Multiple choice

Two real numbers $\alpha$ & $\beta$ are such that $\alpha+\beta=3, \alpha-\beta=4$, then $\alpha$ & $\beta$ are the roots of the quadratic

  1. $4x^{2}-12x-7=0$
  2. $4x^{2}-12x+7=0$
  3. $4x^{2}-12x-25=0$
  4. $None\ of\ these$
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A Correct answer
Explanation

Given alpha + beta = 3 and alpha - beta = 4. Adding: 2*alpha = 7 => alpha = 3.5. Subtracting: 2*beta = -1 => beta = -0.5. The quadratic is (x - 3.5)(x + 0.5) = 0 => x^2 - 3x + 0.5x - 1.75 = 0 => x^2 - 3x - 1.75 = 0. Multiplying by 4: 4x^2 - 12x - 7 = 0.

AI explanation

By adding the two given equations, a + b = 3 and a - b = 4, we find that 2a = 7, so a = 3.5. Substituting this back into the first equation gives b = -0.5. A quadratic equation with roots a and b can be formed using the sum and product of the roots, where the sum is 3 and the product is -1.75. Multiplying the monic polynomial x^2 - 3x - 1.75 by 4 to clear the decimals results in 4x^2 - 12x - 7 = 0.

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