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Standard XII

Mathematics

Higher Order Equations

The sum of in...

Question

The sum of integeral roots of the equation $(x^{2} + 3 x)^{2} - (x^{2} + 3 x) - 6 = 0$ is

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Solution

The correct option is D $- 3$
Put $x^{2} + 3 x = y$
$\Rightarrow y^{2} - y - 6 = 0 \Rightarrow y = - 2, 3$
Now
Case $I :$ $x^{2} + 3 x = - 2 \Rightarrow x = - 1, - 2$
Case $I I :$ $x^{2} + 3 x = 3 \Rightarrow x = \frac{1}{2} (- 3 \pm \sqrt{21})$
So the sum of the integral roots of the equation is $- 3$

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The sum of integeral roots of the equation (x2+3x)2−(x2+3x)−6=0 is

The correct option is D −3Put x2+3x=y ⇒y2−y−6=0⇒y=−2, 3 Now Case I: x2+3x=−2⇒x=−1, −2 Case II: x2+3x=3⇒x=12(−3±√21) So the sum of the integral roots of the equation is −3

The sum of integeral roots of the equation $(x^{2} + 3 x)^{2} - (x^{2} + 3 x) - 6 = 0$ is