1. ## solve for x

i need a help solving this algebraic equation
4(x-3)^2 +1=√((x-1)/4) +3

plz i need some quick answers
i would really appreciate whoever solves this question
thanks

2. ## Yeesh

4(x-3)^2 +1=√((x-1)/4) +3
4(x-3)^2 - 2 = √((x-1)/4) = 1/2 * √(x-1)
8(x-3)^2 - 4 = √(x-1)
8 x^2 - 48 x + 68 = √(x-1) = t
8 (t^2 + 1)^2 - 48 (t^2 + 1) + 68 = t (because x = t^2 + 1)
8 t^4 - 32 t^2 - t + 28 = 0

I am stuck here.

3. Originally Posted by hawtmonkey
i need a help solving this algebraic equation
4(x-3)^2 +1=√((x-1)/4) +3

$4(x-3)^2+1= \sqrt{\frac{x-1}{4}} + 3$

$4(x^2-6x+9) +1 = \sqrt{\frac{x-1}{4}} +3$

$4x^2- 24x +36 + 1 = \sqrt{\frac{x-1}{4}} +3$

$4x^2 -24x + 34 = \sqrt{\frac{x-1}{4}}$

$(4x^2 -24x + 34)^2 = \frac{x-1}{4}$

$16x^4 -192x^3 +848x^2-1632x + 1156 = \frac{x-1}{4}$

$64x^4-768x^3 + 3392x^2 -6528x +4624 = x -1$

$64x^4-768x^3 + 3392x^2 -6529x +4625 = 0$

which factors out to

$(4x^2-25x +37)(16x^2-92x+125) = 0$

according to TI89

so

$(4x^2-25x +37)=0$

$(16x^2-92x+125)=0$

You should be able to finish up from here.

4. thanks for the solution
thanks a lot

5. can you show me how you factored 64x^4-768x^3+3392x^2-6529x+4625=0

to:

(4x^2-25x=37)(16x^2-92+125)=0