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CAT 2022 Slot 2 — QA Question 12

Higher Degree PolynomialsEasy

Let r and c be real numbers. If r and -r are roots of 5x3 + cx2 - 10x + 9 = 0, then c equals

Answer & solution

Correct answer: -9/2

  • A

    4

  • -9/2

  • C

    -4

  • D

    9/2

Solution

Easy

The roots are r,−r,γr,-r,\gamma. Use Vieta's formulas for a cubic to turn the symmetric sums of roots into equations in cc and γ\gamma, then eliminate.

1

Write Vieta's relations for 5x3+cx2−10x+9=05x^3+cx^2-10x+9=0 with roots r,−r,γr,-r,\gamma:

r+(−r)+γ=−c5 ⇒ γ=−c5r(−r)+(−r)γ+γr=−105=−2 ⇒ −r2=−2 ⇒ r2=2r(−r)γ=−95 ⇒ −r2γ=−95\begin{aligned} &r+(-r)+\gamma=-\tfrac{c}{5}\ \Rightarrow\ \gamma=-\tfrac{c}{5}\\ &r(-r)+(-r)\gamma+\gamma r=\tfrac{-10}{5}=-2\ \Rightarrow\ -r^2=-2\ \Rightarrow\ r^2=2\\ &r(-r)\gamma=-\tfrac{9}{5}\ \Rightarrow\ -r^2\gamma=-\tfrac{9}{5} \end{aligned}
2

Find γ\gamma using r2=2r^2=2 in the product relation:

−2γ=−95 ⇒ γ=910-2\gamma=-\tfrac{9}{5}\ \Rightarrow\ \gamma=\tfrac{9}{10}
3

Equate the two expressions for γ\gamma:

−c5=910 ⇒ c=−92-\tfrac{c}{5}=\tfrac{9}{10}\ \Rightarrow\ c=-\tfrac{9}{2}
c=−92c=\mathbf{-\tfrac{9}{2}}

Since r,−rr,-r are roots, plug both in: f(r)+f(−r)=0f(r)+f(-r)=0 kills the odd-degree terms, leaving 2(cr2+9)=0⇒c=−9/r22(cr^2+9)=0\Rightarrow c=-9/r^2. The middle Vieta sum gives r2=2r^2=2 instantly, so c=−9/2c=-9/2.

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CAT 2022 Slot 2 QA Q12 — Higher Degree Polynomials solution | TheCATExam