Question

Find the value of k for which the roots of the quadratic equation (k – 5)x2 + 2(k – 5) x + 2 = 0 are equal.

Answer 1

Sol:
The quadratic equation (k-5)x2 + 2(k-5)x + 2 =0 have equal roots.
⇒ Discriminant (b2 - 4ac) = 0
⇒ [2(k-5)]2 - 4(k-5)(2) = 0
⇒ 4(k2 - 10k + 25) -(8k - 40) = 0
⇒ 4k2 - 40k + 100 - 8k + 40 = 0
⇒ 4k2 - 48k + 140 = 0
⇒ k2 - 12k + 35 = 0
⇒ (k - 7)(k - 5) = 0
⇒ k = 7 or 5.

Answer 2

Question:

Find the value of k for which the quadratic equation (k-5)x² + 2(k-5)x + 2 = 0 has equal roots.

Answer:

k = 5 , 7

Note:

• An equation of degree 2 is know as quadratic equation .

• Roots of an equation is defined as the possible values of the unknown (variable) for which the equation is satisfied.

• The maximum number of roots of an equation will be equal to its degree.

• A quadratic equation has atmost two roots.

• The general form of a quadratic equation is given as , ax² + bx + c = 0 .

• The discriminant of the quadratic equation is given as , D = b² - 4ac .

• If D = 0 , then the quadratic equation would have real and equal roots .

• If D > 0 , then the quadratic equation would have real and distinct roots .

• If D < 0 , then the quadratic equation would have imaginary roots .

Solution:

The given quadratic equation is ;

(k-5)x² + 2(k-5)x + 2 = 0

Clearly , we have ;

a = k-5

b = 2(k-5)

c = 2

We know that ,

The quadratic equation will have equal roots if its discriminant is equal to zero .

=> D = 0

=> [2(k-5)]² - 4•(x-5)•2 = 0

=> 4(k-5)² - 4•2(k-5) = 0

=> 4(k-5)•(k-5-2) = 0

=> (k-5)(k-7) = 0

=> k = 5 , 7

Hence,

The required values of k are 5 and 7 .