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