Multiple choice

If the equation $a{ x }^{ 2 }+x+b=0$ and ${ x }^{ 2 }+bx+a=0$ have a common root and the second equation has equal roots, then $2{ a }^{ 2 }+b=$

  1. 0

  2. -1

  3. 1

  4. 2
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A Correct answer
Explanation

Second equation x^2 + bx + a = 0 has equal roots, so D = b^2 - 4a = 0, meaning b^2 = 4a. If they have a common root r, then ar^2 + r + b = 0 and r^2 + br + a = 0. Solving these leads to 2a^2 + b = 0.

AI explanation

Since the quadratic equation x^2 + bx + a = 0 has equal roots, its discriminant must be zero, meaning b^2 - 4a = 0, so b^2 = 4a. The equal root of this equation is r = -b/2. Because the two equations share this common root, substituting r into the first equation a x^2 + x + b = 0 gives a(-b/2)^2 + (-b/2) + b = 0. Substituting b^2 = 4a into this yields (b^2/4)(b^2/4) - b/2 + b = 0, which simplifies to b^4/16 + b/2 = 0. Factoring out b/16 gives b(b^3 + 8) = 0, so b = 0 or b = -2. If b = 0, then a = 0, making the expression 2a^2 + b equal to 0. If b = -2, then a = 1, making the expression 2(1)^2 + (-2) = 0. In both cases, the result is 0.

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