1. ## No Real Solutions

Hey,

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

Sam

2. ## Re: No Real Solutions

You're talking about the equation, but what's $f(x)$? Is $f(x)=0$?

3. ## Re: No Real Solutions

Yes, $f(x) = 0$:

$\begin{array}{l} f(x) = - 3({x^2} + 6){({x^2} + 4)^2}\\ f(x) = 0 = - 3({x^2} + 6){({x^2} + 4)^2}\\ 0 \ne - 3\\ 0 = {x^2} + 6\\ 0 = {({x^2} + 4)^2} \end{array}$

4. ## Re: No Real Solutions

Originally Posted by ArcherSam
Hey,

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

Sam
Originally Posted by ArcherSam
Yes, $f(x) = 0$:

$\begin{array}{l} f(x) = - 3({x^2} + 6){({x^2} + 6)^2}\\ f(x) = 0 = - 3({x^2} + 6){({x^2} + 6)^2}\\ 0 \ne - 3\\ 0 = {x^2} + 6\\ 0 = {({x^2} + 6)^2} \end{array}$
You've changed from $(x^2+4)^2$ in post 1 to $(x^2+6)^2$ in post 3. Which is correct?

In the case of post 3 write $f(x) = -3(x^2+6)(x^2+6)^2 = -3(x^2+6)^3$

5. ## Re: No Real Solutions

Originally Posted by e^(i*pi)
You've changed from $(x^2+4)^2$ in post 1 to $(x^2+6)^2$ in post 3. Which is correct?

In the case of post 3 write $f(x) = -3(x^2+6)(x^2+6)^2 = -3(x^2+6)^3$
${({x^2} + 4)^2}$ is correct. Sorry about the error.

6. ## Re: No Real Solutions

suppose that:

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

which real numbers have negative squares?

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

$x < 0 \implies -x > 0,\ x^2 = (1)(x^2) = (-1)^2x^2 = (-x)^2 > 0$

7. ## Re: No Real Solutions

LOL, the non-negative property of even exponents has conquered me again. Thanks Deveno.

8. ## Re: No Real Solutions

When findin the zeros for. $x^2+6 = 0$
$x^2 = -6$
$x^2 = {sqrt}{-6}$

*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.

9. ## Re: No Real Solutions

When findin the zeros for. $x^2+6 = 0$
$x^2 = -6$
$x = {sqrt}{-6}$

*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.