Find the roots of the following equation $3x^4\,-\,13x^2\,+\,10\,=\,0$, then
- $\displaystyle\,x\,=\,\pm\,\sqrt\frac{10}{3},\,\pm\,3$
- $\displaystyle\,x\,=\,\pm\,\sqrt\frac{10}{7},\,\pm\,1$
- $\displaystyle\,x\,=\,\pm\,\sqrt\frac{10}{3},\,\pm\,1$
- $\displaystyle\,x\,=\,\pm\,\sqrt\frac{10}{3},\,\pm\,2$
Reveal answer
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C
Correct answer
Explanation
Let y = x^2. The equation becomes 3y^2 - 13y + 10 = 0. Factoring: (3y - 10)(y - 1) = 0. So y = 10/3 or y = 1. Therefore, x^2 = 10/3 or x^2 = 1. Roots are x = +/- sqrt(10/3) and x = +/- 1.
AI explanation
Treat 3x^4 - 13x^2 + 10 = 0 as a quadratic in x^2 and split the middle term to get 3x^4 - 3x^2 - 10x^2 + 10 = 0. Factoring by grouping yields (x^2 - 1)(3x^2 - 10) = 0. Equating the factors to zero gives x^2 = 1 or x^2 = 10/3, so the roots are x = plus or minus 1 and x = plus or minus the square root of (10/3).