Hey,

I came upon a problem where I cannot understand why $\displaystyle {x^2} + 6$ has no real solutions for the polynomial equation $\displaystyle f(x) = - 3({x^2} + 6){({x^2} + 4)^2}$ . Any help is greatly appreciated.

Sam

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- Nov 8th 2011, 09:26 AMArcherSamNo Real Solutions
Hey,

I came upon a problem where I cannot understand why $\displaystyle {x^2} + 6$ has no real solutions for the polynomial equation $\displaystyle f(x) = - 3({x^2} + 6){({x^2} + 4)^2}$ . Any help is greatly appreciated.

Sam - Nov 8th 2011, 09:37 AMSironRe: No Real Solutions
You're talking about the equation, but what's $\displaystyle f(x)$? Is $\displaystyle f(x)=0$?

- Nov 8th 2011, 09:47 AMArcherSamRe: No Real Solutions
Yes, $\displaystyle f(x) = 0$:

$\displaystyle \begin{array}{l} f(x) = - 3({x^2} + 6){({x^2} + 4)^2}\\ f(x) = 0 = - 3({x^2} + 6){({x^2} + 4)^2}\\ 0 \ne - 3\\ 0 = {x^2} + 6\\ 0 = {({x^2} + 4)^2} \end{array}$ - Nov 8th 2011, 09:51 AMe^(i*pi)Re: No Real Solutions
- Nov 8th 2011, 09:53 AMArcherSamRe: No Real Solutions
- Nov 8th 2011, 09:54 AMDevenoRe: No Real Solutions
suppose that:

$\displaystyle x^6 + 6 = 0\ \implies x^2 = -6$

which real numbers have negative squares?

$\displaystyle x = 0 \implies x^2 = 0,\ x > 0 \implies x^2 > 0$

$\displaystyle x < 0 \implies -x > 0,\ x^2 = (1)(x^2) = (-1)^2x^2 = (-x)^2 > 0$ - Nov 8th 2011, 09:56 AMArcherSamRe: No Real Solutions
LOL, the non-negative property of even exponents has conquered me again. Thanks Deveno.

- Nov 8th 2011, 09:59 AMBobBaliRe: No Real Solutions
When findin the zeros for. $\displaystyle x^2+6 = 0$

$\displaystyle x^2 = -6$

$\displaystyle x^2 = {sqrt}{-6}$

*Not possible to have square root of a negative number, hence

no real roots. Also if you graph the parabola for the above

Quad equation, you will see it is a transaltion of 6units vertically

up the yaxis, and the graph doesn't cut the xaxis; hence no solutions. - Nov 8th 2011, 10:03 AMBobBaliRe: No Real Solutions
When findin the zeros for. $\displaystyle x^2+6 = 0$

$\displaystyle x^2 = -6$

$\displaystyle x = {sqrt}{-6}$

*Not possible to have square root of a negative number, hence

no real roots. Also if you graph the parabola for the above

Quad equation, you will see it is a transaltion of 6units vertically

up the yaxis, and the graph doesn't cut the xaxis; hence no solutions.