Find the value of a such that the sum of the squares of the roots of the equation x2−(a−2)x−(a+1)=0 is least.
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Step-by-Step Solution
Step 1: Identify coefficients and root properties
For a quadratic equation Ax2+Bx+C=0, the sum of the roots is given by α+β=−B/A and the product of the roots is given by αβ=C/A. In our given equation, A=1, B=−(a−2), and C=−(a+1).
Step 2: Express sum and product of roots
Using the formulas for the sum and product of roots, we substitute the coefficients from our equation. The sum of the roots α+β becomes a−2, and the product of the roots αβ becomes −(a+1).
Step 3: Formulate sum of squares of roots
We want to find the value of a for which the sum of the squares of the roots, α2+β2, is least. We can express α2+β2 in terms of the sum and product of the roots using the identity (α+β)2=α2+β2+2αβ.
Step 4: Substitute and simplify the expression
Substitute the expressions for (α+β) and αβ into the formula for α2+β2. Expand and simplify the resulting quadratic expression in terms of a.
Step 5: Find the minimum value of the quadratic
The expression for the sum of the squares of the roots is a quadratic function of a, f(a)=a2−2a+6. This is a parabola opening upwards, so its minimum value occurs at the vertex. We find the vertex by setting the first derivative with respect to a to zero.
Step 6: Verify minimum using second derivative
The second derivative of f(a) is 2, which is positive. This confirms that a=1 corresponds to a local minimum.