find the value of k for which the values are real and equal
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Hi !
(k+1)x² + 2(k+3)x + ( k+8) = 0
roots are real and equal , hence , b² - 4ac = 0
{2(k+3)}² - 4(k+1)(k+8) = 0
4(k+3)² - 4(k² + 9k + 8) = 0
4(k² + 6k + 9) - 4k² - 36k - 32 = 0
4k² + 24k + 36 - 4k² - 36k - 32 = 0
-12k + 4 = 0
4 = 12k
k = 4/12
= 1/3
(k+1)x² + 2(k+3)x + ( k+8) = 0
roots are real and equal , hence , b² - 4ac = 0
{2(k+3)}² - 4(k+1)(k+8) = 0
4(k+3)² - 4(k² + 9k + 8) = 0
4(k² + 6k + 9) - 4k² - 36k - 32 = 0
4k² + 24k + 36 - 4k² - 36k - 32 = 0
-12k + 4 = 0
4 = 12k
k = 4/12
= 1/3
vaishu24:
how u got 4(k²+9k+8)
k^2 + 8k+ k + 8 = k^2 + 9k + 8
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