = 0 => q = 0 or p = 1.

When q = 0 then p +q = -p => 2p = -q = 0 => p = 0.

When p = 1, then p +q = -p => q = -2 p = -2.

Hence the required solutions are p = q = 0 or p = 1, q = -2.

### Example

If the roots of the equation 2 x² +(k +1) x +(k² -5 k +6) = 0 are of opposite signs then show that 2 < k < 3.

### Solution

Since roots are of opposite signs, roots are real and distinct,

discriminant> 0 and product of roots < 0

(k +1)² -4. 2. (k² -5 k +6) > 0 and (k² -5 k +6)/2 < 0

First condition is always true when second holds. (because (k +1)² = 0)

Hence (k² -5 k +6)/2 < 0 => k² -5 k +6 < 0 => (k -2)(k -3)
< 0

=> 2 < k < 3

### Example

The coefficient of x in the equation x² +p x +q = 0 was taken as 17 in place of 13 and thus its roots were found to be -2 and -15. Find the roots of the original equation.

### Solution

By given conditions, -2 and -15 are roots of the equation x² +17 x +q = 0

Hence product of roots = (-2)(-15) = q/1 => q = 30.

Therefore, the original equation is x² +13 x +30 = 0

=> (x +10)(x +3) = 0 => x = -3, -10

Hence the roots of the original equation are -3, -10.

## Exercise

- Write the sum and product of the roots of the following equations:

(i) x² +9 = 0

(ii) 2 x² -2 x +4 = 0. - If , are the roots of the
equation 3 x² -6 x +4 = 0, evaluate

- If the roots of the equation x² +p x +7 = 0 are denoted by and , and ² +² = 22, find the possible values of p.
- If , be the roots of the
equation a x² +b x +c = 0, find the value of

(i) ² + ²

(ii) / + /

(iii) ³ + ³

(iv) | -

(v) ² - ²

(vi)^{6}+^{6} - If , are roots of x² +k x +12 = 0 and - = 1, find k.
- The sum of the roots of the equation 1/(x+a) + 1/(x+b) = 1/c is zero.

Prove that the product of the roots is - (a² +b²)/2

[**Hint.**Rewrite the equation as x² +x (a +b -2 c) +(ab -bc -ca) = 0.] - Two candidates attempt to solve a quadratic equation of the form x² +p x +q = 0. One starts with a wrong value of p and finds the roots to be 2 and 6. The other starts with a wrong value of q and finds the roots to be 2 and -9. Find the correct roots and the equation.
- Given that and are the roots
of the equation x² = 7 x +4,

(i) Show that ³ = 53 +28

(ii) Find the value of / + / - Form an equation whose roots are

(i) 3, -1/3

(ii) 3, 1/3

(iii) 0, 0 - Form an equation with rational coefficients one of whose roots is

(3 +1)/(3 -1) - Form an equation with real coefficients one of whose roots is

(i) -1 -2 i

(ii) -2 --3

(iii) 1/(2**+**-2) - If , are the roots of the
equation a x² +b x +c = 0, then form an equation whose roots are:

(i) 2, 2

(ii) /2, /2

(iii) / , /

(iv) ( +)², ( -)²

(v) (/) +1, (/) +1

(vi) +k, +k - If , be the roots of the equation 2 x² -3 x +1 = 0, find an equation whose roots are /(2 +3), /(2 +3)

## Answers

**1.**(i) 0, 9 (ii) 1, 2 2

**2.**8

**3.**±6

**4.**(i) (b² -2ac)/a² (ii) (b² -2ac)/ac (iii) (3a b c -b³)/a³ (iv) [ (b² -2ac)]/|a|

(v) ± [b (b² -2ac)]/a² (vi) [b

^{6}+9 a² b² c² -6ab

^{4}c -2c³ a³]/a

^{6}

**5.**±7

**7.**-3, -4; x² +7 x +12 = 0

**8.**(ii) -57/4

**9.**(i) 3 x² -8 x -3 = 0 (ii) 3x² -4x + 3 = 0 (iii) x² = 0

**10.**x² -4 x +1 = 0

**11.**(i) x² +2 x +5 = 0 (ii) x² +4 x +7 = 0 (iii) 6 x² -4 x +1 = 0

**12.**(i) a x² +2 b x +4 c = 0

(ii) 4 a x² +2 b x +c = 0

(iii) a c x² -(b² -2 a c) x +a c = 0

(iv) a 4 x² -2 a² (b² -2 a c) x +b²(b² -4 ac) = 0

(v) a c x² +b (a +c) x +(c + a)² = 0

(vi) a x² -(2 a k -b) x +(a k² -b k +c) = 0

**13.**40 x² -14 x +1 = 0