Question

find k. 9x²+8kx+16=0 has two equal real roots​

Answer 1

Answer:

Answer is 6√3

Step-by-step explanation:

We know that two. equal root has

b²-4ac=0

therefore a= 9,b=8k , c=16

(8k)²-4(9)(16)=0

8k²-576=0

8k²=576

k²=576/8

k²=72

k= 6√3

itz helpful or not comment plz

Answer 2

Step-by-step explanation:

$given \\ \\f(x) \colon\rightarrow(k + 4)x {}^{2} + (k + 1) + 1 = 0 \\here \: \: in \: \: f(x) \: \: a = (k + 4) \: and\: \: b = (k + 1) \: and \: \: c = 1 \\ let \: \: the \: \: roots \: \: be \: \: \beta \\ formula \: \: to \: \: be \: \: \: applied \\ \beta = \frac{ - b \: ± \: \sqrt{b {}^{2} - 4ac }}{2a} \\ now \\ \frac{ \cancel{ - b} \: + \: \sqrt{b {}^{2} - 4ac} }{ \cancel{2a}} = \frac{ \cancel{ - b} - \sqrt{b {}^{2} - 4ac } } {\cancel{2a} } \\ 2 \sqrt{b {}^{2} - 4ac } = 0 \\ 4(b {}^{2} - 4ac) = 0 \\ b {}^{2} = 4ac \\ (k + 1) {}^{2} = 4(k + 4)(1) \\ k {}^{2} + 1 + 2k = 4k + 16 \\ k {}^{2} - 2k - 15 = 0 \\ k { }^{2} - 5k + 3k - 15 = 0 \\ k(k - 5) + 3(k - 15) = 0 \\ (k + 3)(k - 1) = 0 \\ k = 1 \: \: and \: \: - 3 \: \: \: Ans$