If \alpha and \beta are the distinct roots of the equation x^2 + (3)^1/4x + 3^1/2 = 0 , then the value of \alpha ^96(\alpha ^12 - 1) + \beta ^96(\beta ^12 - 1) is equal to :

Quadratic Equation and Inequalities PYQ — JEE Main 2021 · 20 Jul · Shift 1 · Q25 with Solution

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Correct answer: C

Given quadratic and its roots

The equation is $x^2+3^{1/4}x+3^{1/2}=0.$

Let its distinct roots be $\alpha,\beta$ .

By Vieta's formulas, $\alpha+\beta=-3^{1/4},\qquad \alpha\beta=3^{1/2}.$

We need to evaluate $\alpha^{96}(\alpha^{12}-1)+\beta^{96}(\beta^{12}-1).$

Expand it as

= (\alpha^{108}+\beta^{108})-(\alpha^{96}+\beta^{96}).$$ So define $$S_n=\alpha^n+\beta^n.$$ Then the required value is $$S_{108}-S_{96}.$$ --- 2. **Find a convenient form of the roots** The quadratic is $$x^2+3^{1/4}x+3^{1/2}=0.$$ Its discriminant is $$D=(3^{1/4})^2-4\cdot 3^{1/2}=3^{1/2}-4\cdot 3^{1/2}=-3\cdot 3^{1/2}=-3^{3/2}.$$ So the roots are complex. Now, $$\alpha,\beta=\frac{-3^{1/4}\pm i\sqrt{3^{3/2}}}{2} =\frac{-3^{1/4}\pm i\,3^{3/4}}{2} =\frac{3^{1/4}}{2}(-1\pm i\sqrt{3}).$$ Since $$-1\pm i\sqrt{3}=2\left(\cos\frac{2\pi}{3}\pm i\sin\frac{2\pi}{3}\right),$$ we get $$\alpha=3^{1/4}\,\text{cis}\frac{2\pi}{3}, \qquad \beta=3^{1/4}\,\text{cis}\frac{4\pi}{3},$$ where $\text{cis }\theta=\cos\theta+i\sin\theta$. Thus both roots have modulus $3^{1/4}$ and arguments $\frac{2\pi}{3},\frac{4\pi}{3}$. --- 3. **Compute $S_n=\alpha^n+\beta^n$** Using De Moivre's theorem, $$\alpha^n=3^{n/4}\,\text{cis}\frac{2n\pi}{3}, \qquad \beta^n=3^{n/4}\,\text{cis}\frac{4n\pi}{3}.$$ Therefore, $$S_n=3^{n/4}\left(\text{cis}\frac{2n\pi}{3}+\text{cis}\frac{4n\pi}{3}\right).$$ Now for $n$ divisible by $3$, both angles are multiples of $2\pi$, so each cis term becomes $1$. Since $96$ and $108$ are both divisible by $3$, $$S_{96}=2\cdot 3^{96/4}=2\cdot 3^{24},$$ $$S_{108}=2\cdot 3^{108/4}=2\cdot 3^{27}.$$ Hence required value is $$S_{108}-S_{96}=2\cdot 3^{27}-2\cdot 3^{24} =2\cdot 3^{24}(3^3-1) =2\cdot 3^{24}(27-1) =52\cdot 3^{24}.$$ So the expression equals $$52\times 3^{24}.$$ --- 4. **Match with the options** This is exactly **Option C**. $$\boxed{52\times 3^{24}}$$ --- 5. **Comparison with stored correct answer** Stored correct answer: **C** Our derived answer: **C** They agree.

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View written solutionFree

Correct answer: C

Given quadratic and its roots

The equation is $x^2+3^{1/4}x+3^{1/2}=0.$

Let its distinct roots be $\alpha,\beta$ .

By Vieta's formulas, $\alpha+\beta=-3^{1/4},\qquad \alpha\beta=3^{1/2}.$

We need to evaluate $\alpha^{96}(\alpha^{12}-1)+\beta^{96}(\beta^{12}-1).$

Expand it as

= (\alpha^{108}+\beta^{108})-(\alpha^{96}+\beta^{96}).$$ So define $$S_n=\alpha^n+\beta^n.$$ Then the required value is $$S_{108}-S_{96}.$$ --- 2. **Find a convenient form of the roots** The quadratic is $$x^2+3^{1/4}x+3^{1/2}=0.$$ Its discriminant is $$D=(3^{1/4})^2-4\cdot 3^{1/2}=3^{1/2}-4\cdot 3^{1/2}=-3\cdot 3^{1/2}=-3^{3/2}.$$ So the roots are complex. Now, $$\alpha,\beta=\frac{-3^{1/4}\pm i\sqrt{3^{3/2}}}{2} =\frac{-3^{1/4}\pm i\,3^{3/4}}{2} =\frac{3^{1/4}}{2}(-1\pm i\sqrt{3}).$$ Since $$-1\pm i\sqrt{3}=2\left(\cos\frac{2\pi}{3}\pm i\sin\frac{2\pi}{3}\right),$$ we get $$\alpha=3^{1/4}\,\text{cis}\frac{2\pi}{3}, \qquad \beta=3^{1/4}\,\text{cis}\frac{4\pi}{3},$$ where $\text{cis }\theta=\cos\theta+i\sin\theta$. Thus both roots have modulus $3^{1/4}$ and arguments $\frac{2\pi}{3},\frac{4\pi}{3}$. --- 3. **Compute $S_n=\alpha^n+\beta^n$** Using De Moivre's theorem, $$\alpha^n=3^{n/4}\,\text{cis}\frac{2n\pi}{3}, \qquad \beta^n=3^{n/4}\,\text{cis}\frac{4n\pi}{3}.$$ Therefore, $$S_n=3^{n/4}\left(\text{cis}\frac{2n\pi}{3}+\text{cis}\frac{4n\pi}{3}\right).$$ Now for $n$ divisible by $3$, both angles are multiples of $2\pi$, so each cis term becomes $1$. Since $96$ and $108$ are both divisible by $3$, $$S_{96}=2\cdot 3^{96/4}=2\cdot 3^{24},$$ $$S_{108}=2\cdot 3^{108/4}=2\cdot 3^{27}.$$ Hence required value is $$S_{108}-S_{96}=2\cdot 3^{27}-2\cdot 3^{24} =2\cdot 3^{24}(3^3-1) =2\cdot 3^{24}(27-1) =52\cdot 3^{24}.$$ So the expression equals $$52\times 3^{24}.$$ --- 4. **Match with the options** This is exactly **Option C**. $$\boxed{52\times 3^{24}}$$ --- 5. **Comparison with stored correct answer** Stored correct answer: **C** Our derived answer: **C** They agree.

Quadratic Equation and Inequalities question

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