Quadratic Equations Test

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Quadratic Equations Test - 44

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Weekly Quiz Competition

Question 1
1 / -0

Given that $$r$$ and $$s$$ are constants , the solution of the quadratic equation, $$r{x}^{2}=\dfrac{1}{s}x+3$$ , is:

A
$$x=\dfrac { 1 }{ 2sr } \pm \dfrac { \sqrt { \dfrac { 1 }{ { s }^{ 2 } } +12r } }{ 2r } $$

B
$$x=\dfrac { 1 }{ 2sr } \pm \dfrac { \sqrt { -\dfrac { 1 }{ { s }^{ 2 } } -12r } }{ 2sr } $$

C
$$x=\dfrac { s }{ 2r } \pm \dfrac { \sqrt { \dfrac { 1 }{ { s }^{ 2 } } -12r } }{ 2r } $$

D
$$x=\dfrac { s }{ 2r } \pm \dfrac { \sqrt { { s }^{ 2 }-12sr } }{ 2sr } $$

Solution

Given, $$rx^{2}=\dfrac{1}{s}x+3$$

$$\Rightarrow rx^{2}-\dfrac{1}{s}x-3=0$$

WE know that in equation $$ax^{2}+bx+c=0$$

The roots of equation

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

In given equation $$ rx^{2}-\dfrac{1}{s}x-3=0$$

compare with above equation we get $$a=r$$, $$b=-\dfrac{1}{s}$$ and $$c=-3$$

Then $$x=\dfrac{-\dfrac{-1}{s}\pm \sqrt{\left (- \dfrac{1}{s} \right )^{2}-4\times r\times -3}}{2r}$$

$$x=\dfrac{\dfrac{1}{s}\pm \sqrt{\dfrac{1}{s^{2}+12r}}}{2r}$$

$$\Rightarrow x=\dfrac{1}{2sr}\pm\dfrac{\sqrt{\dfrac{1}{s^{2}}+12r}}{2r}$$
Question 2
1 / -0

The equation $$\displaystyle 9y^{2}(m+3)+6(m-3)y+(m+3)=0 $$, where $$m$$ is real has real roots then

A
$$\displaystyle m< 0$$

B
$$\displaystyle m> 0$$

C
$$\displaystyle m\leq 0$$

D
$$\displaystyle m\geq 0$$

Solution

For an equation $$ a{x}^{2} + bx + c = 0 $$, the discriminant $$ \triangle = {b}^{2} -4ac $$ helps us understand the nature of the roots.

When $$ \triangle \ge 0 $$ then roots are real and equal.

$$ {[6(m-3)]}^{2} -4 \times 9(m+3) \times (m + 3) \ge 0 $$
$$ 36({m}^{2} + 9 +-6m) - 36( {m}^{2} + 9 + 6m) \ge 0 $$
$$ 36{m}^{2} + 324 - 216m - 36{m}^{2} -324 -216m \ge 0 $$
$$ -432m \ge 0 $$
$$\therefore m \le 0 $$
Question 3
1 / -0

A ray emanating from (6,2) is incident on ellipse $$\displaystyle \frac{(x-1)^{2}}{45}+\frac{(y-2)^{2}}{20}=1$$ at (4.6) The equation of reflected ray (after 1st reflection ) is

A
x - 2y + 8 =0

B
x + 2y + 8 =0

C
x + 2y - 8 = 0

D
x - 2y - 8 =0
Question 4
1 / -0

The sum of all roots of the roots of the equations $$\displaystyle |x-1|^{2}-5|x-1|+6=0$$ is

A
5

B
4

C
2

D
6

Solution

Given equation is $$ \left| x-1 \right| ^{ 2 }-5\left| x-1 \right| +6=0 $$

We know that $$ \left| x-1 \right| ^{ 2 } $$ will be a positive value since it is a square.

But $$ \left| x-1 \right| $$ can take both values of $$ (x-1) $$ or $$ -(x-1) = 1-x $$

Hence, $$ \left| x-1 \right| ^{ 2 }-5\left| x-1 \right| +6=0 $$ can be written as both
$${(x-1)}^{ 2 }-5(x-1) +6=0 $$ and $${(x-1)}^{ 2 }-5(1-x) +6=0 $$

On simplifying $$ {(x-1)}^{ 2 }-5(x-1) +6=0 $$ we get $$ {x}^{2} -7x + 12 = 0 $$ whose sum of roots $$ = -\frac {-7}{1} = 7 $$

And on simplifying $$ {(x-1)}^{ 2 }-5(x-1) +6=0 $$ we get $$ {x}^{2} + 3x + 2 = 0 $$ whose sum of roots $$ = -\frac {3}{1} = -3 $$

Thus, sum of all 4 roots $$ 7 - 3 = 4 $$
Question 5
1 / -0

Solve the equation: $$3+\dfrac {30}{{x}^{2}}=-\dfrac {21}{x}$$

A
$$x=2,5$$

B
$$x=-2,5$$

C
$$x=-2,-5$$

D
$$x=2,-5$$

Solution

Solve the equation as follows:
$$3+\dfrac {30}{x^2}=-\dfrac {21}{x}$$
$$\Rightarrow \dfrac {30}{x^2}+\dfrac {21}{x}+3=0$$
$$\Rightarrow 30+21x+3x^2=0$$
$$\Rightarrow x^2+7x+10=0$$
Finding the roots of the above equation as:
$$\Rightarrow x^2+5x+2x+10=0$$
$$\Rightarrow x(x+5)+2(x+5)=0$$
$$\Rightarrow x+2=0$$ and $$x+5=0$$
$$\Rightarrow x=-2$$ and $$x=-5$$
Question 6
1 / -0

If $$(1+2x+x^2)^n=\displaystyle\sum^{2n}_{r=0}a_rx^r$$, then $$a_r=$$.

A
$$(^nC_r)^2$$

B
$$^nC_r\cdot {^nC_{r+1}}$$

C
$$^{2n}C_r$$

D
$$^{2n}C_{r-1}$$

Solution
Question 7
1 / -0

If the expression $$x^2+2(a+b+c)x+3(bc+ca+ab)$$ is a perfect square, then

A
$$a=b=c$$

B
$$a=\pm b=\pm c$$

C
$$a=b\neq c$$

D
None of the above

Solution

$${ x }^{ 2 }+2(a+b+c)x+3(ab+bc+ac)\quad is\quad a\quad perfect\quad square\\ \therefore D=0\\ 4{ (a+b+c) }^{ 2 }-4.3(ab+bc+ac)=0\\ { a }^{ 2 }+b^{ 2 }+{ c }^{ 2 }+2(ab+bc+ac)-3(ab+bc+ac)=0\\ { a }^{ 2 }+b^{ 2 }+{ c }^{ 2 }-ab-bc-ac=0\\ \therefore \cfrac { 1 }{ 2 } [{ (a-b) }^{ 2 }+{ (b-c) }^{ 2 }+{ (c-a) }^{ 2 }]=0\\ \therefore a=b=c\\ As\quad sum\quad of\quad squares\quad is\quad zero\quad only\quad if\quad square\quad are\quad zero$$
Question 8
1 / -0

Number of possible value(s) of integer '$$a$$' for which the quadratic equation $$x^2+ax+16=0$$ has equal roots, is

A
$$4$$

B
$$6$$

C
$$2$$

D
none of these

Solution

Given the quadratic equation $$x^2+ax+16=0$$
D $$=a^2 -4 \times 16 =0$$
$$\Longrightarrow a=+8,-8$$
Question 9
1 / -0

Equation $$x^2 - x + q = 0$$ has imaginary roots if

A
$$q > \dfrac{1}{4}$$

B
$$q < \dfrac{1}{4}$$

C
$$q^2 < \dfrac{1}{4}$$

D
None of these

Solution

$$x^2-x+q=0$$
It has imaginary roots, if
$$D<0$$
$$b^2-4ac<0$$
$$(1)^2-4(1)(q)<0$$
$$1<4q$$
$$\cfrac{1}{4}<q$$
$$\therefore q>\cfrac{1}{4}$$
Question 10
1 / -0

Discriminant is_________ for the quadratic equation $$x+\dfrac { 1 }{ x } =-5$$.

A
29

B
21

C
20

D
6

Solution

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Quadratic Equations Test - 44

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Quadratic Equations Test - 44

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Question 1

$$x=\dfrac { 1 }{ 2sr } \pm \dfrac { \sqrt { \dfrac { 1 }{ { s }^{ 2 } } +12r } }{ 2r } $$

$$x=\dfrac { 1 }{ 2sr } \pm \dfrac { \sqrt { -\dfrac { 1 }{ { s }^{ 2 } } -12r } }{ 2sr } $$

$$x=\dfrac { s }{ 2r } \pm \dfrac { \sqrt { \dfrac { 1 }{ { s }^{ 2 } } -12r } }{ 2r } $$

$$x=\dfrac { s }{ 2r } \pm \dfrac { \sqrt { { s }^{ 2 }-12sr } }{ 2sr } $$

Solution

Question 2

$$\displaystyle m< 0$$

$$\displaystyle m> 0$$

$$\displaystyle m\leq 0$$

$$\displaystyle m\geq 0$$

Solution

Question 3

x - 2y + 8 =0

x + 2y + 8 =0

x + 2y - 8 = 0

x - 2y - 8 =0

Question 4

5

4

2

6

Solution

Question 5

$$x=2,5$$

$$x=-2,5$$

$$x=-2,-5$$

$$x=2,-5$$

Solution

Question 6

$$(^nC_r)^2$$

$$^nC_r\cdot {^nC_{r+1}}$$

$$^{2n}C_r$$

$$^{2n}C_{r-1}$$

Solution

Question 7

$$a=b=c$$

$$a=\pm b=\pm c$$

$$a=b\neq c$$

None of the above

Solution

Question 8

$$4$$

$$6$$

$$2$$

none of these

Solution

Question 9

$$q > \dfrac{1}{4}$$

$$q < \dfrac{1}{4}$$

$$q^2 < \dfrac{1}{4}$$

None of these

Solution

Question 10

29

21

20

6

Solution

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