No Real Solutions

• Nov 8th 2011, 09:26 AM
ArcherSam
No 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 AM
Siron
Re: 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 AM
ArcherSam
Re: 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 AM
e^(i*pi)
Re: No Real Solutions
Quote:

Originally Posted by ArcherSam
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

Quote:

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

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

You've changed from $\displaystyle (x^2+4)^2$ in post 1 to $\displaystyle (x^2+6)^2$ in post 3. Which is correct?

In the case of post 3 write $\displaystyle f(x) = -3(x^2+6)(x^2+6)^2 = -3(x^2+6)^3$
• Nov 8th 2011, 09:53 AM
ArcherSam
Re: No Real Solutions
Quote:

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

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

$\displaystyle {({x^2} + 4)^2}$ is correct. Sorry about the error.
• Nov 8th 2011, 09:54 AM
Deveno
Re: 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 AM
ArcherSam
Re: No Real Solutions
LOL, the non-negative property of even exponents has conquered me again. Thanks Deveno.
• Nov 8th 2011, 09:59 AM
BobBali
Re: 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 AM
BobBali
Re: 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.