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CAT 2023 Slot 3 — QA Question 3

Forming a Quadratic Equation and Relation between roots and coefficientsEasy

If x is a positive real number such that x8 + (1x)8 = 47, then the value of x9 + (1x)9 is

Answer & solution

Correct answer: 34√5

  • 34√5

  • B

    36√5

  • C

    40√5

  • D

    30√5

Solution

Easy

Repeatedly "add 2 and take a square root" to climb down the ladder x8+1x8→x4+1x4→x2+1x2→x+1xx^8+\tfrac1{x^8}\to x^4+\tfrac1{x^4}\to x^2+\tfrac1{x^2}\to x+\tfrac1x. Then cube your way back up to reach x9+1x9x^9+\tfrac1{x^9}, using (t+1t)3=t3+1t3+3(t+1t)\left(t+\tfrac1t\right)^3=t^3+\tfrac1{t^3}+3\left(t+\tfrac1t\right).

1

Climb down to x4+1x4x^4+\tfrac1{x^4}. Add 22 to complete a square:

x8+1x8+2=47+2=49⇒ (x4+1x4)2=49⇒ x4+1x4=7(positive, as x>0)\begin{aligned} &x^8+\frac1{x^8}+2=47+2=49\\ &\Rightarrow\ \left(x^4+\frac1{x^4}\right)^2=49\\ &\Rightarrow\ x^4+\frac1{x^4}=7 \quad(\text{positive, as } x\gt 0) \end{aligned}
2

Down to x2+1x2x^2+\tfrac1{x^2}, then x+1xx+\tfrac1x:

(x2+1x2)2=x4+1x4+2=7+2=9 ⇒ x2+1x2=3(x+1x)2=x2+1x2+2=3+2=5 ⇒ x+1x=5\begin{aligned} &\left(x^2+\frac1{x^2}\right)^2=x^4+\frac1{x^4}+2=7+2=9\ \Rightarrow\ x^2+\frac1{x^2}=3\\ &\left(x+\frac1x\right)^2=x^2+\frac1{x^2}+2=3+2=5\ \Rightarrow\ x+\frac1x=\sqrt5 \end{aligned}
3

Cube up to x3+1x3x^3+\tfrac1{x^3} using t3+1t3=(t+1t)3−3(t+1t)t^3+\tfrac1{t^3}=\left(t+\tfrac1t\right)^3-3\left(t+\tfrac1t\right):

x3+1x3=(5)3−35=55−35=25x^3+\frac1{x^3}=(\sqrt5)^3-3\sqrt5=5\sqrt5-3\sqrt5=2\sqrt5
4

Cube once more with t=x3t=x^3 to reach x9+1x9x^9+\tfrac1{x^9}:

x9+1x9=(x3+1x3)3−3(x3+1x3)=(25)3−3(25)=405−65=345\begin{aligned} x^9+\frac1{x^9}&=\left(x^3+\frac1{x^3}\right)^3-3\left(x^3+\frac1{x^3}\right)\\ &=(2\sqrt5)^3-3(2\sqrt5)\\ &=40\sqrt5-6\sqrt5=34\sqrt5 \end{aligned}

x9+1x9=345x^9+\dfrac1{x^9}=\mathbf{34\sqrt5}.

The handy identity throughout: (t+1t)2=t2+1t2+2\left(t+\tfrac1t\right)^2=t^2+\tfrac1{t^2}+2 (going down) and (t+1t)3=t3+1t3+3(t+1t)\left(t+\tfrac1t\right)^3=t^3+\tfrac1{t^3}+3\left(t+\tfrac1t\right) (going up).

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