Question

(Q:13) If P and Q are the roots of equation F(x) = 6x2 + x - 2, Find the value of P/Q+Q/P​

Answer 1

Step-by-step explanation:

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Answer 2

According to data given in the above question.

We have to find the Pand Q are the two roots of equation.

Step-by-step explanation:

Given,

$F(x) = 6x² + x - 2$

Step-1:

The roots of given equation by splitting method:

The general equation is:

$ax²+bx+c \: \: \: \: ...(1)$

The given equation:

$6x²+x-2 \: \: \: \: ...(2)$

Compare the equation (1) and (2),we get

$a= 6, b=1 \: and \: c= -2$

so,

$a×c= 6×-2=-12$

The factors are 3 and 4

The 2nd term becomes :

$b= 4x-3x$

Now ,

STEP-2:

F(x)=0

The equation becomes :

$F(x) = 6x²+4x-3x-2$

$F(x) = (6x²+4x) -(3x + 2)$

Take out the 2x common from 1st and 2nd term , we get

$F(x) = 2x(3x+2)-1(3x + 2)$

Take out the (3x+2) we get ,

$F(x) = (3x + 2)(2 x - 1)$

STEP-3:

Put F(x)=0

$(3x + 2)(2 x - 1) = 0$

$x = \frac{ - 2}{3} \: and \: \frac{1}{2}$

Or ,

$P= \frac{ - 2}{3} \: and \: Q= \frac{1}{2}$

STEP-4:

The value of

$\frac{P}{Q} + \frac{Q}{P}$

Put the values of P and Q from above , we get

$\frac{ \frac{ - 2}{3} }{ \frac{1}{2} } + \frac{ \frac{1}{2} }{ \frac{ - 2}{3} }$

simplify the terms

$\frac{ - 2}{3} \times \frac{2}{1} + \frac{1}{2} \times \frac{3}{ - 2}$

Multiply the terms and divide the terms which get eliminated by division:

$\frac{ - 4}{3} + \frac{3}{ - 4}$

$\frac{ - 4}{3} - \frac{3}{ 4}$

Take LCM of 3 and 4 , we get 12 as LCM

$\frac{ (- 4 \times 4) -( 3 \times 3)}{12}$

Multiply the terms in numerator

$\frac{ - 16 - 9}{12}$

$\frac{ - 25}{12}$

So the answer is

$\frac{P}{Q} + \frac{Q}{P} = \frac{ - 25}{12}$

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