Question

Find k for which the equation (k-4)x square+2(k-4)x+4=0 has real, equal roots

Answer 1

Answer:

k= 4,k=8

Step-by-step explanation:

Condition for equal and real roots ,D=b^2 - 4ac =0

D= b^2 - 4 ac=0

a= k-4

b= 2k-8

c=4

Application: D=b^2-4ac

0= b^2-4ac

0=[2k-8]^2 - 4* 2k-4* 4

0= 4k^2 - 32k +64 - 16k +64

0= 4k^2 - 48k+128

Dividing the whole equation by 4

0= k^2 - 12k+ 32

Factorising the equation by splitting the middle term, we get:

k^2 - 8k - 4k + 32=0

k(k-8) -4(k- 8)=0

=> k= 4

k=8