Question

Solve the equation x^4-2x^3+4x^2+6x-21=0 the sum of two of the roots being zero

Answer 1

GIVEN :

Solve the equation $x^4-2x^3+4x^2+6x-21=0$ the sum of two of the roots being zero

TO FIND :

Solve the equation $x^4-2x^3+4x^2+6x-21=0$ the sum of two of the roots being zero

SOLUTION :

Given equation is  $x^4-2x^3+4x^2+6x-21=0$

Also give that the sum of two of the roots being zero

Let $\alpha,\beta,\gamma,\delta$ be the roots of given equation

Since sum of two roots is zero.

ie., $\alpha+\beta= 0$

Now we have that $\alpha+\beta+\gamma+\delta=2$

⇒ $0+\gamma+\delta = 2$

$\gamma+\delta = 2$

Let $\alpha \beta= p$, $\gamma \delta= q$

The quadratic equation having the roots $\alpha,\beta$ is $x^2 -(\alpha+\beta) x +\alpha \beta = 0$

∴ $x^2+0+ p = 0$

$x^2+p=0$

The equation having the roots $\gamma,\delta$ is $x^2 - (\gamma+\delta)x + \gamma \delta = 0$

∴ $x^2 - 2x + q = 0$ (since $\gamma+\delta = 2$)

∴ $x^4-2x^3 + 4x^2 + 6x - 21 = (x^2 + p) (x^2-2x + q)$

$= x^4-2x^3 + x^2(p + q) - 2px + pq$

$x^4-2x^3+4x^2+6x-21=x^4-2x^3 + x^2(p + q) - 2px + pq$

Equating the coefficients we get

p+q=4          -2p=6

-3+q=4          $p=\frac{6}{-2}$

q=4+3           p=-3

q=7

⇒ q=7 and p=-3

Now for the equation $x^2+p=0$

$x^2-3=0$

$x^2=3$

$x=\pm \sqrt{3}$

$x=\sqrt{3}$ and $x=-\sqrt{3}$ are the roots

Now for the equation $x^2 - 2x + q = 0$

$x^2-2x+7=0$

For a quadratic equation $ax^2+bx+c=0$ the solution is

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

From this we have a=1 , b=-2 and c=7

$x=\frac{-(-2)\pm \sqrt{(-2)^2-4(1)(7)}}{2(1)}$

$=\frac{2\pm \sqrt{4-28}}{2}$

$=\frac{2\pm \sqrt{-24}}{2}$

$=\frac{2\pm \sqrt{24i^2}}{2}$ (∵ $i^2=-1$)

$=\frac{2\pm 2i\sqrt{6}}{2}$

$x=1\pm i\sqrt{6}$

$x=1+i\sqrt{6}$ and $x=1-i\sqrt{6}$ are the roots

∴ $\sqrt{3}$ $-\sqrt{3}$ , $1+i\sqrt{6}$ and $1-i\sqrt{6}$ are the roots for the given equation.

Answer 2

Given:

$x^4-2x^3+4x^2+6x-21=0$

To Find:

Two roots?

Solution:

Since, here it is given that sum of two of the roots being zero.

        ⇒ roots are equal in magnitude but opposite in signs.

       so, let k and -k be two roots of given equation.

then, $x=k$ satisfy the above equation:-

   so,$k^{4} -2k^{3} +4k^{2} +6k-21=0$ ........(1)

  also, $x= -k$ also satisfies the equation:-

       $(-k)^{4} - 2(-k)^{3} +4(-k)^{2} +6(-k)-21=0\\k^{4} + 2k^{3} +4k^{2} -6k-21=0$........(2)

  now, equation (1) - equation (2), we have

  ⇒  $-4k^{3}+12k=0\\ -4k(k^{2} +3)=0$

  ⇒$-4k=0$  or $k^{2} +3=0$

  ⇒$k=0$  or $k^{2} = -3$

 but, $k\neq 0$ because roots have equal and opposite signs.

 therefore, $k=\pm\sqrt{3}$

Hence, $x = \sqrt{3}$ and $x = -\sqrt{3}$ are two roots.