Find the value of k the quadratic equation (k-3)x^2+4(k-3)x+4=0 has real and equal roots
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FOR REAL AND EQUAL ROOT
DISCRIMINANT = 0
b²-4ac = 0
(4(k-3) )²- 4 (k-3) *4= 0
4( k²+9-6k) - 16k + 48=0
4k² +36 -24k - 16k +48=0
4k² -40k + 84 =0
4k ² - 28 k - 12 k + 84=0
4k(k -7)-12(k-7) =0
(4k-12)=0 or( k-7)=0
k=12/4 or k=7
k=3 or k=7.
DISCRIMINANT = 0
b²-4ac = 0
(4(k-3) )²- 4 (k-3) *4= 0
4( k²+9-6k) - 16k + 48=0
4k² +36 -24k - 16k +48=0
4k² -40k + 84 =0
4k ² - 28 k - 12 k + 84=0
4k(k -7)-12(k-7) =0
(4k-12)=0 or( k-7)=0
k=12/4 or k=7
k=3 or k=7.
Answered by
0
the value of k = 3 or 7
your answer is 3 or 7
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