Question

find the value of K for which the roots of quadratic equation are equal (k-4) x^2 + 2(k-4)x +2.

Answer 1

(k+4)x2+(k+1)x+1=0

D=b2-4ac

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

=k2+2k+1-4k-16

=k2-2k-15

For equal roots, D=0

D=0

K2-2k-15=0

k2-5k+3k-15=0

k(k-5)+3(k-5)=0

(k+3)(k-5)=0

k+3=0 OR k-5=0

k=-3 , k=5

Answer 2

Given:

A quadratic equation (k-4)x² + 2(k-4)x + 2 =0.

To Find:

The value of k for which the roots of the equation are equal.

Solution:

The given problem can be solved using the concepts of quadratic equations.

1. The given quadratic equation is (k-4)x² + 2(k-4)x + 2 =0.

2. For a quadratic equation to have equal roots, the value of the discriminant must be equal to zero.

3. Consider a quadratic equation ax² + b x + c =0, the discriminant of this equation is given by the formula,

=> Discriminant ( D ) = $\sqrt{b^2-4ac}$,

=> For an equation to have equal roots, the value of the discriminant is equal to 0. ( D = 0).

4. Since the given equation has equal roots the value of the discriminant is equal to 0.

=> [(2)(k-4)]² - 4 (k-4) (2) = 0,

=> 4(k² -8k + 16) - 8k + 32 = 0,

=> 4k²  -32k + 64 -8k + 32 = 0,

=> 4k²  - 40k + 96 = 0,

=> 4k² -16k -24k + 96 = 0,

=> 4k(k -6) -16(k -6) = 0,

=> (4k-16) (k-6) = 0,

=> k = 6 (OR) k =4.

5. For K =4 the equation is not quadratic, hence K =4 is omitted.

Therefore, the value of K for which the roots of the given equation is equal is 6.