Question

(iii) (k - 3)x^2 + 4(k - 3)x + 4 = 0
for which the roots are real and equal for the equation ​

Answer 1

SOLUTION:-

Given:

(k-3)x² + 4(k-3)x +4=0

Therefore,

The roots of the quadratic equation;

Ax² + Bx + C =0 compared with.

So,

⚫A= k-3

⚫B= 4(k-3)

⚫C= 4

Discriminate, D= b² -4ac=0

Now,

=) {4(k-3)}² - 4(k-3)(4)=0

=) 4(k)² +(3)² -2×k×3 - 16(k-3) =0

=) 4(k² +9-6k) -16k + 48 =0

=) 4k² + 36 -24k -16k +48 =0

=) 4k² -40k +84 =0

=) 4k² -28k -12k +84 =0

=) 4k(k-7) -12(k-7)=0

=) (k-7)(4k-12)=0

=) k-7 =0 or 4k -12 =0

=) k= 7 or 4k= 12

=) k= 7 or k= 12/4

=) k= 7 or k= 3

Hope it helps ☺️