You visited us 1 times! Enjoying our articles? Unlock Full Access!

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

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

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

Open in App

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 $e$
$e = ((0)^{2} + 7 (0) + 3)^{2} - (0 - 6) (0 - 1) (0 - 9) - 5 = 149$
$∴ Product of the root = e = 149$

Suggest Corrections

Similar questions

\sqrt{3} x^{2} - 2 x + \frac{1}{2} = 0

x^{2} + \frac{1}{x^{2}} = 5

x - \frac{3}{x} = x^{2}

2 x^{2} - \sqrt{3 x} + 9 = 0

x^{2} - 2 x - \sqrt{x} - 5 = 0

x + \frac{1}{x} = 1

{(x + \frac{1}{x})}^{2} = 3 (x + \frac{1}{x}) + 4

x + \frac{1}{x} = x^{2}, x \neq 0

x^{2} - 6 x + 6 = 0

x^{2} + 6 x - 6 = 0

x^{2} - 6 x - 6 = 0

x^{2} + 6 x + 6 = 0

∣ ∣ ∣ \frac{(x + 3) (2 + x)}{(x - 1) (x - 5)} ∣ ∣ ∣ > \frac{x^{2} + 5 x + 6}{x^{2} - 6 x + 5}

x^{2} - 6 x - \sqrt{x^{2} - 6 x - 3} = 5

Join BYJU'S Learning Program