Question

for what value of k does (k-12)x2+2(k-12)x+12=0 have equal roots

Answer 1

Answer:

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The equation ax^2+bx+c=0 has equal roots when,

b^2–4ac = 0

Putting values from the given equation,

(k+12)^2–4(k+12)(-2) =0

k^2+24k+144+8k+96=0

k^2+32k+240=0

Solving this quadratic we get,

k= (-20, -12)

This is not the final answer.

If we put these values of k in the quadratic once again we can see that, for k=-12, the equation doesn’t satisfy.

So, the value of k for which the roots of given equation are equal is only (-20).

Answer 2

Answer:

k=24

Step-by-step explanation:

For a quadratic polynomial ax²+bx+c=0   <- General Form

Determinant , D = b²-4ac

Comaring given equation with General Form, we get

a= k-12

b= 2(k-12) = 2k-24

c= 12

For equal roots,

Determinant=0

=>b²-4ac=0

=>b²=4ac

Putting values of a,b and c, we get

(2k-24)²=4×(k-12)×12

=>4k² + 576 - 96k = 48k - 576

=>4k²-144k+1152=0

=>k²-36k+288=0

=>k²-24k-12k+288=0

=>k(k-24)-12(k-24)=0

=>(k-24)(k-12)=0

=>Either k-24=0 or k-12=0

=>Either k=24 or k=12

But for k=12,

k-12 becomes zero which means equation does not remains quadratic.

Hence,

k=24

and equation becomes

12x²+24x+12=0

=>x²+2x+1=0

HOPE IT HELPS,

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