DETERMINE THE MISSING VALUE IN QUADRATIC EQUATION

Problem 1 :

If the sum of the roots of

4x² + kx – 7 = 0 is 3,

Find the value of k.

Solution :

4x² + kx – 7 = 0

Sum of roots (α + β) = -b/a

α + β = 3

-k/4 = 3

-k = 12

k = -12

Problem 2 :

Find the value of K if the sum of the roots of equation

(2k – 1)x² + (4k – 1) x + (K + 3) = 0 is 5/2

Solution :

(2k – 1)x² + (4k – 1) x + (K + 3) = 0

a = 2k – 1, b = 4k – 1 and c = K + 3

Sum of roots = -b/a

-(4k - 1)/(2k - 1) = 5/2

-2(4k - 1) = 5(2k - 1)

-8k + 2 = 10k - 5

-8k - 10k = -5 - 2

-18k = - 7

k = 7/18

Problem 3 :

For what value of k the roots of the following equations are equal :

(i)  kx² + 4x + 3 = 0       (ii)  2x² + 5x + k = 0

Solution :

(i)  kx² + 4x + 3 = 0

a = k, b = 4 and c = 3

Sum of roots = -4/k

Product of roots = 3/k

Here α = β

(ii)  2x² + 5x + k = 0

Solution :

a = 2, b = 5 and c = k

Sum of roots = -5/2

Product of roots = k/2

Here α = β

Problem 4 :

If the difference of the roots of 6x² – 23x + c = 0 is 5/6, find the value of c

Solution :

Difference of roots = α - β = √(α + β)² - 4αβ

α - β = 5/6

α + β = -(-23)/6 ==> 23/6

α β = c/6

α - β = √(23/6)² - 4(c/6)

(5/6)² = 529/36 - (2c/3)

(2c/3) = (529/36) - (25/36)

(2c/3) = (529 - 25)/36

c = (504/36) x (3/2)

c = 21

Problem 5 :

If the difference of the roots of x² – 7x + k – 4 = 0 is 5, find the value of k and the roots.

Solution :

x² – 7x + k – 4 = 0 ----(1)

α - β = 5

α + β = -(-7)/1  ==> -7

α β = (k -4)/1  ==> k - 4

α - β = √(α + β)² - 4αβ

5 = √(-7)² - 4(k - 4)

5² = 49 - 4k + 16

25 = 65 - 4k

4k = 65 - 25

4k = 40

k = 10

By applying the value of k in (1), we get

x² – 7x + 10 – 4 = 0

x² – 7x + 6 = 0

(x - 1)(x - 6) = 0

x = 1 and x = 6

So, the roots are 1 and 6.

Problem 6 :

Find the value of k given that if one root of

9x² – 15x + k = 0

exceeds the other by 3. Also find the roots.

Solution :

9x² – 15x + k = 0

α = β + 3

α + β = -(-15)/9 ==> 5/3

α + β = 5/3

α β = k/9

Applying the value of α in (1), we get

β + 3 + β = 5/3

2β + 3 = 5/3

2β = (5/3) - 3

2β = -4/3

β = -2/3

α = (-2/3) + 3

α = 7/3

So, the roots are -2/3 and 7/3.

Problem 7 :

The product of the roots of the equation

(k + 1)x² + (4k + 3)x + (k – 1) = 0

is 7/2

Solution :

(k + 1)x² + (4k + 3)x + (k – 1) = 0

a = k + 1, b = 4k + 3 and c = k - 1

Product of roots = c/a

(k - 1)/(k + 1) = 7/2

2(k - 1) = 7(k + 1)

2k - 2 = 7k + 7

2k - 7k = 7 + 2

-5k = 9

k = -9/5

Problem 8 :

If one zero of the quadratic polynomial x² + 3x + k is 2, then the value of k is

a)  10     b) -10    c) 5    d)  -5

Solution :

Let f(x) = x² + 3x + k

When one zero of the polynomial is 2, then x = 2

f(2) = 2² + 3(2) + k

0 = 4 + 6 + k

0 = 10 + k

k = -10

So, the value of k is -10, option b is correct.

Problem 9 :

If α, β are the zeroes of polynomial f(x) = x² - p(x + 1) - c then (α + 1)(β + 1) = 

a)  c - 1     b)  1 - c     c)  c     d) 1 + c

Solution :

f(x) = x² - p(x + 1) - c

= x² - px - p - c

= x² - px - (p + c)

(α + 1)(β + 1) = α β + α + β + 1

a = 1, b = -p and c = -(p + c)

Sum of roots = -b/a

α + β = -(-p)/1

= p

Product of roots = c/a

α β = -(p + c)/1

= -p - c

Applying the values of sum and product of roots, we get

α β + α + β + 1 = -p - c + p + 1

= 1 - c

So, option b is correct.

Problem 10 :

If α, β are the zeroes of the polynomial f(x) = x² + x + 1, then 1/α + 1/β = 

a)  1     b)  -1    c) 0    d) none

Solution :

1/α + 1/β = (α + β)/αβ

f(x) = x² + x + 1

a = 1, b = 1 and c = 1

Sum of roots α + β = -b/a

α + β = -1

Product of roots α β = c/a

α β = 1

Applying sum and product of roots, we get

= -1/1

= -1

So, answer is option b.

Problem 11 :

If one of the zeroes of the quadratic polynomial (k - 1) x² + kx + 1 is -3, then the value of k is

a)  4/3     b)  -4/3    c)  -2/3    d)  2/3

Solution :

p(x) = (k - 1) x² + kx + 1

Since -3 is one of the zeroes, then x = -3

p(-3) = (k - 1) (-3)² + k(-3) + 1

0 = 9(k - 1) - 3k + 1

0 = 9k - 9 - 3k + 1

6k - 8 = 0

6k = 8

k = 8/6

k = 4/3

So, option a is correct.