Quadratic Equations Test

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Quadratic Equations Test - 42

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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
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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
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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
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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
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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
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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
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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
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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,
$$\Rightarrow$$ $$((a+c)(b+d)-2(ac+bd)+(a-c)(b-d))((a+c)(b+d)-2(ac+bd)-(a-c)(b-d))<0$$
$$\Rightarrow$$ $$(ab+ad+bc+cd-2ac-2bd+ab-ad-bc+cd)(ab+ad+bc+cd-2ac-2bd-ab+ad+bc-cd)<0$$
$$\Rightarrow$$ $$(ab+cd-ac-bd)(ad+bc-ac-bd)<0$$
$$\Rightarrow$$ $$(a-d)(b-c)(a-b)(d-c)<0$$
$$\Rightarrow$$ $$a<b<c<d$$

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Re-Attempt Weekly Quiz Competition

Quadratic Equations Test - 42

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Quadratic Equations Test - 42

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SHARING IS CARING

Question 1

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

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

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

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

Solution

Question 2

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

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

$$n - 1 = 2n$$

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

Solution

Question 3

$${n}^{2}$$

$$n(n+2)$$

$$n(n+1)$$

none of these

Solution

Question 4

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

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

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

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

Solution

Question 5

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

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

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

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

Solution

Question 6

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

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

$$n - 1 = 2n$$

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

Solution

Question 7

$$10$$

$$64$$

$$46$$

$$30$$

Solution

Question 8

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

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

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

none of these

Question 9

$$(0,6)$$

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

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

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

Solution

Question 10

$$a < b < c < d$$

$$a > b < c < d$$

$$a < b > c < d$$

$$a < b = c < d$$

Solution

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