Sum of the non-real roots of x 2- x -2 x 2- x -3-12 isA- -1B- 1C- -6D- 6

The correct option is A 1 nbsp;(x^2+x 2)(x^2+x 3)=12 Put x^2 +x =y Then (y 2) (y 3)=12 ⇒ y^2 5y 6=0 ⇒ (y 6)(y+1)=0 ⇒ y= 1,6 y=6 ⇒ x^2 +x 6=0 ⇒ (x+3) (x ...

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Standard XII

Mathematics

Nature of Roots

Sum of the no...

Question

Sum of the non-real roots of $(x^{2} + x - 2) (x^{2} + x - 3) = 12$ is

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Solution

The correct option is A $- 1$
$(x^{2} + x - 2) (x^{2} + x - 3) = 12$
Put $x^{2} + x = y$
Then $(y - 2) (y - 3) = 12$
$\Rightarrow y^{2} - 5 y - 6 = 0$
$\Rightarrow (y - 6) (y + 1) = 0$
$\Rightarrow y = - 1, 6$

$y = 6 \Rightarrow x^{2} + x - 6 = 0$
$\Rightarrow (x + 3) (x - 2) = 0$
$\Rightarrow x = - 3, 2$

$y = - 1 \Rightarrow x^{2} + x + 1 = 0$
$D = - 3 < 0$
So, this equation has non-real roots whose sum is $- 1$

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Q. Sum of non-real roots of

(x^{2} + x - 2) (x^{2} + x - 3) = 12

Q. The product of non-real roots of the equation

(x^{2} + x - 2) (x^{2} + x - 3) = 12

Q. Find the sum of the non-real roots of the equation

(x^{2} + x - 2) (x^{2} + x - 3) = 12

Q. Sum of the real roots of

(x^{2} + x - 2) (x^{2} + x - 3) = 12

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Sum of the non-real roots of (x2+x−2)(x2+x−3)=12 is

The correct option is A −1 (x2+x−2)(x2+x−3)=12 Put x2+x=y Then (y−2)(y−3)=12 ⇒y2−5y−6=0 ⇒(y−6)(y+1)=0 ⇒y=−1,6 y=6⇒x2+x−6=0 ⇒(x+3)(x−2)=0 ⇒x=−3,2 y=−1⇒x2+x+1=0 D=−3<0 So, this equation has non-real roots whose sum is −1

Sum of the non-real roots of $(x^{2} + x - 2) (x^{2} + x - 3) = 12$ is