Question

Quadratic Equations Of 4x^4-4x^3-7x^2-4x-4 = 0

Answer 1

Given:

Quadratic Equations Of 4x^4 - 4x^3 - 7x^2 - 4x - 4 = 0

To find:

The roots of equation 4x^4 - 4x^3 - 7x^2 - 4x - 4 = 0

Solution:

From given, we have an equation 4x^4 - 4x^3 - 7x^2 - 4x - 4 = 0

Let f(x) = 4x^4 - 4x^3 - 7x^2 - 4x - 4

when x = 2, we get,

f(2) = 4(2)^4 - 4(2)^3 - 7(2)^2 - 4(2) - 4

= 64 - 32 - 28 - 8 + 4

= 0

∴ (x - 2) is a factor of 4x^4 - 4x^3 - 7x^2 - 4x - 4.

So, when we divide f(x) by (x - 2), we get remaining factors.

$\dfrac{\left(4x^4-4x^3-7x^2-4x-4\right)}{x-2}$

$=4x^3+\dfrac{4x^3-7x^2-4x-4}{x-2}$

$=4x^3+4x^2+\dfrac{x^2-4x-4}{x-2}$

$=4x^3+4x^2+x+\dfrac{-2x-4}{x-2}$

$=4x^3+4x^2+x-2+\dfrac{-8}{x-2}$

$=4x^3+4x^2+x-2-\dfrac{8}{x-2}$

So, f(x) = (x - 2) (4x³ + 4x² + x - 2)

Again for x = 1/2, f(1/2) = 0

∴ (x - 1/2) is a factor of 4x³ + 4x² + x - 2

⇒ (2x - 1)  is a factor of 4x³ + 4x² + x - 2

Again divide 4x³ + 4x² + x - 2 by (2x - 1)

$\dfrac{4x^3+4x^2+x-2}{2x-1}$

$=\dfrac{\left(2x-1\right)\left(2x^2+3x+2\right)}{2x-1}$

= 2x² + 3x + 2

∴ 2x² + 3x + 2 is a factor of f(x)

Now consider a quadratic equation, 2x² + 3x + 2

2x² + 3x + 2 = 0

$x=-\dfrac{3 \pm i \sqrt 7}{4}$

Therefore, we have,

f(x) = (x - 2) (2x - 1) (2x² + 3x + 2)

Hence the roots are 2, 1/2 and - (3 ± i√7)/4