Complex Numbers and Quadratic Equations Test 15

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Complex Numbers and Quadratic Equations Test 15

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Weekly Quiz Competition

Question 1
1 / -0

The roots of $$2{x}^{2}-6x+3=0$$ are

A
rational

B
equal

C
irrational

D
imaginary

Solution

$$2x^2-6x+3=0$$
$$x=\dfrac{6\pm\sqrt{36-24}}{2}$$
$$x=\dfrac{6\pm\sqrt{12}}{2}$$
Question 2
1 / -0

For what values of $$k$$ will the quadratic equation : $$\displaystyle { 2x }^{ 2 }-kx+1=0$$ have real and equal roots?

A
$$\displaystyle \pm 2\sqrt { 2 } $$

B
$$\displaystyle \pm \sqrt { 2 } $$

C
$$\displaystyle \pm 3\sqrt { 2 } $$

D
$$\displaystyle \pm \sqrt { 3 } $$

Solution

Given equation is: $$\displaystyle { 2x }^{ 2 }-kx+1=0$$
Hence, $$\displaystyle a=2,b=-k.c=1$$
$$\displaystyle D={ b }^{ 2 }-4ac={ \left( -k \right) }^{ 2 }-4\times 2\times 1-{ k }^{ 2 }-8$$
We know that a quadratic equation has real and equal roots, if
$$\displaystyle D=0\Leftarrow { k }^{ 2 }-8=0\Rightarrow { k }^{ 2 }=8$$
$$\displaystyle k=\pm \sqrt { 8 } =\pm 2\sqrt { 2 } $$
Question 3
1 / -0

Find the value of 'm' so that the equation $$\displaystyle { 9x }^{ 2 }-8mx-9=0$$ has one root as the negative of the other.

A
0

B
1

C
2

D
None of these

Solution

Given equation is: $$9x^2-8mx-9=0$$
Comparing with standard equation $$ax^{2}+bx+c=0$$, we get
$$a=9, b=-8m, c=-9$$
Let $$\displaystyle \alpha ,\beta $$ be the roots
Then, $$ \alpha +\beta =-\dfrac{b}{a}=\dfrac { 8m }{ 9 } $$
But, $$\alpha =-\beta $$
Thus, $$\alpha +\beta =0$$
$$\Rightarrow \dfrac{8m}{9}=0$$
$$\Rightarrow m=0$$
Question 4
1 / -0

Given that $$kx(x-2)+6=0$$ has real and equal roots, the root is

A
$$2$$

B
$$-1$$

C
$$1$$

D
$$\cfrac{1}{2}$$

Solution

Given equation is: $$kx(x-2)+6=0$$
$$kx^{2}-2kx+6=0$$
Then $$a=k ,b=-2k ,c=6$$
Then $$b^{2}-4ac=0$$
So $$(-2k)^{2}-4k(6)=0$$
$$4k^{2}-24k=0$$
$$4k(k-6)=0$$
$$ k=0$$ and $$k=6$$
Put value of $$k $$
Then $$6x^{2}-12x+6=0$$
$$6x^{2}-6x-6x+6=0$$
$$ 6x(x-1)-6(x-1)=0$$
$$\therefore x=1$$
Question 5
1 / -0

The value of $$k$$ for which $$4{x}^{2}-4\sqrt {3}x+k=0$$ is satisfied by only one real value of $$x$$ is

A
$$\sqrt 6$$

B
$$-3$$

C
$$3$$

D
$$\cfrac{1}{\sqrt {3}}$$

Solution

We know that $$(a-b)^{2}=a^{2}-2ab+b^{2}$$
Given Equation is $$4x^{2}-4\sqrt{3}x+k$$
Co-pairing both equations, we get
$$a=2x$$ and $$ k=3$$
Question 6
1 / -0

$$kx^2-2\sqrt 5x +4 = 0$$
For what value of $$k$$ will the quadratic equation have real and equal roots ?

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

B
$$\displaystyle \frac { 4 }{ 5 } $$

C
$$\displaystyle \frac { 5 }{ 4 } $$

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

Solution

Given equation is $$\displaystyle { kx }^{ 2 }-2\sqrt { 5 } +4=0$$
$$\displaystyle a=k,b=-2\sqrt { 5 } ,c=4$$
$$\displaystyle \therefore D={ b }^{ 2 }-4ac={ \left( -2\sqrt { 5 } \right) }^{ 2 }-4\times k\times 4=20-16$$
We know that a quadratic equation has real and equal roots, if $$D=0$$.
$$\displaystyle \Rightarrow 20-16k=0$$
$$\displaystyle \Rightarrow -16k=-20$$
$$\displaystyle \Rightarrow k=\frac { 20 }{ 16 } =\frac { 5 }{ 4 } $$
Question 7
1 / -0

For what values of k will the quadratic equation : $$\displaystyle { 3x }^{ 2 }+kx+3=0$$ have real equal roots?

A
$$\displaystyle \pm \sqrt { 3 } $$

B
$$\displaystyle \pm \sqrt { 6 } $$

C
$$\displaystyle \pm 6$$

D
$$\displaystyle \pm 3$$

Solution

Given equation is: $$\displaystyle { 3x }^{ 2 }+kx+3=0$$
Here, $$\displaystyle a=3,b=k,c=3$$
Now, $$\displaystyle D={ b }^{ 2 }-4ac={ k }^{ 2 }-4\times 3\times 3{ k }^{ 2 }-36$$
We know that a quadratic equation has real and equal roots,
If $$\displaystyle D=0\Rightarrow { k }^{ 2 }-36=0\Rightarrow { k }^{ 2 }=36$$
$$\displaystyle \therefore \quad k=\pm \sqrt { 36 } =\pm 6$$
Question 8
1 / -0

For what value of 'k' the equation $$\displaystyle \left( k+3 \right) { x }^{ 2 }-\left( 5-k \right) x+1=0$$ has coincident roots ?

A
$$1, 13$$

B
$$1, 12$$

C
$$3, 13$$

D
None of these

Solution

For coincident roots, $$\displaystyle D=0$$
Now, $$\displaystyle { \left[ -\left( 5-k \right) \right] }^{ 2 }-4\times \left( k+3 \right) \times 1=0$$
$$\displaystyle \Rightarrow \quad 25+{ k }^{ 2 }-10k-4k-12=0$$
$$\displaystyle \Rightarrow \quad { k }^{ 2 }-14k+13=0$$
$$\displaystyle \Rightarrow \quad \left( k-13 \right) \left( k-1 \right) =0$$
$$\displaystyle \therefore \quad k=13,1$$
Question 9
1 / -0

Find the value of $$k$$ for which the equation
$${x}^{2}+kx+81=0$$ and $${x}^{2}-6\sqrt {2}x+k=0$$ has both real roots, $$k> 0$$

A
$$k=9$$

B
$$k=18$$

C
$$k=3\sqrt {2}$$

D
No such value

Solution

For, $${x}^{2}+kx+81=0$$ to have real roots

$${k}^{2}-324 \ge 0$$ $$\Rightarrow$$ $$k\ge 18.......(1)$$

For $${x}^{2}-6\sqrt {2}x+k=0$$ to have real roots

$$72-4k\ge 0$$ $$\Rightarrow$$ $$k\le 18..........(2)$$

For both equations to have real roots $$k=18$$
Question 10
1 / -0

For what value of $$k$$ will the quadratic equation $$\displaystyle { 2x }^{ 2 }+3x+k=0$$ have real equal roots ?

A
$$\displaystyle \frac { 3 }{ 8 } $$

B
$$\displaystyle \frac { 8 }{ 3 } $$

C
$$\displaystyle \frac { 9 }{ 8 } $$

D
$$\displaystyle \frac { 8 }{ 9 } $$

Solution

Given equation is $$\displaystyle 2{ x }^{ 2 }+3x+k=0$$
Here, $$\displaystyle a=2,b=3,c=k$$
$$\displaystyle \because D={ b }^{ 2 }-4ac={ \left( 3 \right) }^{ 2 }-4\times 2\times k=9-8k$$.
We know that a quadratic equation has real and equal roots if $$D = 0$$.
$$\displaystyle \Rightarrow 9-8k=0\Rightarrow -8k=-9$$
$$\displaystyle \Rightarrow k=\frac { 9 }{ 8\\ } $$

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Complex Numbers and Quadratic Equations Test 15

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Question 1

rational

equal

irrational

imaginary

Solution

Question 2

$$\displaystyle \pm 2\sqrt { 2 } $$

$$\displaystyle \pm \sqrt { 2 } $$

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

$$\displaystyle \pm \sqrt { 3 } $$

Solution

Question 3

0

1

2

None of these

Solution

Question 4

$$2$$

$$-1$$

$$1$$

$$\cfrac{1}{2}$$

Solution

Question 5

$$\sqrt 6$$

$$-3$$

$$3$$

$$\cfrac{1}{\sqrt {3}}$$

Solution

Question 6

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

$$\displaystyle \frac { 4 }{ 5 } $$

$$\displaystyle \frac { 5 }{ 4 } $$

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

Solution

Question 7

$$\displaystyle \pm \sqrt { 3 } $$

$$\displaystyle \pm \sqrt { 6 } $$

$$\displaystyle \pm 6$$

$$\displaystyle \pm 3$$

Solution

Question 8

$$1, 13$$

$$1, 12$$

$$3, 13$$

None of these

Solution

Question 9

$$k=9$$

$$k=18$$

$$k=3\sqrt {2}$$

No such value

Solution

Question 10

$$\displaystyle \frac { 3 }{ 8 } $$

$$\displaystyle \frac { 8 }{ 3 } $$

$$\displaystyle \frac { 9 }{ 8 } $$

$$\displaystyle \frac { 8 }{ 9 } $$

Solution

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