Question

if alpha and 1/alpha are the zeroes of 6x^2+11x-(k-2).find the value of k?

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Answer 1

Step-by-step explanation:

P(X)=6x²+11x-(k-2)

I AM WRITING ALPHA =@. BECAUSE THERE IS NO ALPHA.SO,U CAN WRITE ALPHA VALUE

6(@)²+11X-K+2

6@²+11@-K+2

6@²+11@+2=K

6@²-1@-12@+2

1(6@+1)-2(6@+1)

1-2(6@+1)

@=1/6

Answer 2

AnsWer :

k = -4.

Given :

• Zeros above polynomial be alpha and 1/alpha.
• polynomial 6x² + 11x - (k-2).

Concepts Required :

$\tt\blacksquare \: Sum \: of \: Zeros \\ \tt\alpha . \beta = \frac{-b}{a} = \frac{Coefficient \: of \: x}{coefficient \: of \: {x}^{2} }$

$\tt\blacksquare \: product \: of \: zeros \\ \tt\alpha . \beta = \frac{c}{a} = \frac{constant \: term}{coefficient \: of \: {x}^{2} }$

Solution :

We have,

$\tt \dagger \: \: \: 6 {x}^{2} + 11x - (k - 2).$

So,

$\tt\longmapsto \alpha \beta = \frac{c}{a}$

$\rule{90}2$

Here,

• alpha = alpha.
• beta = 1/ alpha.

$\tt \longmapsto \alpha . \frac{1}{ \alpha } = \frac{ - (k - 2)}{6}$

$\tt\longmapsto 1 = \frac{ - k + 2}{6}$

$\tt\longmapsto 6 - 2= - k$

$\tt \longmapsto - k = 4$

$\tt\longmapsto k = - 4.$

Therefore, the value of k be -4.