Question

I find the value of k for which (k+1) x ²
+2 (k+3) x+(k+8) = 0 has equal roots​

Answer 1

Given:-

→(k+1)x²+2(k+3)x+(k+8)=p(x)

$\rule{200}{1}$

To Find:-

→The Value of k?

$\rule{200}{1}$

AnsWer:-

→(k+1)x²+2(k+3)x+(k+8)

•a=k+1,b=2(k+3),c=k+8

→d=b²-4ac=0

•As roots are Equal 'd'=0•

★Putting the Values★

→0=(2k+6)²-4×(k+1)×(k+8)

→0=(2k)²+(6)²+2×2k×6-4×(k²+8k+k+8)

→0=4k²+36+24k-4k²-36k-32

→-12k+4=0

→-12k=-4

→k=$\frac{\cancel{-4}}{\cancel{-12}}$

→k=$\frac{1}{3}$

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Answer 2

$\tt{\huge{\orange {Hello!!! }}} A.L$

QUESTION :

I find the value of k for which (k+1) x ²

I find the value of k for which (k+1) x ²+2 (k+3) x+(k+8) = 0 has equal roots

SOLUTION :

Since they have equal roots ,

D = 0

=> { D } ^ 2 = 0

=> { b } ^ 2 = 4 ac

=> { 2 ( k + 3 ) } ^ 2 = 4 ( k + 1 ) ( k + 8 )

4 { K^2 + 6 k + 3 } = 4 { K^2 + 9 K + 8 ]

=> 6 k + 3 = 9 k + 8

=> 3 k = -5

=> K = - 5 / 3........( A)