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how do i solve for all real values of x in this one? thanks.. 6 (x + 2)^4 - 11 (x + 2)^2 = -4
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i would also change a nicer looking variable for that x+2
then, same as before, synthetic division, those 4 and 6 terms are shouting to be divided
let u = (x + 2)^2 then rewriting the equation gives 6u^2 - 11u + 4 = 0 solve the quadratic (2u -1)(3u - 4) = 0 u = 1/2 and 4/3 then replacing u with (x + 2)^2 (x + 2)^2 = 1/2 solve for x \[x + 2 = \pm \sqrt{1/2} \] \[x = -2\pm \sqrt{1/2}\] I'll let you solve the other solution of 4/3
thanks, dears! :)
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