Quadratic Equations Test

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Quadratic Equations Test - 29

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Weekly Quiz Competition

Question 1
1 / -0

Find the roots of each of the following quadratic equations by the method of completing the squares
$$2x^2 - 5x + 3 = 0$$

A
$$x=2,\,\,x=-7$$

B
$$x=-1,\,\,x=3$$

C
$$x=\displaystyle 1,x= \frac{3}{2}$$

D
$$x=\displaystyle 1, x=\frac{1}{2}$$

Solution

Dividing the entire equation by 2, we get
$$x^{2}-\dfrac{5x}{2}+\dfrac{3}{2}=0$$

$$(x-\dfrac{5}{4})^{2}-\dfrac{25}{16}+\dfrac{24}{16}=0$$

$$(x-\dfrac{5}{4})^{2}=\dfrac{1}{16}$$

$$x-\dfrac{5}{4}=\pm\dfrac{1}{4}$$

$$x=\dfrac{5\pm1}{4}$$
Hence
$$x=\dfrac{3}{2}$$ and $$x=1$$
Question 2
1 / -0

Find the solutions of $$3x^2 - 2\sqrt 6x + 2 = 0$$ by the method of completing the squares when $$x$$ is an irrational number.

A
$$x=\pm\sqrt{\dfrac{7}{2}}$$

B
$$x=\pm\sqrt{\dfrac{2}{3}}$$

C
$$x=\sqrt{\dfrac{2}{3}}$$

D
$$x=\sqrt{\dfrac{7}{2}}$$

Solution

$$3x^{2}-2\sqrt{6}x+2=0$$
Dividing the entire equation, by $$ 3$$, we get
$$x^{2}-\dfrac{2\sqrt{6}}{3}+\dfrac{2}{3}=0$$
$$\left (x-\dfrac{\sqrt{6}}{3}\right)^{2}-\dfrac{6}{9}\dfrac{2}{3}=0$$
$$\left (x-\dfrac{\sqrt{6}}{3}\right)^{2}=0$$
$$x=\dfrac{\sqrt{6}}{3}$$
Hence, no rational roots.
Question 3
1 / -0

Solve the equation $$x^2 - (\sqrt 3 + 1) x + \sqrt 3 = 0$$ by the method of completing the square.

A
$$x = \sqrt 1, 3$$

B
$$x = \sqrt 5, 1$$

C
$$x = \sqrt 10, 2$$

D
$$x = \sqrt 3, 1$$

Solution

Given equation is $$x^{2}-(\sqrt{3}+1)x+\sqrt{3}=0$$
$$\left (x-\dfrac{\sqrt{3}+1}{2}\right)^{2}+\sqrt{3}-(\dfrac{\sqrt{3}+1}{2})^2=0$$
$$\left (x-\dfrac{\sqrt{3}+1}{2}\right)^{2}+\sqrt{3}-\dfrac{4+2\sqrt{3}}{4}=0$$
$$\left (x-\dfrac{\sqrt{3}+1}{2}\right)^{2}+\sqrt{3}-1-\dfrac{\sqrt{3}}{2}=0$$
$$\left (x-\dfrac{\sqrt{3}+1}{2}\right)^{2}=1-\dfrac{\sqrt{3}}{2}$$
$$\left (x-\dfrac{\sqrt{3}+1}{2}\right)^{2}=(\dfrac{\sqrt{3}-1}{2})^2$$
$$x-\dfrac{\sqrt{3}+1}{2}=\pm\dfrac{\sqrt{3}-1}{2}$$
$$x=\sqrt{3},1$$
Question 4
1 / -0

Find $$x$$ by solving the given equation:
$$\displaystyle \frac{x + 3}{x + 2} = \frac{3x - 7}{2x - 3}$$

A
$$5,-1$$

B
$$-5,-1$$

C
$$5,1$$

D
$$-5,1$$

Solution

Given,
$$\dfrac{x + 3}{x + 2} = \dfrac{3x - 7}{2x - 3}$$
$$\Rightarrow 3x^2-7x+6x-14=2x^2-3x+6x-9$$
$$\Rightarrow 3x^2-2x^2-7x+6x+3x-6x-14+9=0$$
$$\Rightarrow x^2-4x-5=0$$
$$\Rightarrow x^2-5x+x-5=0$$
$$\Rightarrow x(x-5)+1(x-5)=0$$
$$\Rightarrow (x-5)(x+1)=0$$
$$\Rightarrow x-5=0$$ and $$x+1=0$$
$$\Rightarrow x=5$$ and $$x=-1$$
Question 5
1 / -0

Solve for $$y$$: $$ \sqrt 7 y^2 - 6y - 13 \sqrt 7 = 0$$

A
$$\sqrt 7, 2 \sqrt 7$$

B
$$\displaystyle 3, \frac{2}{\sqrt 7}$$

C
$$\displaystyle \frac{13}{\sqrt 7}, -\sqrt 7$$

D
None of these

Solution

Given quadratic equation is $$\sqrt 7 y^2 - 6y - 13 \sqrt 7 = 0 $$
Comparing it with the standard form of equation $$ax^2+bx+c$$ we get $$a=7,b=6,c=13\sqrt 7$$
Therefore, $$x= \dfrac { -b\pm \sqrt { b^ 2-4ac } }{ 2a } $$
$$= \dfrac { -(-6)\pm \sqrt { (-6)^{ 2 }-4\times \sqrt { 7 } \times \left( -13\sqrt { 7 } \right) } }{ 2\sqrt { 7 } } $$
$$=\dfrac { 6\pm \sqrt { 36+364 } }{ 2\sqrt { 7 } } $$
$$= \dfrac { 6\pm \sqrt { 400 } }{ 2\sqrt { 7 } } $$
$$=\dfrac { 6\pm 20 }{ 2\sqrt { 7 } } $$
$$=\dfrac { 6- 20 }{ 2\sqrt { 7 } } $$
$$=\dfrac { \\- 14 }{ 2\sqrt { 7 } } $$
$$=\dfrac { \\ -7 }{ \sqrt { 7 } } $$
$$=\dfrac { \\ -7\times \sqrt { 7 } }{ \sqrt { 7 } \times \sqrt { 7 } } $$
$$=-\sqrt 7$$
$$=\dfrac { 6+ 20 }{ 2\sqrt { 7 } } $$
$$=\dfrac { \\ 26 }{ 2\sqrt { 7 } } $$
$$=\dfrac { \\ 13 }{ \sqrt { 7 } } $$
Question 6
1 / -0

Find the roots of each of the following quadratic equations by the method of completing the sqaures:
$$x^2 - 6x + 4 = 0$$

A
$$ 9\pm \sqrt 8$$

B
$$6 \pm \sqrt 7$$

C
$$3 \pm \sqrt 5$$

D
$$ 1\pm \sqrt 11$$

Solution

Given, $$x^2 - 6x + 4 = 0$$
$$\Rightarrow x^2-2\times x\times 3+3^2-5=0$$
$$\Rightarrow (x-3)^2=5$$
$$\Rightarrow x-3=\pm \sqrt 5$$
$$\Rightarrow x=3\pm \sqrt 5$$
Question 7
1 / -0

Find the roots of the following quadratic equations, if they exist, by the method of completing the square. $$\displaystyle 2x^2 - 7x + 3 = 0$$

A
The roots of the given quadratic equation are $$\displaystyle \frac{1}{3}$$ and $$1$$.

B
The roots of the given quadratic equation are $$\displaystyle \frac{2}{5}$$ and $$5$$.

C
The roots of the given quadratic equation are $$\displaystyle \frac{1}{2}$$ and $$3$$.

D
The roots of the given quadratic equation are $$\displaystyle \frac{2}{7}$$ and $$4$$.

Solution

Given equation is $$2x^2 - 7x + 3 = 0$$
$$\Rightarrow \left(x^2 - \dfrac{7}{2}x - \dfrac{7}{2}x + \left(- \dfrac{7}{2}\right)^2 \right ) - \left(-\dfrac{7}{2} \right)^2 + 6 = 0$$
$$\Rightarrow \left( x-\dfrac { 7 }{ 2 } \right) ^{ 2 }-\dfrac { 49 }{ 4 } +6=0$$
$$\Rightarrow \left( x-\dfrac { 7 }{ 2 } \right) ^{ 2 }-\dfrac { 25 }{ 4 } =0$$
$$\Rightarrow \left( 2x-\dfrac { 7 }{ 2 } \right) ^{ 2 }=\dfrac { 25 }{ 4 }$$
$$\Rightarrow 2x-\dfrac { 7 }{ 2 } =\pm \sqrt { \dfrac { 25 }{ 4 } } $$
$$\Rightarrow 2x - \dfrac{7}{2} = \pm \dfrac{5}{2}$$
$$\Rightarrow 2x-\dfrac { 7 }{ 2 } =\dfrac { 5 }{ 2 }$$ or $$2x-\dfrac { 7 }{ 2 } =-\dfrac { 5 }{ 2 } $$
$$\Rightarrow 2x=\dfrac { 5 }{ 2 } +\dfrac { 7 }{ 2 }$$ or $$2x=-\dfrac { 5 }{ 2 } +\dfrac { 7 }{ 2 } $$
$$\Rightarrow 2x=6$$ or $$2x=1$$
$$\Rightarrow x=3$$ or $$x=\dfrac{1}{2}$$
Thus, the roots are $$3$$ and $$\dfrac{1}{2}$$.
Question 8
1 / -0

Determine the value of $$k$$ for which the quadratic equation $$4x^2 - 3kx + 1 = 0$$ has equal roots.

A
$$\displaystyle \pm \frac{2}{3} $$

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

C
$$\displaystyle \pm 4$$

D
$$\displaystyle \pm 6$$

Solution

Nature of the roots for a quadratic equation $$ax^2+bx+c=0$$ , can be determined by its discriminant
$$ D=b^2-4ac$$
Here, given quadratic equation is $$4x^2 - 3kx + 1 = 0$$
Putting the value of $$a,b,c$$, we get
$$D=(-3k)^2-4\times 4\times 1$$
$$\Rightarrow 9k^2-16=0$$
$$\Rightarrow 9k^2=16$$
$$\Rightarrow k^2=\cfrac{16}{9}$$
$$\Rightarrow k=\pm \cfrac{4}{3}$$
Question 9
1 / -0

Determine k such that the quadratic equation $$x^2 + 7(3 + 2k)+(1 + 3k)x = 0$$ has equal roots

A
$$2, 7$$

B
$$7, 5$$

C
$$2, $$ $$\displaystyle - \frac{10}{9}$$

D
None of these

Solution
Question 10
1 / -0

Which of the following is a quadratic equation?

A
$$\displaystyle x^{\frac{1}{2}}+2x+3=0$$

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

C
$$\displaystyle x^{4}-x+5=0$$

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

Solution

The first option is not a quadratic equation, since the degree of $$x$$ is $$1$$.
Considering second option, we get
$$(x-1)(x+4)=x^{2}+1$$
$$x^{2}+3x-4=x^{2}+1$$
$$3x-5=0$$ .... a linear equation is one variable.
$$x^{4}-x+5=0$$ is not a quadratic equation since the degree is $$4$$.

Considering the last option
$$(2x+1)(3x-4)=2x^{2}+3$$
$$6x^{2}-5x-4=2x^{2}+3$$
$$4x^{2}-5x-7=0$$ ..... Hence a quadratic equation.

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Quadratic Equations Test - 29

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Question 1

$$x=2,\,\,x=-7$$

$$x=-1,\,\,x=3$$

$$x=\displaystyle 1,x= \frac{3}{2}$$

$$x=\displaystyle 1, x=\frac{1}{2}$$

Solution

Question 2

$$x=\pm\sqrt{\dfrac{7}{2}}$$

$$x=\pm\sqrt{\dfrac{2}{3}}$$

$$x=\sqrt{\dfrac{2}{3}}$$

$$x=\sqrt{\dfrac{7}{2}}$$

Solution

Question 3

$$x = \sqrt 1, 3$$

$$x = \sqrt 5, 1$$

$$x = \sqrt 10, 2$$

$$x = \sqrt 3, 1$$

Solution

Question 4

$$5,-1$$

$$-5,-1$$

$$5,1$$

$$-5,1$$

Solution

Question 5

$$\sqrt 7, 2 \sqrt 7$$

$$\displaystyle 3, \frac{2}{\sqrt 7}$$

$$\displaystyle \frac{13}{\sqrt 7}, -\sqrt 7$$

None of these

Solution

Question 6

$$ 9\pm \sqrt 8$$

$$6 \pm \sqrt 7$$

$$3 \pm \sqrt 5$$

$$ 1\pm \sqrt 11$$

Solution

Question 7

The roots of the given quadratic equation are $$\displaystyle \frac{1}{3}$$ and $$1$$.

The roots of the given quadratic equation are $$\displaystyle \frac{2}{5}$$ and $$5$$.

The roots of the given quadratic equation are $$\displaystyle \frac{1}{2}$$ and $$3$$.

The roots of the given quadratic equation are $$\displaystyle \frac{2}{7}$$ and $$4$$.

Solution

Question 8

$$\displaystyle \pm \frac{2}{3} $$

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

$$\displaystyle \pm 4$$

$$\displaystyle \pm 6$$

Solution

Question 9

$$2, 7$$

$$7, 5$$

$$2, $$ $$\displaystyle - \frac{10}{9}$$

None of these

Solution

Question 10

$$\displaystyle x^{\frac{1}{2}}+2x+3=0$$

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

$$\displaystyle x^{4}-x+5=0$$

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

Solution

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