Let α₁,α₂ and β₁,β₂ be the roots of ax²+bx+c=0 and px²+qx+r=0 respectively. If the system of equations α₁y+α₂z=0 and β₁y+β₂z=0 has a non-trivial solution, then

Correct Answer is : b2q2=acprb²/q²=ac/prq2b2=prac

Correct Answer is : b2q2=acprb²/q²=ac/prq2b2=prac

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BITSAT 2016 Paper

Section: Mathematics

Question : 96 of 150

Marks: +1, -0

Let

\alpha_1,\alpha_2

\beta_1,\beta_2

ax^2+bx+c=0

px^2+qx+r=0

\alpha_1y+\alpha_2z=0

\beta_1y+\beta_2z=0

has a non-trivial solution, then

Solution:

Since

\alpha_1,\alpha_2

and

\beta_1,\beta_2

are the roots of

ax^2+bx+c=0

and

px^2+qx+r=0

respectively, therefore

a_1+\alpha_2=\frac{-b}{a},\alpha_1,\alpha_2=\frac{c}{a}

...(1)

and

\beta_1+\beta_2=\frac{-q}{p},\beta_1\beta_2=\frac{r}{p}

.....(2)

Since the given system of equation has a non-trivial solution

\therefore \begin{vmatrix} \alpha_1 & \alpha_2 \\ \beta_1 & \beta_2 \end{vmatrix}=0

i.e

\alpha_1\beta_2-\alpha_2\beta_1=0

\;\;\frac{\alpha_1}{\beta_1}=\;\;\frac{\alpha_2}{\beta_2}=\;\;\frac{\alpha_1+\alpha_2}{\beta_1+\beta_2}=\sqrt{\;\;\frac{\alpha_1\alpha_2}{\beta_1\beta_2}}

\Rightarrow \;\;\frac{pb}{qa}=\sqrt{\;\;\frac{pc}{ra}}\Rightarrow \;\;\frac{b^2}{q^2}=\;\;\frac{ac}{pr}

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