Question

Find the roots of the equation (x^2+3x)^2-(x^2+3x)-6=0

Answer 1

Answer:

Answer given below v v v v v

Step-by-step explanation:

$(x^2+3x)^2-(x^2+3x)-6=0$

According to quadratic formula,

$x^2+3x=\frac{-(-1)\pm\sqrt{(-1)^2-4(1)(-6)}}{2(1)} \\=\frac{1\pm \sqrt{1+24}}{2} \\\\=\frac{1\pm \sqrt{25}}{2} \\\\=\frac{1\pm 5}{2}=[3,-2]$

If x²+3x=3,   then x=[-3±√(9-(-12))]/2*1=[-3±√21]/2

                   

If x²+3x=-2 then x=[-3±√(9-8)]/2*1=(-3±1)/2={-1,-2}

So,

$x=[\frac{-3\pm \sqrt{21}}{2},-1,-2]$

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