Solve the equation p^4+40=14p^2 - The solutions are p=pm square and p=pm square

To solve the equation $p^{4}+40=14p^{2}$, we can rewrite it as $p^{4}-14p^{2}+40=0$.<br /><br />Let $x=p^{2}$, then the equation becomes $x^{2}-14x+40=0$.<br /><br />We can solve this quadratic equation by factoring or using the quadratic formula. The quadratic formula is given by:<br /><br />$x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}$<br /><br />In this case, $a=1$, $b=-14$, and $c=40$. Plugging these values into the quadratic formula, we get:<br /><br />$x=\frac{14\pm \sqrt{196-160}}{2}$<br />$x=\frac{14\pm \sqrt{36}}{2}$<br />$x=\frac{14\pm 6}{2}$<br /><br />So, the solutions for $x$ are $x=10$ and $x=4$.<br /><br />Since $x=p^{2}$, we can find the values of $p$ by taking the square root of $x$:<br /><br />$p=\pm \sqrt{10}$ and $p=\pm \sqrt{4}$<br /><br />Therefore, the solutions to the equation $p^{4}+40=14p^{2}$ are $p=\pm \sqrt{10}$ and $p=\pm 2$.