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

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Answer 1 · 2015-04-30T17:57:03

In this way:

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

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

$Delta=b^2-4ac=26^2-4*1*25=676-100=576=24^2$

$t_(1,2)=(-b+-sqrtDelta)/(2a)=(26+-24)/2$

$t_1=(26-24)/2=1$

$t_2=(26+24)/2=50/2=25$ .

So:

$x^2=1rArrx=+-1$

$x^2=25rArrx=+-5$ .

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

$Delta/4=(b/2)^2-ac=169-25=144=12^2$

$t_(1,2)=(-b/2+-sqrt(Delta/2))/a=(13+-12)/1$

with, obviously, the same two solutions!

How do you solve using the quadratic formula x^4 - 26x^2 +25x4−26x2+25x^4 - 26x^2 +25?