Quadratic Equations Test

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Quadratic Equations Test - 38

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Weekly Quiz Competition

Question 1
1 / -0

Given $$\alpha $$ and $$\beta $$ are the roots of the equation $${x^2} - 4x + k = 0(k \ne 0)$$. If $$\alpha \beta ,\ \alpha {\beta ^2} + {\alpha ^2}\beta ,\ {\alpha ^3} + {\beta ^3}$$ are in geometric progression then the value of $$'k'$$ equals

A
$$4$$

B
$$\dfrac {16}{7}$$

C
$$\dfrac {3}{7}$$

D
$$12$$

Solution

If $$ax^2+bx+c=0 $$ is a quadratic equation, then the relation between it's roots is given as :
Sum of the roots : $$\displaystyle \alpha +\beta =\frac{-b}{a} $$
Product of the roots: $$\displaystyle \alpha \beta =\frac{c}{a} $$
$$\begin{array}{l}{x^2} - 4x + k = 0\\\alpha + \beta = 4\\\alpha \beta = k\\\alpha \beta ,\alpha {\beta ^2} + {\alpha ^2}\beta ,{\alpha ^3} + {\beta ^3},are,in,GP\\ \Rightarrow {\left( {\alpha {\beta ^2} + {\alpha ^2}\beta } \right)^2} = \alpha \beta \left( {{\alpha ^3} + {\beta ^3}} \right)\\ \Rightarrow [{\alpha \beta}(\alpha + \beta)]^2 = \alpha \beta(\alpha + \beta )({\alpha ^2} - \beta \alpha + {\beta ^2})\\ \Rightarrow k^2(16) = k \cdot 4\left[ {{{(\alpha + \beta )}^2} - 3\alpha \beta } \right] \end{array}$$
$$ \Rightarrow 4k = 16 - 3k\\ \Rightarrow k = \dfrac{{16}}{7}$$
Question 2
1 / -0

The value of $$\sqrt {6 + \sqrt {6 + \sqrt {6 + \ldots \ldots \ldots \ldots \infty } } } $$ is

A
$$5$$

B
$$4$$

C
$$3$$

D
$$3.5$$

Solution

Let $$\sqrt {6+\sqrt{6+\sqrt{6 +..........................\infty }}} = x$$

$$\therefore \sqrt {6+x} = x$$

$$6+x = x^2 $$
$$x^2 -x-6 = 0$$
$$(x-3)(x+2) = 0\implies x = 3$$ as x can only be positive.
Question 3
1 / -0

The equation $$ 2 sin^2 x-(p+3) sin x+2p-2=0 $$ posses a real solution, if :

A
$$0 \leqslant p \leqslant 1$$

B
$$ - 1 \leqslant p \leqslant 3$$

C
$$4 \leqslant p \leqslant 6$$

D
$$p \geqslant 6$$

Solution

$$2sin^2x -(p+3) sinx + 2p-2 = 0$$
Using quadratic formula-

$$ sinx = \cfrac{p+3 \pm \sqrt{(p+3)^2 - 4*8*2(p-1)}}{4}$$

$$ = \cfrac{p+3 \pm \sqrt{(p+3)^2 - 64(p-1)}}{4}$$ for this to be real

$$\therefore (p+3)^2 - 64(p-1) \geq 0$$

$$\Rightarrow p^2 -58p-73 \geq 0$$

$$\Rightarrow (p-29-16\sqrt3)(p-29+16\sqrt3) \geq 0$$

Hence, either $$p \leq 29-16\sqrt3$$ or $$ p \geq 29+16\sqrt3$$

$$0 \leq p \leq 1$$
Question 4
1 / -0

Solve by the method of completing the square $$5x^2 - 6x - 2 = 0$$

A
$$\frac{3+\sqrt{19}}{5}$$ $$and$$ $$\frac{3-\sqrt{19}}{5}$$

B
$$\frac{-3+\sqrt{19}}{5}$$ $$and$$ $$\frac{3-\sqrt{19}}{5}$$

C
$$\frac{3+\sqrt{19}}{3}$$ $$and$$ $$\frac{3-\sqrt{19}}{5}$$

D
$$\frac{3+\sqrt{19}}{5}$$ $$and$$ $$\frac{-3-\sqrt{19}}{3}$$

Solution

$$\\5x^2-6x-2=0\\x^2-(\frac{6}{5})x-(\frac{2}{5})=0\\(x-(\frac{3}{5}))^2-(\frac{3}{5})^2-(\frac{2}{5})=0\\(x-(\frac{3}{5}))^2-(\frac{9}{25})-(\frac{2}{5})=0\\(x-(\frac{3}{5}))^2=(\frac{9+10}{25})\\(x-(\frac{3}{5}))=\pm \sqrt {(\frac{19}{25})}\\x=(\frac{3}{5})\pm \sqrt {(\frac{19}{25})}\therefore x=(\frac{3+\sqrt{19}}{5})or(\frac{3-\sqrt{19}}{5})$$
Question 5
1 / -0

Let $$f\left( x \right) ={ x }^{ 3 }+a{ x }^{ 2 }+bx+5\sin ^{ 2 }{ x } $$ be an increasing function in the in the set of real numers $$R$$.Then $$a$$ and $$b$$ satisfy the condition.

A
$$a^{2}-3b-15>0$$

B
$$a^{2}-3b+15 \ge$$

C
$$a^{2}+3b-15<0$$

D
$$a>0$$ & $$b>0$$

Solution

$$f(x)=x^3+ax^2+bx+5\sin^2x$$

Given that $$f(x)$$ is incresing on $$R$$

Therefore, $$\rightarrow f'(x)\geq0$$

$$\Rightarrow f'(x)=3x^2+2ax+b+10\sin x\cos x$$

$$\Rightarrow f'(x)=3x^2+2ax+b+5\sin 2x\geq 0$$

For roots to exist, $$\Delta\geq 0$$

Thereore, $$\Rightarrow(2a)^2-4(3)(b+5\sin2x)\geq 0 $$

$$\Rightarrow 4a^2-4(3)(b+5\sin2x)\geq 0$$

$$\Rightarrow a^2-3b-15\sin2x \geq 0$$

Maximum value of $$\sin2x=1$$

Therefore,$$\Rightarrow a^2-3b-15 \geq 0$$
Question 6
1 / -0

If $$\alpha $$ and $$\beta $$ are the roots of equation $${x^2} - 3x + 1 = 0$$ and $${a_n} = {\alpha ^n} + {\beta ^n},n \in N$$ then the value of $$\dfrac{{{a_7} + {a_5}}}{{{a_6}}}$$

A
$$1$$

B
$$2$$

C
$$3$$

D
$$4$$

Solution

$$x^{2}-3x+1= 0 \rightarrow \alpha , \beta $$ are roots $$\alpha \beta = 1 , \alpha + \beta =3$$.
$$a_{n}= \alpha^{n}+ \beta^{n}.$$
$$\dfrac{a_{7}+a_{5}}{a_{6}}= \dfrac{\alpha^{7}+ \beta^{7}+ \alpha^{5}+ \beta^{5}}{\alpha^{6}+ \beta^{6}}$$
$$=\dfrac{\alpha}{1+ \left( \dfrac{\beta}{ \beta} \right)^{6}} + \dfrac{\beta}{ \left(1+ \left( \dfrac{\beta}{\alpha} \right)^{6} \right) } \dfrac{1}{\alpha \left(1 \left( \dfrac{\beta}{\alpha} \right)^{6} \right)} + \dfrac{1}{\beta \left( 1+ \left( \dfrac{\alpha}{\beta} \right)^{6} \right)}$$
$$\dfrac{\alpha}{ \beta} = \dfrac{\beta}{\alpha } =1$$
$$= \dfrac{\alpha}{2}+ \dfrac{\beta}{2} + \dfrac{1}{2 \alpha} + \dfrac{1}{2 \beta}$$
$$=\dfrac{3}{2}+ \dfrac{\alpha + \beta}{2 \alpha \beta}$$
$$=\dfrac{3}{2}+ \dfrac{3}{2}$$
$$=3$$
$$C$$ is correct.
Question 7
1 / -0

The sum of the roots of the quadratic $$5x^{2}-6x+1=0$$ is

A
$$-\dfrac{6}{5}$$

B
$$\dfrac{1}{5}$$

C
$$-\dfrac{5}{6}$$

D
$$-\dfrac{1}{5}$$

Solution

Consider the following equation.

$$ 5{{x}^{2}}-6x+1=0 $$

$$ x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} $$

$$ =\dfrac{\left( -6 \right)\pm \sqrt{36-20}}{10} $$

$$ =\left( -\dfrac{1}{5}, -1 \right) $$
$$Sum = (-\dfrac{1}{5})+ (-1) = $$ $$(-\dfrac{6}{5})$$

Option A, It is the correct option.

Hence, This is the required answer.
Question 8
1 / -0

If one root of $$2x^{2}+kx+1=0$$ is $$\dfrac{1}{2}$$, then the value of $$k$$ is ____.

A
$$3$$

B
$$5$$

C
$$-3$$

D
$$-5$$

Solution

Given equation is:
$$2{{x}^{2}}+kx+1=0.....(1)$$

It is given that,

$$x=\dfrac{1}{2}$$ is a root of the given equation.

Putting $$x=\dfrac{1}{2}$$ in $$(1)$$,

$$\Rightarrow 2\times{{\left( \dfrac{1}{2} \right)}^{2}}+k\times\dfrac{1}{2}+1=0 $$

$$\Rightarrow 2\times\dfrac{1}{4}+k\times\dfrac{1}{2}+1=0 $$
$$\Rightarrow \dfrac{1}{2}+k\times\dfrac{1}{2}+1=0 $$
$$\Rightarrow \dfrac{k}{2}+\dfrac{3}{2}=0 $$
$$\Rightarrow \dfrac{k}{2}=-\dfrac{3}{2} $$
$$\Rightarrow k=-3 $$

Hence, the value of $$k$$ is $$-3.$$
Question 9
1 / -0

If $$\alpha,\beta$$ are the roots of $${x}^{2}-x+1=0$$ then $$\cfrac { { \left( { \alpha }^{ 2 }-\alpha \right) }^{ 3 }+{ \left( { \beta }^{ 2 }-\beta \right) }^{ 3 } }{ \left( 2-\alpha \right) \left( 2-\beta \right) } $$ is equal to

A
$$0$$

B
$$1$$

C
$$\dfrac{-2}{3}$$

D
$$2$$

Solution

$$\alpha ,\beta $$ are the roots of $$x^{2}-x+1=0$$
$${\alpha}^{2}-\alpha+1=0\implies {\alpha}^{2}-\alpha=-1$$
$${\beta}^{2}-\beta+1=0\implies {\beta}^{2}-\beta=-1$$
$$\alpha+\beta=1,{\alpha}{\beta}=1$$
$$(2-\alpha)(2-\beta)=4-2(\alpha+\beta)+\alpha\beta=4-2+1=3$$
$$\dfrac{({\alpha}^{2}-\alpha)^{3}+({\beta}^{2}-\beta)^{3}}{(2-\alpha)(2-\beta)}=\dfrac{(-1)^{3}+(-1)^{3}}{3}=-\dfrac{2}{3}$$
Question 10
1 / -0

The roots of $$a{x^2} + bx + c = 0$$ where $$a \ne 0$$ and coefficients are real and $$a+c <b$$, then

A
$$4a+c>2b$$

B
$$4a+c<2b$$

C
$$4a+c=2b$$

D
none of the above

Solution

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Re-Attempt Weekly Quiz Competition

Quadratic Equations Test - 38

Result

Quadratic Equations Test - 38

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Rank

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SHARING IS CARING

Question 1

$$4$$

$$\dfrac {16}{7}$$

$$\dfrac {3}{7}$$

$$12$$

Solution

Question 2

$$5$$

$$4$$

$$3$$

$$3.5$$

Solution

Question 3

$$0 \leqslant p \leqslant 1$$

$$ - 1 \leqslant p \leqslant 3$$

$$4 \leqslant p \leqslant 6$$

$$p \geqslant 6$$

Solution

Question 4

$$\frac{3+\sqrt{19}}{5}$$ $$and$$ $$\frac{3-\sqrt{19}}{5}$$

$$\frac{-3+\sqrt{19}}{5}$$ $$and$$ $$\frac{3-\sqrt{19}}{5}$$

$$\frac{3+\sqrt{19}}{3}$$ $$and$$ $$\frac{3-\sqrt{19}}{5}$$

$$\frac{3+\sqrt{19}}{5}$$ $$and$$ $$\frac{-3-\sqrt{19}}{3}$$

Solution

Question 5

$$a^{2}-3b-15>0$$

$$a^{2}-3b+15 \ge$$

$$a^{2}+3b-15<0$$

$$a>0$$ & $$b>0$$

Solution

Question 6

$$1$$

$$2$$

$$3$$

$$4$$

Solution

Question 7

$$-\dfrac{6}{5}$$

$$\dfrac{1}{5}$$

$$-\dfrac{5}{6}$$

$$-\dfrac{1}{5}$$

Solution

Question 8

$$3$$

$$5$$

$$-3$$

$$-5$$

Solution

Question 9

$$0$$

$$1$$

$$\dfrac{-2}{3}$$

$$2$$

Solution

Question 10

$$4a+c>2b$$

$$4a+c<2b$$

$$4a+c=2b$$

none of the above

Solution

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