Question

Two non-integers roots of equation

$(x {}^{2} + 3x) {}^{2} - (x {}^{2} + 3x) - 6 = 0$
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Answer 1

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$\tt\:(x {}^{2} + 3x) {}^{2} - (x {}^{2} + 3x) - 6 = 0 \: \: \: \: \: .......(1)$

$\tt\:put \: \: \: x {}^{2} + 3x = y$

$\tt\:y {}^{2} - y - 6 = 0 \\ \\ (y + 2)(y - 3) = 0 \\ \\ y = \: - 2 \: \: \: (or) \: \: 3$

$\tt\:x {}^{2} + 3x = - 2 \\ \\ x {}^{2} + 3x + 2 = 0 \\ \\ x = - 1$

$\tt\:x \: \: = \: \: \dfrac{1}{2} ( - 3± \sqrt{9 + 12} ) \: \: =\tt \dfrac{1}{2} ( - 3± \sqrt{21} )$

Answer 2

Solution

Two integer root of the equation (x^2 +3x)- (x^2+3x) -6= 0 are-

Let us put

==)x^2+3x=y

==)so, y^2-y-6= (y+2)(y-3)=0

==) Therefore, y= -2 OR y= 3

==)x^2+3x= -2

==) x^2+3x+2=0

==)(x+1) (x+2)=0

==) x= -1 or x= -2

==) Now, y= 3

==) x^2+3x-3=0

==) x= 1/2(-3+_√9+√12)

==) x= 1/2(-3+_√21)

HOPE IT HELP