Question

find the value of k if the given equation has real and equal roots

1. (k-12)x2+2(k-12)x+2=0

Answer 1

Hi friend!!

Given quadratic equation,

$(k - 12) {x}^{2} + 2(k - 12)x + 2 = 0$
For a quadratic equation to have real and equal roots, it's discriminant must be equal to zero.

→ b²-4ac=0

= 4(k-12)²-4(k-12)(2)=0

k²+144-24k-2k+24=0

k²-26k+168=0

k²-12k-14k+168=0

k(k-12)-14(k-12)=0

(k-14)(k-12)=0

k=12,14

I hope this will help you ;)

Answer 2

Discriminant = b²-4ac

=> [2(k-12)]² - 4(k -12)(2)

=> [4(k²+144-24k] -8k + 96

=> 4 [k² + 144 -24k - 2k +24]

We know that equal roots, Discriminant =0


4[k² -26k + 168] = 0

=> k² - 26k + 168 = 0


=> k² - (14 + 12)k + 168 =0

=> k² - 14k - 12k + 168 =0

=> k(k - 14) - 12( k - 14) = 0

=> (k - 14)(k - 12) =0

k = 14 or 12


I hope this will help you

-by ABHAY