Quadratic Equations Test

Result

Quadratic Equations Test - 42

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

If $\alpha$ and $\beta$ are roots of the equation $x^2 + \sin \theta . 2 \sin \theta = 0$ then $\dfrac{\alpha^{12} + \beta^{12}}{(\alpha^{-12} + \beta^{-12})(\alpha - \beta)^{24}}$ is equal to

A
$\dfrac{2^{24}}{(8 + \sin \theta)^{12}}$

B
$\dfrac{2^{12}}{(8 + \sin \theta)^{12}}$

C
$\dfrac{2^{12}}{(8 - \sin \theta)^{12}}$

D
$\dfrac{1}{2^{24}}$

Solution

$x^2 + x\sin \theta - 2 \sin \theta = 0$ , has roots $\alpha$ and $\beta$
$\dfrac{\alpha^{12}+\beta^{12}}{(\alpha^{-12}+\beta^{-12})(\alpha - \beta)^{24}} = \dfrac{\alpha^{12} . \beta^{12}}{(\alpha - \beta)^{24}} = \dfrac{(\alpha \beta)^{12}}{(\alpha - \beta)^{24}}$
$= \dfrac{(-2\sin \theta)^{12}}{(\sqrt{\sin^2 \theta + 8 \sin \theta})^{24}} = \dfrac{2^{12}}{(8 + \sin \theta)^{12}}$
Question 2
1 / -0

Which one of the following is the quadratic equation?

A
$\dfrac{5}{x} - 3 = x^2$

B
$x(x + 5) = 4$

C
$n - 1 = 2n$

D
$\dfrac{1}{x^2} (x + 2) = x$

Solution

Option : [A]
$\dfrac{5}{x} - 3 = x^2$
$\dfrac{5 - 3x}{x} = x^2$
$5 - 3x = x^3$
$x^3 + 3x - 5 = 0$
$\Rightarrow$ This is not a quadratic equation.

Option : [B]
$x (x + 5) = 4$
$x^2 + 5x = 4$
$x^2 + 5x - 4 = 0$
$\Rightarrow$ This is a quadratic equation.

Option : [C]
$n - 1 = 2n$
$\Rightarrow$ This is not a quadratic equation.

Option : [D]
$\dfrac{1}{x^2} (x +2) = x$
$x +2 = x^3$
$x^3 - (x + 2)=0$
$x^3 - x - 2 = 0$
$\Rightarrow$ This is not a quadratic equation.

$\therefore$ Answer is option $[B]$
Question 3
1 / -0

The integral values of $m$ for which the roots of the equation $m x^{2}+(2 m-1) x+(m-2)=0$ are rational are given by the expression [where $n$ is integer ]

A
${n}^{2}$

B
$n(n+2)$

C
$n(n+1)$

D
none of these

Solution

Discriminant $D=(2 m-1)^{2}-4(m-2) m=4 m+1$ must be perfect square. Hence,
$4 m+1=k^{2},$ say for some $k \in I$
$\Rightarrow \quad m=\dfrac{(k-1)(k+1)}{4}$
Clearly, $k$ must be odd. Let $k=2 n+1$
$\therefore \quad m=\dfrac{2 n(2 n+2)}{4}=n(n+1), n \in I$
Question 4
1 / -0

Which of the following is not a quadratic equation?

A
$2\left ( x-1 \right )^{2} = 4x^{2} - 2x + 1$

B
$2x - x^{2} = x^{2} + 5$

C
$\left ( \sqrt{2x} + \sqrt{3} \right )^{2} + x^{2} = 3x^{2} - 5x$

D
$\left ( 2^{2} + 2x \right )^{2} = x^{4} + 3 +4x^{3}$

Solution

The correct answer is option $(C)$ .
Main concept used: An equation will not be a quadratic in which a = 0 in equation of the form $ax^{2}$ + bx + c = 0

(a) Given equation is $2(x-1)^{2} = 4x^{2} - 2x + 1$
$\Rightarrow 2[(x)^{2} + (1)^{2} - 2(x)(1)] - 4x^{2} + 2x - 1 = 0$
$\Rightarrow 2x^{2} + 2 - 4x -4x^{2} + 2x - 1 = 0$
$\Rightarrow -2x^{2} - 2x + 1 = 0$
$\therefore$ Given equation is quadratic as it is of the form $ax^{2} + bx + c = 0$ and $a\neq 0$

(b) The given equation is $2x - x^{2} = x^{2} + 5$
$\Rightarrow 2x - x^{2} - x^{2} - 5 = 0$
$\Rightarrow -2x^{2} + 2x - 5 = 0$
$\Rightarrow 2x^{2} - 2x + 5 = 0$
$\therefore$ Given equation is quadratic, as it is of the form $ax^{2} + bx + c = 0$ and $a\neq0.$

(c) The given equation is $(\sqrt{2}x + \sqrt{3})^{2} + x^{2} = 3x^{2} - 5x$
$\Rightarrow (\sqrt{2}x)^{2} + (\sqrt{3})^{2} + 2(\sqrt{2}x)(\sqrt{3}) + x^{2} - 3x^{2} -5x = 0$
$\Rightarrow 2x^{2} + 3 + 2\sqrt{6}x + x^{2} - 3x^{2} + 5x = 0$
$\Rightarrow 0 + (2\sqrt{6} + 5)x + 3 = 0$
As $a = 0,$ the given equation is not quadratic

(d) Given equation is $(x^{2} + 2x)^{2} = x^{4} + 3 +4x^{3}$
$\Rightarrow (x^{2})^{2} + (2x)^{2} + 2(x^{2})^{2} (2x) - x^{4} - 3 - 4x^{3} = 0$
$\Rightarrow x^{4} + 4x^{2} + 4x^{3} - x^{4} - 3 -4x^{3} = 0$
$\Rightarrow 4x^{2} - 3 = 0$
$\therefore$ Given equation is quadratic, as it is of the form $ax^{2} + bx + c = 0$ and $a\neq0.$
Question 5
1 / -0

Which of the following is a quadratic equation?

A
$x^{2}+2x + 1 = \left ( 4-x \right )^{2} + 3$

B
$-2x^{2} = \left ( 5-x \right ) \left ( 2x - \dfrac{2}{5} \right )$

C
$\left ( k+1 \right )x^{2} + \dfrac{3}{2}x = 7$ ( Where k =-1 )

D
$x^{3}- x^{2} = \left ( x - 1 \right )^{3}$

Solution

The correct answer is option $(D)$
Main concept used : An equation of the form $ax^{2} + bx + c = 0$ where, $a, b, c,$ are numbers and $a\neq0$ , is called a quadratic equation.

$(a)$ $x^{2} + 2x + 1 = \left ( 4-x \right )^{2} + 3$
$\Rightarrow x^{2} + 2x + 1 = (4)^{2} + (x)^{2} -2(4) (x) + 3$
$\Rightarrow 2x + 1 = 16 - 8x + 3$
$\therefore$ Coefficient of $x^{2}$ is zero or a = 0. So, it is not a quadratic equation.

(b) $-2x^{2} = (5 - x) \left ( 2x - \dfrac{2}{5} \right )$
$\Rightarrow -2x^{2} = 10x - 2 -2x^{2} + \dfrac{2}{5}x$
$\Rightarrow -2x^{2} + 2x^{2} = 10x - 2 + \dfrac{2}{2}x$
$\Rightarrow 0 = 10x - 2 + \dfrac{2}{5}x$
As the coefficient of $x^{2}$ in the above equation is zero or $a = 0.$
So, it is not a quadratic equation.

$(c)$ $(k + 1)x^{2} + \dfrac{3}{2}x = 7$ (where $k = -1$ )
$\Rightarrow (-1 + 1)x^{2} + \dfrac{3}{2}x = 7$
So, the coefficient of $x^{2}$ is zero or $a = 0$ . Hence, the equation in not quadratic.

(d) $x^{3} - x^{2} = (x-1)^{3}$
$\Rightarrow x^{3} - x^{2} = (x)^{3} - (1)^{3} - 3 (x)^{2}(1) + 3(x) (1)^{2}$
$\Rightarrow x^{3} - x^{2} = x^{3} - 1 - 3x^{2} + 3x$
$\Rightarrow-x^{2} = -1 - 3x^{2} + 3x$
$\Rightarrow2x^{2} - 3x + 1 = 0$
As the coefficient of $x^{2}$ in the above equation is $2$ or $a = 2$ , so it is a quadratic equation.
Question 6
1 / -0

Which among the below are quadratic equations?

A
$\frac{5} {x} - 3 = x^{2}$

B
$x \left(x + 5\right) = 2$

C
$n - 1 = 2n$

D
$\dfrac{1} {x^{2}} \left(x + 2\right) = 2$

Solution
Question 7
1 / -0

Discriminant of quadratic equation $3\sqrt 3x^2+10x+\sqrt 3=0$

A
$10$

B
$64$

C
$46$

D
$30$

Solution

Comparing $3\sqrt 3x^2+10x+\sqrt 3$ by $ax^2+bx+c=0$ .
$a=\sqrt 3, b=10$ and $c=\sqrt 3$
Discriminant $(D)=b^2-4ac$
$=(10)^2-4\times 3\sqrt 3 \times \sqrt 3$
$=100-4\times 3\times 3$
$=100-36$
$=64$
Hence, option (B) is correct.
Question 8
1 / -0

All solutions of the equation $\displaystyle 4x^2 - 40x + 51 = 0$ lie in the interval

A
$\displaystyle \left(\frac {23}{10}, \frac {83}{10}\right)$

B
$\displaystyle \left(\frac {23}{10}, \frac {15}{2}\right)$

C
$\displaystyle \left(7, \frac {83}{10}\right)$

D
none of these
Question 9
1 / -0

The values of $a$ for which both the roots of the equation $(a-6)x^2=a(x-3)$ are positive and are given by

A
$(0,6)$

B
$\displaystyle \left( 0,\frac { 72 }{ 11 } \right)$

C
$\displaystyle \left( 6,\frac { 72 }{ 11 } \right)$

D
$\displaystyle \left( \frac { 72 }{ 13 } ,6 \right)$

Solution

The given equation is $(a-6){x}^{2}-ax+3a=0$
$\displaystyle D={ a }^{ 2 }-4\left( a-6 \right) .3a>0\Rightarrow 0<a<\frac { 72 }{ 11 }$ ...(i)
Since, both the roots are positive, so sum & product of roots are positive
$\therefore$ $\displaystyle \alpha +\beta =\frac { a }{ a-6 } >0\Rightarrow a<0$ or $a>6$ ...(ii)
and $\displaystyle \alpha \beta =\frac { 3a }{ a-6 } >0\Rightarrow a<0$ or $a>6$
$\displaystyle \therefore 6<a<\frac { 72 }{ 11 }$ $[$ From Eqs.(i) and (ii) $]$

a = $\displaystyle \left( 6,\frac { 72 }{ 11 } \right)$
Question 10
1 / -0

The roots of $(x-a)(x-c)+k(x-b)(x-d)=0$ are real and distinct for all real $k$ if

A
$a < b < c < d$

B
$a > b < c < d$

C
$a < b > c < d$

D
$a < b = c < d$

Solution

The roots of $(x-a)(x-c)+k(x-b)(x-d)=0$ are real and distinct for all real $k$
$f(x)=(1+k){ x }^{ 2 }-(a+c+k(b+d))x+ac+kbd$
As $D>0$ for real roots.
So ,
$\Rightarrow$ ${(a+c+k(b+d)) }^{ 2}-4(1+k)(ac+kbd)>0$
$\Rightarrow$ $(b+d)^{ 2 }{ k }^{ 2}+2(a+c)(b+d)k+(a+c)^{ 2 }-4(bd{ k }^{ 2 }+(ac+bd)k+ac)>0$
$\Rightarrow$ $(b-d)^{ 2 }{ k }^{ 2}+(2(a+c)(b+d)-4(ac+bd))k+(a-c)^{ 2 }>0$
This true for all $k$
$D<0$
$(2(a+c)(b+d)-4(ac+bd))^{ 2 }-4((a-c)(b-d))^{ 2 }<0$
$((a+c)(b+d)-2(ac+bd))^{ 2 }-((a-c)(b-d))^{ 2 }<0$

Applying ${a }^{ 2 }-b^{ 2 }=(a-b)(a+b)$ we get,
$\Rightarrow$ $((a+c)(b+d)-2(ac+bd)+(a-c)(b-d))((a+c)(b+d)-2(ac+bd)-(a-c)(b-d))<0$
$\Rightarrow$ $(ab+ad+bc+cd-2ac-2bd+ab-ad-bc+cd)(ab+ad+bc+cd-2ac-2bd-ab+ad+bc-cd)<0$
$\Rightarrow$ $(ab+cd-ac-bd)(ad+bc-ac-bd)<0$
$\Rightarrow$ $(a-d)(b-c)(a-b)(d-c)<0$
$\Rightarrow$ $a<b<c<d$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Quadratic Equations Test - 42

Result

Quadratic Equations Test - 42

Score

-

Rank

-

SHARING IS CARING

Question 1

224(8+sin⁡θ)12\dfrac{2^{24}}{(8 + \sin \theta)^{12}}(8+sinθ)12224​

212(8+sin⁡θ)12\dfrac{2^{12}}{(8 + \sin \theta)^{12}}(8+sinθ)12212​

212(8−sin⁡θ)12\dfrac{2^{12}}{(8 - \sin \theta)^{12}}(8−sinθ)12212​

1224\dfrac{1}{2^{24}}2241​

Solution

Question 2

5x−3=x2\dfrac{5}{x} - 3 = x^2x5​−3=x2

x(x+5)=4x(x + 5) = 4x(x+5)=4

n−1=2nn - 1 = 2nn−1=2n

1x2(x+2)=x\dfrac{1}{x^2} (x + 2) = xx21​(x+2)=x

Solution

Question 3

n2{n}^{2}n2

n(n+2)n(n+2)n(n+2)

n(n+1)n(n+1)n(n+1)

none of these

Solution

Question 4

2(x−1)2=4x2−2x+12\left ( x-1 \right )^{2} = 4x^{2} - 2x + 12(x−1)2=4x2−2x+1

2x−x2=x2+52x - x^{2} = x^{2} + 52x−x2=x2+5

(2x+3)2+x2=3x2−5x\left ( \sqrt{2x} + \sqrt{3} \right )^{2} + x^{2} = 3x^{2} - 5x(2x​+3​)2+x2=3x2−5x

(22+2x)2=x4+3+4x3\left ( 2^{2} + 2x \right )^{2} = x^{4} + 3 +4x^{3}(22+2x)2=x4+3+4x3

Solution

Question 5

x2+2x+1=(4−x)2+3x^{2}+2x + 1 = \left ( 4-x \right )^{2} + 3x2+2x+1=(4−x)2+3

−2x2=(5−x)(2x−25)-2x^{2} = \left ( 5-x \right ) \left ( 2x - \dfrac{2}{5} \right )−2x2=(5−x)(2x−52​)

(k+1)x2+32x=7\left ( k+1 \right )x^{2} + \dfrac{3}{2}x = 7(k+1)x2+23​x=7 ( Where k =-1 )

x3−x2=(x−1)3x^{3}- x^{2} = \left ( x - 1 \right )^{3}x3−x2=(x−1)3

Solution

Question 6

5x−3=x2\frac{5} {x} - 3 = x^{2}x5​−3=x2

x(x+5)=2x \left(x + 5\right) = 2x(x+5)=2

n−1=2nn - 1 = 2nn−1=2n

1x2(x+2)=2\dfrac{1} {x^{2}} \left(x + 2\right) = 2x21​(x+2)=2

Solution

Question 7

101010

646464

464646

303030

Solution

Question 8

(2310,8310)\displaystyle \left(\frac {23}{10}, \frac {83}{10}\right)(1023​,1083​)

(2310,152)\displaystyle \left(\frac {23}{10}, \frac {15}{2}\right)(1023​,215​)

(7,8310)\displaystyle \left(7, \frac {83}{10}\right)(7,1083​)

none of these

Question 9

(0,6)(0,6)(0,6)

(0,7211 )\displaystyle \left( 0,\frac { 72 }{ 11 } \right) (0,1172​ )

(6,7211 )\displaystyle \left( 6,\frac { 72 }{ 11 } \right) (6,1172​ )

(7213,6)\displaystyle \left( \frac { 72 }{ 13 } ,6 \right) (1372​,6)

Solution

Question 10

a<b<c<da < b < c < da<b<c<d

a>b<c<da > b < c < da>b<c<d

a<b>c<da < b > c < da<b>c<d

a<b=c<da < b = c < da<b=c<d

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.

$\dfrac{2^{24}}{(8 + \sin \theta)^{12}}$

$\dfrac{2^{12}}{(8 + \sin \theta)^{12}}$

$\dfrac{2^{12}}{(8 - \sin \theta)^{12}}$

$\dfrac{1}{2^{24}}$

$\dfrac{5}{x} - 3 = x^2$

$x(x + 5) = 4$

$n - 1 = 2n$

$\dfrac{1}{x^2} (x + 2) = x$

${n}^{2}$

$n(n+2)$

$n(n+1)$

$2\left ( x-1 \right )^{2} = 4x^{2} - 2x + 1$

$2x - x^{2} = x^{2} + 5$

$\left ( \sqrt{2x} + \sqrt{3} \right )^{2} + x^{2} = 3x^{2} - 5x$

$\left ( 2^{2} + 2x \right )^{2} = x^{4} + 3 +4x^{3}$

$x^{2}+2x + 1 = \left ( 4-x \right )^{2} + 3$

$-2x^{2} = \left ( 5-x \right ) \left ( 2x - \dfrac{2}{5} \right )$

$\left ( k+1 \right )x^{2} + \dfrac{3}{2}x = 7$ ( Where k =-1 )

$x^{3}- x^{2} = \left ( x - 1 \right )^{3}$

$\frac{5} {x} - 3 = x^{2}$

$x \left(x + 5\right) = 2$

$n - 1 = 2n$

$\dfrac{1} {x^{2}} \left(x + 2\right) = 2$

$10$

$64$

$46$

$30$

$\displaystyle \left(\frac {23}{10}, \frac {83}{10}\right)$

$\displaystyle \left(\frac {23}{10}, \frac {15}{2}\right)$

$\displaystyle \left(7, \frac {83}{10}\right)$

$(0,6)$

$\displaystyle \left( 0,\frac { 72 }{ 11 } \right)$

$\displaystyle \left( 6,\frac { 72 }{ 11 } \right)$

$\displaystyle \left( \frac { 72 }{ 13 } ,6 \right)$

$a < b < c < d$

$a > b < c < d$

$a < b > c < d$

$a < b = c < d$