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Biquadratic Equations of the form: ax^4+bx^3+cx^2+bx+a=0

The product o...

Question

The product of all roots of the equation $(x^{2} + 7 x + 3)^{2} - (x - 6) (x - 1) (x - 9) = 5$ is

149.0

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149.00

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

149

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

0149

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

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Solution

Given equation is
$(x^{2} + 7 x + 3)^{2} - (x - 6) (x - 1) (x - 9) = 5$
$\Rightarrow (x^{2} + 7 x + 3)^{2} - (x - 6) (x - 1) (x - 9) - 5 = 0$
The given equation will be of form
$x^{4} + b x^{3} + c x^{2} + d x + e = 0$
So, the product of the roots $= e$
Putting $x = 0$ , in the given equation we get the value of $e$
$e = ((0)^{2} + 7 (0) + 3)^{2} - (0 - 6) (0 - 1) (0 - 9) - 5 = 149$
$∴ Product of the root = e = 149$

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