# How do you solve using the quadratic formula x^4 - 26x^2 +25?

Apr 30, 2015

In this way:

let $t = {x}^{2}$, so:

${t}^{2} - 26 t + 25 = 0 \Rightarrow$

$\Delta = {b}^{2} - 4 a c = {26}^{2} - 4 \cdot 1 \cdot 25 = 676 - 100 = 576 = {24}^{2}$

${t}_{1 , 2} = \frac{- b \pm \sqrt{\Delta}}{2 a} = \frac{26 \pm 24}{2}$

${t}_{1} = \frac{26 - 24}{2} = 1$

${t}_{2} = \frac{26 + 24}{2} = \frac{50}{2} = 25$.

So:

${x}^{2} = 1 \Rightarrow x = \pm 1$

${x}^{2} = 25 \Rightarrow x = \pm 5$.

Since the coefficient $b$ of the quadratic equation is even, we could use also the reducted formula.

$\frac{\Delta}{4} = {\left(\frac{b}{2}\right)}^{2} - a c = 169 - 25 = 144 = {12}^{2}$

${t}_{1 , 2} = \frac{- \frac{b}{2} \pm \sqrt{\frac{\Delta}{2}}}{a} = \frac{13 \pm 12}{1}$

with, obviously, the same two solutions!