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
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
$\displaystyle 4(x-3)^2+1= \sqrt{\frac{x-1}{4}} + 3$
$\displaystyle 4(x^2-6x+9) +1 = \sqrt{\frac{x-1}{4}} +3$
$\displaystyle 4x^2- 24x +36 + 1 = \sqrt{\frac{x-1}{4}} +3$
$\displaystyle 4x^2 -24x + 34 = \sqrt{\frac{x-1}{4}} $
$\displaystyle (4x^2 -24x + 34)^2 = \frac{x-1}{4}$
$\displaystyle 16x^4 -192x^3 +848x^2-1632x + 1156 = \frac{x-1}{4}$
$\displaystyle 64x^4-768x^3 + 3392x^2 -6528x +4624 = x -1$
$\displaystyle 64x^4-768x^3 + 3392x^2 -6529x +4625 = 0$
which factors out to
$\displaystyle (4x^2-25x +37)(16x^2-92x+125) = 0$
according to TI89
so
$\displaystyle (4x^2-25x +37)=0$
$\displaystyle (16x^2-92x+125)=0$
You should be able to finish up from here.