SATHEE: Matrices and Determinants 4 Question 27

Matrices and Determinants 4 Question 27

28. Let $α_{1}, α_{2}, β_{1}, β_{2}$ be the roots of $a x^{2} + b x + c = 0$ and $p x^{2} + q x + r = 0$ , respectively. If the system of equations $α_{1} y + α_{2} z = 0$ and $β_{1} y + β_{2} z = 0$ has a non-trivial solution, then prove that $\frac{b^{2}}{q^{2}} = \frac{a c}{p r}$ .

(1987, 3M)

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Answer:

Correct Answer: 28. $x = \frac{4 - 5 k}{7}, y = \frac{13 k - 9}{7}, z = k$

Solution:

Since, $α_{1}, α_{2}$ are the roots of $a x^{2} + b x + c = 0$ .

$\Rightarrow α_{1} + α_{2} = - \frac{b}{a}$ and $α_{1} α_{2} = \frac{c}{a}$

Also, $β_{1}, β_{2}$ are the roots of $p x^{2} + q x + r = 0$ .

$\Rightarrow β_{1} + β_{2} = - \frac{q}{p}$ and $β_{1} β_{2} = \frac{r}{p}$

Given system of equations

$α_{1} y + α_{2} z = 0$

and $β_{1} y + β_{2} z = 0$ , has non-trivial solution.

$∴ | \begin{array}{ll} α_{1} & α_{2} \\ β_{1} & β_{2} \end{array} | = 0 \Rightarrow \frac{α_{1}}{α_{2}} = \frac{β_{1}}{β_{2}}$

Applying componendo-dividendo, $\frac{α_{1} + α_{2}}{α_{1} - α_{2}} = \frac{β_{1} + β_{2}}{β_{1} - β_{2}}$

$\Rightarrow α_{1} + α_{2} β_{1} - β_{2} = α_{1} - α_{2} β_{1} + β_{2}$

$\Rightarrow (α_{1} + α_{2})^{2} (β_{1} + β_{2})^{2} - 4 β_{2} β_{2}$

$= (β_{1} + β_{2})^{2} (α_{1} + α_{2})^{2} - 4 α_{1} α_{2}$

From Eqs. (i) and (ii), we get

$\frac{b^{2}}{a^{2}} \frac{q^{2}}{p^{2}} - \frac{4 r}{p} = \frac{q^{2}}{p^{2}} \frac{b^{2}}{a^{2}} - \frac{4 c}{a}$

$\Rightarrow \frac{b^{2} q^{2}}{a^{2} p^{2}} - \frac{4 b^{2} r}{a^{2} p} = \frac{b^{2} q^{2}}{a^{2} p^{2}} - \frac{4 q^{2} c}{a p^{2}}$

$\Rightarrow \frac{b^{2} r}{a} = \frac{q^{2} c}{p} \Rightarrow \frac{b^{2}}{q^{2}} = \frac{a c}{p r}$

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Matrices and Determinants 4 Question 27

28. Let $α_{1}, α_{2}, β_{1}, β_{2}$ be the roots of $a x^{2} + b x + c = 0$ and $p x^{2} + q x + r = 0$ , respectively. If the system of equations $α_{1} y + α_{2} z = 0$ and $β_{1} y + β_{2} z = 0$ has a non-trivial solution, then prove that $\frac{b^{2}}{q^{2}} = \frac{a c}{p r}$ .

Answer:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

Matrices and Determinants 4 Question 27

28. Let α1,α2,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 prove that b2q2=acpr.

Answer:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

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28. Let $α_{1}, α_{2}, β_{1}, β_{2}$ be the roots of $a x^{2} + b x + c = 0$ and $p x^{2} + q x + r = 0$ , respectively. If the system of equations $α_{1} y + α_{2} z = 0$ and $β_{1} y + β_{2} z = 0$ has a non-trivial solution, then prove that $\frac{b^{2}}{q^{2}} = \frac{a c}{p r}$ .