Let α be a common root of the equations x³-2x-25λ=0,3x³-8x-\;175/3λ=0 and λ>0 . Then λ=

Correct Answer is : 355\;{3}{5{5}}553

Correct Answer is : 355\;{3}{5{5}}553

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TS EAMCET 20-July-2022 Shift 2 Solved Paper

Section: Mathematics

Question : 12 of 160

Marks: +1, -0

Let

\alpha

x^{3}-2x-25\lambda=0,3x^{3}-8x-\;\frac{175}{3}\lambda=0

\lambda>0 .

\lambda=

Solution:

x^3-2x-25\lambda=0

\Rightarrow \;\; \;\frac{x^3-2x}{25}=\lambda

....(i)

-3x^3-8x-\;\frac{175}{3}\lambda=0

\Rightarrow \;\; \;\frac{9x^3-24x}{175}=\lambda

On equating Eqs. (i) and (ii),

\;\frac{x^3-2x}{25}=\;\frac{9x^3-24x}{175}

\Rightarrow \;\; 7x^3-14x=9x^3-24x \Rightarrow 10x=2x^3

\Rightarrow \;\; x^3-5x=0 \Rightarrow x(x^2-5)=0

\Rightarrow \;\; x=0, x=\pm \sqrt{5}

For

x=0

\Rightarrow \lambda=0

[From Eqs. (i) and (ii)]

But

\lambda>0

(given)

For

x=\sqrt{5}

5\sqrt{5}-2\sqrt{5}-25\lambda=0

[from Eq. (i)]

\Rightarrow \;\frac{3\sqrt{5}}{25}=\lambda

\Rightarrow \lambda = \;\frac{3}{5\sqrt{5}}

(\lambda>0)

x=-\sqrt{5}

-5\sqrt{5}+2\sqrt{5}-25\lambda=0

\Rightarrow \;\; \lambda = \;\frac{-3\sqrt{5}}{25} \;\; (\lambda<0)

Therefore,

\lambda = \;\frac{3}{5\sqrt{5}}

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