Two non - integer roots of the equation $(x^{2}+3x)^{2}-(x^{2}+3x)-6= 0$ are
- $\frac{1}{2}(-3+\sqrt{11}),\frac{1}{2}(-3-\sqrt{11})$
- $\frac{1}{2}(-3+\sqrt{7}),\frac{1}{2}(-3-\sqrt{7})$
- $\frac{1}{2}(-3+\sqrt{21}),\frac{1}{2}(-3-\sqrt{21})$
-
none of these
Reveal answer
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C
Correct answer
Explanation
Let u = x^2 + 3x. The equation becomes u^2 - u - 6 = 0, which factors to (u-3)(u+2) = 0. So x^2 + 3x = 3 or x^2 + 3x = -2. Solving x^2 + 3x - 3 = 0 gives x = (-3 +/- sqrt(9 - 4(1)(-3)))/2 = (-3 +/- sqrt(21))/2.
AI explanation
Treating the equation as a quadratic in u = x^2 + 3x gives u^2 - u - 6 = 0. Factoring this gives (u - 3)(u + 2) = 0, resulting in two possible values for u: 3 and -2. Substituting u back gives x^2 + 3x = 3 and x^2 + 3x = -2. The integer roots come from x^2 + 3x + 2 = 0, while the non-integer roots are found by solving x^2 + 3x - 3 = 0 using the quadratic formula. Applying the formula gives x = (-3 +/- sqrt(21)) / 2.