find K for which the roots are real and equal K + 1 X square + 2 into K K + 3 X + K + 8 is equal to zero
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9
for real and equal roots use D=0
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vaishu24:
brother how u got -12+4
by sloving 4k²+24k+36-4k²-32k-4k-32=0
can u plz explain
if u r free
4k^2 and -4k^2 are cancelled
24k-32k-4k= -12k
and 36-32=4
now ok
Answered by
6
(k+1)x² + 2 (k+3) x + k+8 = 0
For the roots of a quadratic equation to be real, the discriminant must be non-negative. If they are equal then the discriminant is 0. So the discriminant:
[ 2² (k+3)² - 4* (k+1) (k+8) ] = 0
4(k² + 6 k + 9) - 4 (k² + 9k + 8) = 0
-12 k + 4 = 0
3 k = 1
k = 1/3 answer
For the roots of a quadratic equation to be real, the discriminant must be non-negative. If they are equal then the discriminant is 0. So the discriminant:
[ 2² (k+3)² - 4* (k+1) (k+8) ] = 0
4(k² + 6 k + 9) - 4 (k² + 9k + 8) = 0
-12 k + 4 = 0
3 k = 1
k = 1/3 answer
:-)
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