Question

For what value of k does the quadratic equation (4-k)x^2+(2k+4)x+8k+1=0 have equal roots

Answer 1

Answer:

roots are equal b^2-4ac=0

(2k+4)^2-4(4-k)(8k+1)=0

4k^2+16+16k-4(32k+4-8k^2-k)=0

4k^2+16+16k-128k-16+32k^2+4k=0

36k^2-108k=0

36k(k-3)=0

36k=0 k-3=0

k=0 k= 3