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
suppose that:
$\displaystyle x^6 + 6 = 0\ \implies x^2 = -6$
which real numbers have negative squares?
$\displaystyle x = 0 \implies x^2 = 0,\ x > 0 \implies x^2 > 0$
$\displaystyle x < 0 \implies -x > 0,\ x^2 = (1)(x^2) = (-1)^2x^2 = (-x)^2 > 0$
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.
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.