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CAT 2020 Slot 1 — QA Question 18

Solving Quadratic EquationsEasy

The number of distinct real roots of the equation (x+1x)2-3(x+1x)+2 = 0 equals

Answer & solution

Answer: 1

Solution

Easy

Substitute y=x+1xy = x+\tfrac1x to turn the equation into a quadratic in yy. Solve for yy, then for each yy-value check which give real xx. Recall that for real xx, the value x+1xx+\tfrac1x is always ≥2\ge 2 (if x>0x>0) or ≤−2\le -2 (if x<0x<0), so ∣y∣≥2|y|\ge 2.

1

Substitute and solve the quadratic. Let y=x+1xy=x+\tfrac1x.

y2−3y+2=0⇒ (y−1)(y−2)=0⇒ y=1 or y=2\begin{aligned} &y^2 - 3y + 2 = 0\\ &\Rightarrow\ (y-1)(y-2)=0\\ &\Rightarrow\ y = 1 \ \text{or}\ y = 2 \end{aligned}
2

Discard the impossible value. For real xx, ∣x+1x∣≥2\left|x+\tfrac1x\right|\ge 2, so y=1y=1 is impossible.

y=1⇒x+1x=1⇒x2−x+1=0, discriminant=−3<0(no real x)⇒ only y=2 survives\begin{aligned} &y = 1 \Rightarrow x+\tfrac1x = 1 \Rightarrow x^2-x+1=0,\ \text{discriminant}=-3<0 \quad\text{(no real }x\text{)}\\ &\Rightarrow\ \text{only } y = 2 \text{ survives} \end{aligned}
3

Solve y=2y=2 for xx.

x+1x=2⇒ x2−2x+1=0⇒ (x−1)2=0⇒x=1(one distinct real root)\begin{aligned} &x+\tfrac1x = 2\\ &\Rightarrow\ x^2 - 2x + 1 = 0\\ &\Rightarrow\ (x-1)^2 = 0 \Rightarrow x = 1 \quad\text{(one distinct real root)} \end{aligned}
number of distinct real roots=1\text{number of distinct real roots} = 1
CAT 2020 Slot 1 QA Q18: The number of distinct real roots of the equation x + 1 x 2 - 3 x + 1 x + 2 = 0 equals — Solution | TheCATExam