Question

solve the equation $p^4 + 40 = 14p^2$. the solutions are $p = pm square$ and $p = pm square$.

Explanation:

Step1: Rearrange to standard quartic form

$p^4 - 14p^2 + 40 = 0$

Step2: Substitute $x=p^2$

$x^2 - 14x + 40 = 0$

Step3: Factor the quadratic

$(x - 4)(x - 10) = 0$

Step4: Solve for $x$

$x=4$ or $x=10$

Step5: Substitute back $x=p^2$

$p^2=4$ or $p^2=10$

Step6: Solve for $p$

$p=\pm2$ or $p=\pm\sqrt{10}$

Answer:

The solutions are $p = \pm 2$ and $p = \pm \sqrt{10}$.