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

α + α = -4/k

2α = -4/k

α = -4/2k

α = -2/k

α ⋅ α = 3/k

α2 = 3/k.

(-2/k)2 = 3/k

4/k2 = 3/k

4/k = 3

k = 4/3

(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

α + α = -5/2

2α = -5/2

α = -5/4

α ⋅ α = k/2

α2 = k/2

(-5/4)2 = k/2

25/16 = k/2

k = (25/16) x 2

k = 25/8

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

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