JEE Main 2023 — Mathematics Algebra

Given equation is:

$$3({x}^{2}+\frac{1}{{x}^{2}})-2(x+\frac{1}{x})+5=0$$3(x2+x21​)−2(x+x1​)+5=0

We know,

$${(x+\frac{1}{x})}^{2}={x}^{2}+2+\frac{1}{{x}^{2}}(\text{using algebraic identities})$$(x+x1​)2=x2+2+x21​(using algebraic identities)

$$\Rightarrow {x}^{2}+\frac{1}{{x}^{2}}={(x+\frac{1}{x})}^{2}-2$$⇒x2+x21​=(x+x1​)2−2

So, we can write the given equation as

$$3[{(x+\frac{1}{x})}^{2}-2]-2(x+\frac{1}{x})+5=0$$3[(x+x1​)2−2]−2(x+x1​)+5=0

Let $x+\frac{1}{x}=t$x+x1​=t

$$\Rightarrow 3({t}^{2}-2)-2t+5=0\Rightarrow 3{t}^{2}-6-2t+5=0$$⇒3(t2−2)−2t+5=0⇒3t2−6−2t+5=0

$$\Rightarrow 3{t}^{2}-2t-1=0\Rightarrow 3{t}^{2}-3t+t-1=0$$⇒3t2−2t−1=0⇒3t2−3t+t−1=0

$$\Rightarrow 3t(t-1)+(t-1)=0\Rightarrow (3t+1)(t-1)=0$$⇒3t(t−1)+(t−1)=0⇒(3t+1)(t−1)=0

Therefore, $t=-\frac{1}{3}\mathrm{or}t=1$t=−31​ort=1

$$\mathrm{Case}I$$CaseI

$$x+\frac{1}{x}=1$$x+x1​=1

$$\Rightarrow {x}^{2}-x+1=0$$⇒x2−x+1=0

$$\text{Here},D=\sqrt{-3}\text{ i.e. no real solution}$$Here,D=−3​ i.e. no real solution

$$\mathrm{Case}\mathrm{II} x+\frac{1}{x}=-\frac{1}{3}$$CaseIIx+x1​=−31​

$$\Rightarrow 3{x}^{2}+x+3=0$$⇒3x2+x+3=0

$$\text{Here},D=\sqrt{-35}\text{i.e. no real solution}$$Here,D=−35​i.e. no real solution

Hence, we don't have real solutions.