Quadratic Equations Questions and answers

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Elementary Mathematics

61). Given that \( \Large \tan \alpha \) and \( \Large \tan \beta \) are the roots of \( \Large x^{2}-px+q=0 \), then the value of \( \Large \sin^{2} \left(\alpha + \beta \right) \) is equal to:

A). \( \Large \frac{p^{2}}{p^{2}+ \left(1-q\right)^{2} } \)

B). \( \Large \frac{p^{2}}{p^{2}+q^{2}} \)

C). \( \Large \frac{q^{2}}{p^{2} \left(1-q\right)^{2} } \)

D). \( \Large \frac{p^{2}}{ \left(p+q\right)^{2} } \)

View Answer

Correct Answer: \( \Large \frac{p^{2}}{p^{2}+ \left(1-q\right)^{2} } \)

We have, \( \Large x^{2}-px+q=0 \)

Here, \( \Large \tan \alpha + \tan \beta = P \) ...(i)

\( \Large \tan \alpha + \tan \beta = q \) ...(ii)

Hence, \( \Large \tan \left( \alpha + \beta \right) = \frac{\tan \alpha + \tan \beta }{1-\tan \alpha \tan \beta } = \frac{P}{1-q} \) ...(iii)

\( \Large \therefore \sin^{2} \left( \alpha + \beta \right) = \frac{1-\cos\left[ 2 \left( \alpha + \beta \right) \right]}{2} \)

= \( \Large \frac{1}{2}\{ 1-\frac{1-\tan^{2} \left( \alpha + \beta \right) }{1+\tan^{2} \left( \alpha + \beta \right) } \} \left[ \because \cos2 \alpha =\frac{1-\tan^{2} \alpha }{1+\tan^{2} \alpha } \right] \)

= \( \Large \frac{1}{2}\left[ 1 - \frac{1- \left(\frac{P}{1-q}\right)^{2} }{1 + \left(\frac{P}{1-q}\right)^{2} } \right] \)

=\( \Large \frac{1}{2}\left[ \frac{ \left(1-q\right)^{2}+P^{2}- \left(1-q\right)^{2}+P^{2} }{ \left(1-q\right)^{2}+P^{2} } \right] \)

= \( \Large \frac{P^{2}}{P^{2}+ \left(1-q\right)^{2} } \)

62). The number of values of K for which the system of equations \( \Large \left(K+1\right)x+84=4K \) and \( \Large Kx+ \left(K+3\right)y=3 K-1 \) has infinitely many solution, is:

A). 0

B). 1

C). 2

D). infinite

View Answer

63). The set of all real numbers x for which \( \Large x^{2}-|x+2|+x>0 \) is:

A). \( \Large \left(-\infty, -2\right) \ \left(2, \infty\right) \)

B). \( \Large \left(-\infty, -\sqrt{2}\right) \ \left(\sqrt{2}, \infty\right) \)

C). \( \Large \left(-\infty, -1\right) \ \left(1, \infty\right) \)

D). \( \Large \left(\sqrt{2}, \infty\right) \)

View Answer

Correct Answer: \( \Large \left(-\infty, -\sqrt{2}\right) \ \left(\sqrt{2}, \infty\right) \)

Given \( \Large x^{2}-|x+2|+x>0 \)

we know that\( \Large |x+2|=\pm \left(x+2\right) \)

If \( \Large |x+2|=-x+2x\ge -2 \)

Equation (i) becomes \( \Large \left(when\ x \ge - 2 \right) \)

\( \Large x^{2}-x-2+x>0 \)

\( \Large x^{2}-2 > 0 \)

=> \( \Large x < -\sqrt{2}\ or\ x > \sqrt{2} \)

\( \Large x \epsilon \left[ -2, -\sqrt{2} \right] \cup \left(\sqrt{2}, \infty\right) \)

If therefore, equation (i) becomes

\( \Large x^{2}+x+2+x>0\ when\ x \le -2 \)

\( \Large x^{2}+2x+2 > 0\ when\ x \le -2 \)

\( \Large \left(x+1\right)^{2}+1 > 0\ when x \le -2 \)

which is true for all x.

\( \Large x < -2\ or\ x \left(-\infty,\ -2\right) \)

from equations (ii) and (iii)

\( \Large x \epsilon \left(-\infty,\ -\sqrt{2} \right) \cup \left(\sqrt{2},\ \infty\right) \)

64). Let a, b, c be positive numbers, the following systems of equations in x, y and z \( \Large \frac{x^{2}}{a^{2}} +\ \frac{y^{2}}{b^{2}}-\frac{z^{2}}{c^{2}}=1; \ \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}} +\ \frac{z^{2}}{c^{2}}=1 and\ \frac{-x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}}+\frac{z^{2}}{c^{2}}=1 \) has;

A). no solutions

B). unique solution

C). infinitely many solution

D). finitely many solution.

View Answer

Correct Answer: finitely many solution.

Given system of equations are

\( \Large \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}-\frac{zc}{c^{2}} = 1, \)

\( \Large \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}+\frac{z^{2}}{c^{2}} = 1 \)

and \( \Large -\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}+\frac{z^{2}}{c^{2}} = 1 \)

On adding all these equations, we get,

\( \Large \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}+\frac{z^{2}}{c^{2}} = 3 \)

\( \Large \frac{2z^{2}}{c^{2}}=2,\ \frac{2y^{2}}{b^{2}}=2,\ \frac{ax^{2}}{a^{2}} = 2 \)

On subtracting (i) from (iv), (ii) equation from (iv) and (iii) from (iv) we get

=> \( \Large z = \pm \ c,\ y = \pm \ b,\ x = \pm \ a \)

65). If \( \Large 2 \sin^{2}\frac{ \pi }{8} \) is a root of equation\( \Large x^{2}+ax+b=0 \), where a and b are rational numbers, then a - b is equal to

A). \( \Large -\frac{5}{2} \)

B). \( \Large -\frac{3}{2} \)

C). \( \Large -\frac{1}{2} \)

D). \( \Large \frac{1}{2} \)

View Answer

Correct Answer: \( \Large -\frac{5}{2} \)

\( \Large 2\sin^{2}\frac{ \pi }{8}=1-\cos \frac{ \pi }{4}=1-\frac{1}{\sqrt{2}}=\frac{\sqrt{2-1}}{\sqrt{2}} \)

(irrational root) so, other roots \( \Large \frac{\sqrt{2}+1}{\sqrt{2}} \)

Sum of roots \( \Large -a = 1-\frac{1}{\sqrt{2}}+1+\frac{1}{\sqrt{2}}=2 => a = -2 \)

products of roots \( \Large 1-\frac{1}{2} = \frac{1}{2} = b \)

so, \( \Large a-b = -2-\frac{1}{2} = -\frac{5}{2} \)

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