Multiple choice

If the sum of the squares of the roots of the equation ${ x }^{ 2 }-\left( a-2 \right) x-\left( a+1 \right) =0$ is least, then the value of $a$ is

  1. $-1$
  2. $1$
  3. $2$
  4. $-2$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let roots be a, b. a+b = a-2, ab = -(a+1). Sum of squares = (a+b)^2 - 2ab = (a-2)^2 + 2(a+1) = a^2 - 4a + 4 + 2a + 2 = a^2 - 2a + 6. This is a parabola opening upward, minimum at a = -(-2)/(2*1) = 1.

AI explanation

Let the roots of the quadratic equation be m and n, where their sum is a minus 2 and their product is minus a minus 1. The sum of the squares of the roots is found using the identity m squared plus n squared equals the square of the sum of the roots minus two times their product. Substituting the values gives the quantity a minus 2 squared minus 2 times the quantity negative a minus 1, which simplifies to a squared minus 2a plus 6. To find the minimum value of this expression, we complete the square to get the quantity a minus 1 squared plus 5. The minimum occurs when a equals 1.

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