Complex Numbers and Quadratic Equations Test 16

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Complex Numbers and Quadratic Equations Test 16

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Weekly Quiz Competition

Question 1
1 / -0

The values of $$b$$ for which the equation $$\displaystyle { 2bx }^{ 2 }-40x+25=0$$ has equal root is :

A
$$5$$

B
$$6$$

C
$$7$$

D
$$8$$

Solution

Given equation is $$2bx^2-40x+25=0$$
$$\therefore a=2b, b=-40, c=25$$
For equal roots, $$D = 0$$
$$\therefore b^2-4ac=0$$
$$\displaystyle \Rightarrow { \left( 40 \right) }^{ 2 }-4\times 25\times 2b=0$$
$$\displaystyle \Rightarrow 200b=1600$$
$$\displaystyle \Rightarrow b=8$$
Question 2
1 / -0

If $$z_1, z_2$$ and $$z_3$$ are complex numbers such that $$|z_1| = |z_2| = |z_3| = \left | \displaystyle \frac{1}{z_1} + \frac{1}{z_2} + \frac{1}{z_3} \right | = 1,$$ then $$|z_1 + z_2 + z_3|$$ is

A
equal to 1

B
less than 1

C
greater than 3

D
equal to 3

Solution

$$|z_1| = |z_2| = |z_3| = 1$$ (given)
Now $$|z_1| = 1 \Rightarrow |z_r|^2 = 1$$
$$\Rightarrow z_1 \bar z_1 = 1$$
Similarly $$z_2 \bar z_2 = 1, z_3 \bar z_3 = 1$$
Now $$\displaystyle \left | \dfrac{1}{z_1} + \dfrac{1}{z_2} + \dfrac{1}{z_3} \right | = 1$$
$$\Rightarrow |\bar z_1 + \bar z_2 + \bar z_3| = 1$$
$$\Rightarrow |\overline{z_1 + z_2 + z_3}| = 1 \Rightarrow |z_1 + z_2 + z_3| = 1$$
Question 3
1 / -0

For what values of k, will quadratic equation $$\displaystyle { 9x }^{ 2 }+3kx+4=0$$ have real and equal roots?

A
$$\displaystyle \pm 4$$

B
$$\displaystyle \pm 3$$

C
$$\displaystyle \pm 2$$

D
$$\displaystyle \pm 1$$

Solution

Given equation is Comparing with $$\displaystyle { ax }^{ 2 }+bx+c=0$$
$$\displaystyle \Rightarrow a=9,b=3k,c=4$$, we get
$$\displaystyle D={ b }^{ 2 }-4ac={ \left( 3k \right) }^{ 2 }-4\left( 9 \right) \left( 4 \right) =9{ k }^{ 2 }-144$$
Since the quadratic equation has real and equal roots.
Therefore, $$\displaystyle D=0$$
$$\Rightarrow \displaystyle { 9k }^{ 2 }-144=0$$
$$\Rightarrow \displaystyle { k }^{ 2 }=\frac { 144 }{ 9 } =16$$
$$\Rightarrow \displaystyle k=\pm \sqrt { 16 } =\pm 4$$
$$\Rightarrow \displaystyle k=4$$ or $$ -4$$
Question 4
1 / -0

Find the value of P for which the following equation has equal roots $$\displaystyle { px }^{ 2 }-8x+2p=0$$

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

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

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

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

Solution

$$\displaystyle { px }^{ 2 }-8x+2p=0$$
here, $$\displaystyle a=p,b=-8,c=2p$$
For equal roots $$\displaystyle { b }^{ 2 }-4ac=0$$
or, $$\displaystyle { \left( -8 \right) }^{ 2 }-4\left( p \right) \left( 2p \right) =0$$
or, $$\displaystyle 64-8{ p }^{ 2 }=0$$
or, $$\displaystyle -8{ p }^{ 2 }=-64$$
or, $$\displaystyle 8{ p }^{ 2 }=64$$
or, $$\displaystyle { p }^{ 2 }=\frac { 64 }{ 8 } =8$$
or, $$\displaystyle p=\pm \sqrt { 8 } =\pm 2\sqrt { 2 } $$
Question 5
1 / -0

Which of the following equations has no solution for $$a$$ ?

A
$${a}^{2}-6a+7=0$$

B
$${a}^{2}+6a-7=0$$

C
$${a}^{2}+4a+3=0$$

D
$${a}^{2}-4a+3=0$$

E
$${a}^{2}-4a+5=0$$

Solution

You can determine which of the equations has no solution by determining which equation has a negative discriminant:
(A) $${b}^{2}-4ac={(-6)}^{2}-4(1)(7)=36-28=8$$
(B) $${b}^{2}-4ac={(6)}^{2}-4(1)(-7)=36+28=64$$
(C) $${b}^{2}-4ac={(4)}^{2}-4(1)(3)=16-12=4$$
(D) $${b}^{2}-4ac={(-4)}^{2}-4(1)(3)=16-12=4$$
(E) $${b}^{2}-4ac={(-4)}^{2}-4(1)(5)=16-20=-4$$
Question 6
1 / -0

The roots of the equation $${2x^{2 }+ 3x + 2} = 0$$ are

A
Real, rational, and equal

B
Real,rational and unequal

C
Real, irrational, and unequal

D
non real (imaginary)

Solution

Given quadratic is $$2x^2+3x+2=0$$
Comparing it with standard from of quadratic equation $$ax^2+bx+c=0$$, we get $$a=2, b=3, c=2$$
Thus discriminant of the given quadratic is
$$D=b^2-4ac=9-16=-7<0$$
Hence roots of given quadratic are imaginary.
Thus choice $$D$$ is correct
Question 7
1 / -0

Let $${ X }_{ n }=\left\{ z=x+iy:{ \left| z \right| }^{ 2 } \le \dfrac { 1 }{ n } \right\} $$ for all integers $$n\ge 1$$. Then, $$\displaystyle\bigcap _{ n=1 }^{ \infty }{ { X }_{ n } } $$ is

A
A singleton set

B
Not a finite set

C
An empty set

D
A finite set with more than one element

Solution

Given, $${ X }_{ n }=\left\{ z=x+iy:{ \left| z \right| }^{ 2 }\le \dfrac { 1 }{ n } \right\} $$
$$=\left\{ { x }^{ 2 }+{ y }^{ 2 }\le \dfrac { 1 }{ n } \right\} $$
$$\therefore { X }_{ 1 }=\left\{ { x }^{ 2 }+{ y }^{ 2 }\le 1 \right\} $$
$${ X }_{ 2 }=\left\{ { x }^{ 2 }+{ y }^{ 2 }\le \dfrac { 1 }{ 2 } \right\} $$
$${ X }_{ 3 }=\left\{ { x }^{ 2 }+{ y }^{ 2 }\le \dfrac { 1 }{ 3 } \right\} $$
.............................................
.............................................
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$${ X }_{ \infty }=\left\{ { x }^{ 2 }+{ y }^{ 2 }\le 0 \right\} $$
$$\therefore \bigcap _{ n=1 }^{ \infty }{ { X }_{ n } } ={ X }_{ 1 }\cap { X }_{ 2 }\cap { X }_{ 3 }\cap \dots \cap { X }_{ \infty }$$
$$=\left\{ { x }^{ 2 }+{ y }^{ 2 }=0 \right\} $$
Hence, $$\bigcap _{ n=1 }^{ \infty }{ { X }_{ n } } $$ is a singleton set.
Question 8
1 / -0

The modulus of $$\dfrac { 1-i }{ 3+i } +\dfrac { 4i }{ 5 } $$ is

A
$$\sqrt { 5 } $$ unit

B
$$\dfrac { \sqrt { 11 } }{ 5 } $$ unit

C
$$\dfrac { \sqrt { 5 } }{ 5 } $$ unit

D
$$\dfrac { \sqrt { 12 } }{ 5 } $$ unit

Solution

Let $$z=\dfrac { 1-i }{ 3+i } +\dfrac { 4i }{ 5 } $$
$$=\dfrac { 5-5i+12i-4 }{ 5\left( 3+i \right) } =\dfrac { 1+7i }{ 5\left( 3+i \right) } $$
$$=\dfrac { \left( 1+7i \right) \left( 3-i \right) }{ 5\left( 9+1 \right) } $$
$$=\dfrac { 10+20i }{ 50 } =\dfrac { 1+2i }{ 5 } $$
$$\therefore \left| z \right| =\sqrt { { \left( \dfrac { 1 }{ 5 } \right) }^{ 2 }+{ \left( \dfrac { 2 }{ 5 } \right) }^{ 2 } } =\dfrac { 1 }{ 5 } \sqrt { 1+4 } =\dfrac { \sqrt { 5 } }{ 5 } $$
Question 9
1 / -0

In the complex numbers, where $$i = \sqrt {-1}$$, the conjugate of any value $$a + bi$$ is $$a -ib$$. What is the result when you multiply $$2 + 7i$$ by its conjugate?

A
$$45$$

B
$$-45$$

C
$$45i$$

D
$$53$$

E
$$53i$$

Solution

Conjugate of $$ 2 + 7i $$ is $$ 2-7i$$
$$ (2 + 7i) \times (2-7i) = 2^2 - (7i)^2$$
$$ = 4 -49(-1) $$
$$= 4 + 49 $$
$$= 53 $$
Question 10
1 / -0

In the complex numbers, where $$i^{2} = -1$$, what is the value of $$5 + 6i$$ multiplied by $$3- 2i$$?

A
$$27$$

B
$$27i$$

C
$$27 + 8i$$

D
$$15 + 8i$$

E
$$15 - 18i$$

Solution

We know, $$i^2=-1$$
The required product is $$ (5 + 6i)(3-2i) $$
$$= 15-10i+18i+12 $$
$$= 27 + 8i $$

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Complex Numbers and Quadratic Equations Test 16

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Question 1

$$5$$

$$6$$

$$7$$

$$8$$

Solution

Question 2

equal to 1

less than 1

greater than 3

equal to 3

Solution

Question 3

$$\displaystyle \pm 4$$

$$\displaystyle \pm 3$$

$$\displaystyle \pm 2$$

$$\displaystyle \pm 1$$

Solution

Question 4

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

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

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

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

Solution

Question 5

$${a}^{2}-6a+7=0$$

$${a}^{2}+6a-7=0$$

$${a}^{2}+4a+3=0$$

$${a}^{2}-4a+3=0$$

$${a}^{2}-4a+5=0$$

Solution

Question 6

Real, rational, and equal

Real,rational and unequal

Real, irrational, and unequal

non real (imaginary)

Solution

Question 7

A singleton set

Not a finite set

An empty set

A finite set with more than one element

Solution

Question 8

$$\sqrt { 5 } $$ unit

$$\dfrac { \sqrt { 11 } }{ 5 } $$ unit

$$\dfrac { \sqrt { 5 } }{ 5 } $$ unit

$$\dfrac { \sqrt { 12 } }{ 5 } $$ unit

Solution

Question 9

$$45$$

$$-45$$

$$45i$$

$$53$$

$$53i$$

Solution

Question 10

$$27$$

$$27i$$

$$27 + 8i$$

$$15 + 8i$$

$$15 - 18i$$

Solution

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