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CAT 2017 Slot 2 — QA Question 24

Forming a Quadratic Equation and Relation between roots and coefficientsEasy

The minimum possible value of the squares of the roots of the equation: x2 + (a + 3)x – (a + 5) = 0 is

Answer & solution

Correct answer: 3

  • A

    1

  • B

    2

  • 3

  • D

    4

Solution

Easy

Let the roots be α,β\alpha,\beta. Use Vieta's formulas to write the sum of the squares α2+β2\alpha^2+\beta^2 as a function of aa, then complete the square to find its minimum.

1

Apply Vieta's formulas. For x2+(a+3)x−(a+5)=0x^2+(a+3)x-(a+5)=0, the sum and product of the roots are:

α+β=−(a+3)αβ=−(a+5)\begin{aligned} &\alpha+\beta=-(a+3)\\ &\alpha\beta=-(a+5) \end{aligned}
2

Express the sum of squares in terms of aa. Use α2+β2=(α+β)2−2αβ\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta.

α2+β2=(α+β)2−2αβ⇒ α2+β2=(a+3)2−2(−(a+5))(from step 1)⇒ α2+β2=a2+6a+9+2a+10⇒ α2+β2=a2+8a+19\begin{aligned} &\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta\\ &\Rightarrow\ \alpha^2+\beta^2=(a+3)^2-2\big(-(a+5)\big) \quad\text{(from step 1)}\\ &\Rightarrow\ \alpha^2+\beta^2=a^2+6a+9+2a+10\\ &\Rightarrow\ \alpha^2+\beta^2=a^2+8a+19 \end{aligned}
3

Complete the square and minimise. Write the quadratic in aa as a perfect square plus a constant.

α2+β2=(a2+8a+16)+3⇒ α2+β2=(a+4)2+3\begin{aligned} &\alpha^2+\beta^2=(a^2+8a+16)+3\\ &\Rightarrow\ \alpha^2+\beta^2=(a+4)^2+3 \end{aligned}

Since (a+4)2≥0(a+4)^2\ge 0, the minimum occurs at a=−4a=-4, giving (a+4)2=0(a+4)^2=0.

(α2+β2)min⁡=0+3=3\begin{aligned} &\big(\alpha^2+\beta^2\big)_{\min}=0+3=3 \end{aligned}
α2+β2=3\alpha^2+\beta^2=3

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CAT 2017 Slot 2 QA Q24: The minimum possible value of the squares of the roots of the equation: x 2 + (a + 3)x – (a + 5) = 0 is — Solution | TheCATExam