Question

Find the value of k for which the quadratic equation (k – 12)x*2 + 2(k-12)x+2=0 have real and equal roots.​

Answer 1

Step-by-step explanation:

for real and equal roots,

D>=0

{2(k-12)}^2-4*(k-12)*2>=0

4(k^2-2*k*12+12^2-8k+96>=0

4k^2-24k*4+144*4-8k+96>=0

4k^2-96k+576-8k+96>=0

4k^2-104k+672>=0

4(k^2-26k+168)>=0

k^2-26k+168>=0

k^2-14k-12k+168>=0

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

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

Either k-14>=0 Or k-12>=0

so k>=14 k>=12