Multiple choice

If the roots of the equation $\displaystyle \left ( a^{2}+b^{2} \right )x^{2}-2b\left ( a+c \right )x+\left ( b^{2}+c^{2} \right )=0 $ are equal then

  1. $2b = ac$
  2. $\displaystyle b^{2}=ac $
  3. $\displaystyle b=\frac{2ac}{a+c} $

  4. b = ac
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For equal roots, the discriminant D = b^2 - 4ac must be zero. Applying this to the given equation: (-2b(a+c))^2 - 4(a^2+b^2)(b^2+c^2) = 0. Expanding and simplifying leads to 4b^2(a^2+2ac+c^2) - 4(a^2b^2 + a^2c^2 + b^4 + b^2c^2) = 0, which reduces to b^2 = ac.

AI explanation

For the equation to have equal roots, its discriminant must equal zero, so we set [-2b(a + c)]^2 - 4(a^2 + b^2)(b^2 + c^2) = 0. Expanding both squares gives 4b^2(a^2 + 2ac + c^2) - 4(a^2b^2 + a^2c^2 + b^4 + b^2c^2) = 0. Dividing by 4 and canceling common terms leaves 2ab^2c - a^2c^2 - b^4 = 0, which factors perfectly as (ac - b^2)^2 = 0. Taking the square root of both sides yields b^2 = ac.

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