Solve the following quadratic equation by factorization method: $\sqrt{5}x^2+2x-3\sqrt{5}=0$
- $\left \{ -\sqrt{5}, \dfrac{3}{\sqrt{5}} \right \}$
- $\left \{ \sqrt{5}, \dfrac{3}{\sqrt{5}} \right \}$
- $\left \{ -\sqrt{5}, \dfrac{-3}{\sqrt{5}} \right \}$
-
None of these
Reveal answer
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A
Correct answer
Explanation
The equation is sqrt(5)x^2 + 2x - 3sqrt(5) = 0. Using the quadratic formula or factoring: sqrt(5)x^2 + 5x - 3x - 3sqrt(5) = 0, which is sqrt(5)x(x + sqrt(5)) - 3(x + sqrt(5)) = 0. This gives (sqrt(5)x - 3)(x + sqrt(5)) = 0, resulting in x = -sqrt(5) and x = 3/sqrt(5).
AI explanation
To solve sqrt(5)x^2 + 2x - 3sqrt(5) = 0 by factorization, we find two numbers that multiply to -15 and add to 2, which are 5 and -3. We split the middle term to get sqrt(5)x^2 + 5x - 3sqrt(5)x - 3sqrt(5) = 0, and factor by grouping to obtain (sqrt(5)x - 3)(x + sqrt(5)) = 0. Setting both factors to zero gives the roots 3/sqrt(5) and -sqrt(5).