Quadratic Equations Test

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Quadratic Equations Test - 28

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Weekly Quiz Competition

Question 1
1 / -0

Find the value of discriminant for the following equation.
$$2x^{2}\, +\, x\, +\, 1\, =\, 0$$

A
$$7$$

B
$$-7$$

C
$$9$$

D
$$-9$$

Solution

The value of discriminant of $$2x^2 + x + 1 =0 $$ is
$$D = b^2 - 4ac$$
$$a=2, b=1, c=1$$
$$\therefore D = (1)^2 - 4\times2\times1 $$
$$= 1 - 8$$
$$= -7$$
Question 2
1 / -0

Find $$m$$, if the quadratic equation $$(m\, -\, 1)\, x^{2}\, -\, 2\, (m\, -\, 1)\, x\, +\, 1\, =\, 0$$ has real equal roots.

A
$$1$$

B
$$2$$

C
$$3$$

D
$$4$$

Solution

For the quadratic equation, $$(m\, -\, 1)\, x^{2}\, -\, 2\, (m\, -\, 1)\, x\, +\, 1\, =\, 0$$
Discriminant = $${(-2(m-1))}^2 - 4\times(m-1)= 0$$...... Since roots are real equal roots
$$4m^2 -8m + 4 - 4m + 4 = 0$$
$$4m^2 - 12m + 8 =0 $$
$$m^2 - 3m + 2 = 0$$
$$\therefore m = 1, 2 $$
Neglect $$m= 1$$ as the equation will not be quadratic anymore.
Hence, $$m = 2$$
Question 3
1 / -0

Find c, if the quadratic equation $$x^{2}\, -\, 2\, (c\, +\, 1)\, x\, +\, c^{2}\, =\, 0 $$ has real and equal roots.

A
$$c=\cfrac{1}{2}$$

B
$$c=-\cfrac{1}{2}$$

C
$$c=2$$

D
$$c=-2$$

Solution

Since, the roots are real and equal, $$D = 0$$
$$D=\sqrt {b^2-4ac=0}$$
$$(-2(c+1))^2 - 4c^2 = 0$$
$$4(c^2 + 2c + 1) - 4c^2 = 0 $$
$$8c + 4 = 0 $$
$$\therefore c = \displaystyle \frac{-1}{2}$$
Question 4
1 / -0

Find the value of discriminant for the following equation.
$$4x^{2}\, -\, kx\, +\, 2\, =\, 0$$

A
$$\Delta\, =\, k^{2}\, -16$$

B
$$\Delta\, =\, k^{2}\, + 32$$

C
$$\Delta\, =\, k^{2}\, -32$$

D
$$\Delta\, =\, k^{2}\,+16$$

Solution

Given equation is:
$$4 x^2-kx+2=0$$
Discriminant = $$b^2 - 4ac$$
$$a=4, b=-k, c=2$$
$$\therefore D= (k)^2 - 4(2)(4)$$
$$= k^2 - 32$$
$$\therefore \Delta= k^2-32$$
Question 5
1 / -0

Solve the equation using formula.
$$2x^{2}\, +\, \displaystyle \frac{x\, -\, 1}{5}\, =\, 0$$

A
$$x\, =\, \displaystyle \frac{-1\, \pm\, \sqrt{10}}{4}$$

B
$$x\, =\, \displaystyle \frac{-1\, \pm\, \sqrt{41}}{20}$$

C
$$x\, =\, \displaystyle \frac{1\, \pm\, \sqrt{41}}{20}$$

D
$$x\, =\, \displaystyle \frac{1\, \pm\, \sqrt{10}}{4}$$

Solution

Given equation is $$2x^{2}\, +\, \displaystyle \frac{x\, -\, 1}{5}\, =\, 0$$
$$\therefore 10x^2 + x - 1 = 0$$
Applying qudratic formula,
$$x = \displaystyle \frac{-b \pm \sqrt{b^2-4ac}}{2a}$$
$$a=10, b=1, c=-1$$

$$\therefore x =\displaystyle \frac{-1 \pm \sqrt{1 + 40}}{20}$$

$$\therefore x =\displaystyle \frac{ -1 \pm \sqrt{41}}{20}$$
$$\therefore$$ roots of the given equation are $$\displaystyle \frac {-1 \pm \sqrt {41}}{20}$$
Question 6
1 / -0

Which constant should be added and subtracted to solve the quadratic equation $$4{ x }^{ 2 }-\sqrt { 3 } x-5=0$$ by the method of completing the square?

A
$$\displaystyle\frac{9}{10}$$

B
$$\displaystyle\frac{3}{16}$$

C
$$\displaystyle\frac{3}{4}$$

D
$$\displaystyle\frac{\sqrt{3}}{4}$$

Solution

We have,
$$4{ x }^{ 2 } - \sqrt { 3 } x - 5 = 0$$
Divide whole equation by $$4$$,
$$\therefore { x }^{ 2 } - \displaystyle\frac { \sqrt { 3 } }{ 4 } x - \displaystyle\frac { 5 }{ 4 } = 0$$
$$\therefore { \left( x -\displaystyle\frac { \sqrt { 3 } }{ 8 } \right) }^{ 2 } - \displaystyle\frac { 5 }{ 4 } = \displaystyle\frac { 3 }{ 64 }$$
$$\Rightarrow 4{ \left( x -\displaystyle\frac { \sqrt { 3 } }{ 8 } \right) }^{ 2 } - 5 - \displaystyle\frac { 3 }{ 16 } = 0$$
$$\therefore \displaystyle \frac { 3 }{ 16 }$$ must be added to make $$4{ x }^{ 2 } - \sqrt { 3 } x - 5 = 0$$ a complete square
Question 7
1 / -0

The value of $$p$$ for which the equation $$x^2+4=(P+2)x $$ has equal roots?

A
$$2,-6$$

B
$$-2, 6$$

C
$$-2,-6$$

D
$$2,6$$

Solution

Given Equation is:
$$x^{2}+4=(P+2)x\Rightarrow x^{2}-(P+2)x+4=0\Rightarrow a=1, b=-(P+2),C=4$$
Since given equation has equal roots
Therefore $$b^{2}-4ac=0\Rightarrow [-P+2]^{2}-4(4)=0$$
$$P+2=\pm 4\Rightarrow P=2,-6$$
Question 8
1 / -0

The ratio of the roots of the equation $$a{ x }^{ 2 }+bx+c=0$$ is same as the ratio of the roots of the equation $$p{ x }^{ 2 }+qx+r=0$$. If $${ D }_{ 1 }$$ and $${ D }_{ 2 }$$ are the discriminants of $$a{ x }^{ 2 }+bx+c=0$$ and $$p{ x }^{ 2 }+qx+r=0$$ respectively, then $${ D }_{ 1 }:{ D }_{ 2 }$$ is equal to

A
$$\displaystyle \frac { { a }^{ 2 } }{ { p }^{ 2 } } $$

B
$$\displaystyle \frac { { b }^{ 2 } }{ { q }^{ 2 } } $$

C
$$\displaystyle \frac { { c }^{ 2 } }{ { r }^{ 2 } } $$

D
None of these

Solution

Let $$\displaystyle { \alpha }_{ 1 },{ \beta }_{ 1 }$$ be the roots of $$a{ x }^{ 2 }+bx+c=0$$ and $${ \alpha }_{ 2 },{ \beta }_{ 2 }$$ be the roots of $$p{ x }^{ 2 }+qx+r=0$$. Then

$$\displaystyle \frac { { \alpha }_{ 1 } }{ { \beta }_{ 1 } } =\frac { { \alpha }_{ 2 } }{ { \beta }_{ 2 } } $$ (given)

$$\displaystyle \frac { { \alpha }_{ 1 }+{ \beta }_{ 1 } }{ { \alpha }_{ 1 }-{ \beta }_{ 1 } } =\frac { { \alpha }_{ 2 }+{ \beta }_{ 2 } }{ { \alpha }_{ 2 }-{ \beta }_{ 2 } } $$ (Applying componendo and dividendo)

$$\displaystyle \frac { { \left( { \alpha }_{ 1 }+{ \beta }_{ 1 } \right) }^{ 2 } }{ { \left( { \alpha }_{ 1 }-{ \beta }_{ 1 } \right) }^{ 2 } } =\frac { { \left( { \alpha }_{ 2 }+{ \beta }_{ 2 } \right) }^{ 2 } }{ { \left( { \alpha }_{ 2 }-{ \beta }_{ 2 } \right) }^{ 2 } } $$

$$\displaystyle\frac { { \left( { \alpha }_{ 1 }+{ \beta }_{ 1 } \right) }^{ 2 } }{ { \left( { \alpha }_{ 1 }+{ \beta }_{ 1 } \right) }^{ 2 }-4{ \alpha }_{ 1 }{ \beta }_{ 1 } } =\frac { { \left( { \alpha }_{ 2 }+{ \beta }_{ 2 } \right) }^{ 2 } }{ { \left( { \alpha }_{ 2 }+{ \beta }_{ 2 } \right) }^{ 2 }-4{ \alpha }_{ 2 }{ \beta }_{ 2 } } $$

$$\displaystyle \dfrac { \dfrac { { b }^{ 2 } }{ { a }^{ 2 } } }{ \dfrac { { b }^{ 2 }-4ac }{ { a }^{ 2 } } } =\dfrac { \dfrac { { q }^{ 2 } }{ { p }^{ 2 } } }{ \dfrac { { q }^{ 2 }-4rp }{ { p }^{ 2 } } } $$
$$\Rightarrow \dfrac { { b }^{ 2 } }{ { D }_{ 1 } } =\dfrac { { q }^{ 2 } }{ { D }_{ 2 } }$$
$$ \Rightarrow \dfrac { { D }_{ 1 } }{ { D }_{ 2 } } =\dfrac { { b }^{ 2 } }{ { q }^{ 2 } } $$
Question 9
1 / -0

The given equation $$\displaystyle (x + 1)^2 = 2 (x - 3)$$ is

A
linear

B
quadratic

C
cubic

D
none of these

Solution

$$(x+1)^{2}=2(x-3)$$
Hence
$$x^{2}+2x+1=2x-6$$
$$x^{2}+7=0$$
$$x^{2}+(0)x+7=0$$
Hence it is a quadratic equation.
Question 10
1 / -0

Check whether the following is a quadratic equation.
$$\displaystyle (x - 2) (x + 1) = (x - 1) (x + 3)$$

A
Yes

B
No

C
Ambiguous

D
Data insufficient

Solution

Given, $$(x-2)(x+1)=(x-1)(x+3)$$
$$\Rightarrow x\left( x+1 \right) -2(x+1)=x\left( x+3 \right) -1\left( x+3 \right) $$
$$\Rightarrow { x }^{ 2 }+x-2x-2={ x }^{ 2 }+3x-x-3$$
$$\Rightarrow 2x-2x-2-3x+3=0$$
$$\Rightarrow -3x+1=0$$
$$\Rightarrow -3x=-1$$
Hence, $$x=\dfrac{1}{3}$$
A quadratic equation is of the form $$ax^2+bx+c=0$$
Thus the above equation is not a quadratic equation.

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Re-Attempt Weekly Quiz Competition

Quadratic Equations Test - 28

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Quadratic Equations Test - 28

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SHARING IS CARING

Question 1

$$7$$

$$-7$$

$$9$$

$$-9$$

Solution

Question 2

$$1$$

$$2$$

$$3$$

$$4$$

Solution

Question 3

$$c=\cfrac{1}{2}$$

$$c=-\cfrac{1}{2}$$

$$c=2$$

$$c=-2$$

Solution

Question 4

$$\Delta\, =\, k^{2}\, -16$$

$$\Delta\, =\, k^{2}\, + 32$$

$$\Delta\, =\, k^{2}\, -32$$

$$\Delta\, =\, k^{2}\,+16$$

Solution

Question 5

$$x\, =\, \displaystyle \frac{-1\, \pm\, \sqrt{10}}{4}$$

$$x\, =\, \displaystyle \frac{-1\, \pm\, \sqrt{41}}{20}$$

$$x\, =\, \displaystyle \frac{1\, \pm\, \sqrt{41}}{20}$$

$$x\, =\, \displaystyle \frac{1\, \pm\, \sqrt{10}}{4}$$

Solution

Question 6

$$\displaystyle\frac{9}{10}$$

$$\displaystyle\frac{3}{16}$$

$$\displaystyle\frac{3}{4}$$

$$\displaystyle\frac{\sqrt{3}}{4}$$

Solution

Question 7

$$2,-6$$

$$-2, 6$$

$$-2,-6$$

$$2,6$$

Solution

Question 8

$$\displaystyle \frac { { a }^{ 2 } }{ { p }^{ 2 } } $$

$$\displaystyle \frac { { b }^{ 2 } }{ { q }^{ 2 } } $$

$$\displaystyle \frac { { c }^{ 2 } }{ { r }^{ 2 } } $$

None of these

Solution

Question 9

linear

quadratic

cubic

none of these

Solution

Question 10

Yes

No

Ambiguous

Data insufficient

Solution

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