Multiple choice

On Solving the following quadratic equation by factorization, the roots are $\dfrac{a^2 \, + \, b^2}{3}, \, \dfrac{b^2 \, - \, a^2}{3}$ : $9x^2 \, - \, 6b^2x \, - \, (a^4 \,- \, b^4) \, = \, 0$

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The equation is 9x^2 - 6b^2x - (a^4 - b^4) = 0. Using the quadratic formula, x = [6b^2 +/- sqrt(36b^4 + 36(a^4 - b^4))] / 18 = [6b^2 +/- sqrt(36a^4)] / 18 = [6b^2 +/- 6a^2] / 18 = (b^2 +/- a^2) / 3. The roots are (b^2+a^2)/3 and (b^2-a^2)/3.

AI explanation

To verify the roots for 9x^2 - 6b^2x - (a^4 - b^4) = 0, we split the middle term to get 9x^2 - 3(a^2 + b^2)x + 3(a^2 - b^2)x - (a^2 - b^2)(a^2 + b^2) = 0. Factoring by grouping gives 3x(3x - (a^2 + b^2)) + (a^2 - b^2)(3x - (a^2 + b^2)) = 0, which becomes (3x + (a^2 - b^2))(3x - (a^2 + b^2)) = 0. Setting each factor to zero gives x = (a^2 + b^2)/3 and x = (b^2 - a^2)/3, confirming the statement is True.

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