Complex Numbers and Quadratic Equations Test -6

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Complex Numbers and Quadratic Equations Test -6

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Weekly Quiz Competition

Question 1
1 / -0

Which of the following equations has two distinct real roots ?

A
$$2x^2-3\sqrt 2 x+\dfrac 94=0$$

B
$$x^{2}+x-5=0$$

C
$$x^{2}+3x+2\sqrt{2}=0$$

D
$$5x^{2}-3x+1=0$$

Solution

An equation is said to have two distinct and real roots if the discriminant $$b^2-4ac>0$$
Case (i): For equation: $$2x^2-3\sqrt 2 x+\dfrac 94=0$$.
Here $$a=2, b=-3\sqrt 2, c=\dfrac 94$$
The discrimant is $$(-3\sqrt 2)^2-4(2)\left(\dfrac 94\right)=18-18=0$$
Hence this equation has equal real roots

Case (ii): For equation: $$x^2+x-5=0$$.
Here $$a=1, b=1, c=-5$$
The discrimant is $$1^2-4(1)(-5)=1+20=21>0$$
Hence this equation has two distinct real roots

Case (iii): For equation: $$x^2+3x+2\sqrt 2=0$$.
Here $$a=1, b=3, c=2\sqrt 2$$
The discrimant is $$3^2-4(1)(2\sqrt2)=9-8\sqrt 2<0$$
Hence this equation has no real roots

Case (iv): For equation: $$5x^2-3x+1=0$$.
Here $$a=5, b=-3, c=1$$
The discrimant is $$(-3)^2-4(5)(1)=9-20<0$$
Hence this equation has no real roots
Question 2
1 / -0

A quadratic equation $$ax^2 + bx+c=0$$ has two distinct real roots, if

A
$$a=0$$

B
$$b^2-4ac = 0$$

C
$$b^2-4ac < 0$$

D
$$b^2-4ac > 0$$

Solution

If $$a=0$$, it becomes linear equation.
If $${ b }^{ 2 }-4ac=0$$, then there will be real and equal roots.
If $${ b }^{ 2 }-4ac<0$$, then the roots will be unreal.
Only if $${ b }^{ 2 }-4ac>0$$, we will get two real distinct roots.
Option D is correct!
Question 3
1 / -0

The roots of the equation $$3x^{2} - 4x + 3 = 0$$ are :

A
real and unequal

B
real and equal

C
imaginary

D
none of these

Solution

Given equation is $$3x^2-4x+3=0$$
To find, the nature of the roots of the equation
An equation is said to have
(i) two distinct and real roots if the discriminant $$b^2-4ac>0$$
(ii) equal real roots if $$b^2-4ac=0$$
(iii) no real roots or imaginary roots if $$b^2-4ac<0$$
In the given equation $$a=3, b=-4, c=3$$
Hence the discriminant is $$(-4)^2-4(3)(3)=16-36=-20<0$$
Therefore the roots of the given equation are imaginary in nature.
Question 4
1 / -0

Determine the nature of roots of the equation $$x^2 + 2x\sqrt{3}+3=0$$.

A
Real and distinct

B
Non-real and distinct

C
Real and equal

D
Non-real and equal

Solution

The nature of the roots can be determined from the discriminant $$b^2-4ac$$
$$\therefore b^2-4ac=(2\sqrt {3})^2-(4\times 1\times 3)$$
$$\Rightarrow b^2-4ac=12-12$$
$$\Rightarrow b^2-4ac=0$$
$$\because b^2-4ac=0$$
There are two real and equal roots.
Question 5
1 / -0

The roots of the equation $$ax^{2}+x+b=0 $$ are equal if :

A
$$ b^{2}=4a$$

B
$$ b^{2}<4a$$

C
$$ b^{2}>4a$$

D
$$ab=\dfrac{1}{4}$$

Solution

Given: quadratic equation $$ax^2+x+b=0$$
If the discriminant ($$b^2-4ac=0$$) of a quadratic function is equal to zero, that function has exactly one real root or two roots are equal.
In the given quadratic equation, $$a=a, b=1, c=b$$
Therefore the discriminant becomes,
$$1-4ab=0\implies ab=\dfrac 14$$
Question 6
1 / -0

For $$a<0$$, arg $$a=$$

A
$$\dfrac{\pi }{2}$$

B
$$\dfrac{-\pi }{2}$$

C
$$\pi $$

D
$$-\pi $$

Solution

It is given that $$a<0$$
Thus, 'a' is purely real.
Hence $$arg(a)$$ will be $$0$$ or, $$\pi$$
But it is given that $$a<0$$
Hence $$cos(arg(a))=-1$$
Or
$$arg(a)=\pi$$.
Question 7
1 / -0

If the square of $$(a + ib)$$ is real, then $$ ab=$$

A
$$0$$

B
$$1$$

C
$$-1$$

D
$$2$$

Solution

$$(a+ib)^{2}=a^{2}-b^{2}+2iab$$ is given to be real
$$\Rightarrow ab=0$$
Question 8
1 / -0

If $$x^{2}-2px+8p-15=0$$ has equal roots, then $$p=$$

A
$$3$$ or $$-5$$

B
$$3$$ or $$5$$

C
$$-3$$ or $$5$$

D
$$-3$$ or $$-5$$

Solution

Given equation is $$x^2-2px+8p-15=0$$
Here $$a=1, b=-2p, c=8p-15$$
We know for equal roots, $$b^{2}-4ac=0$$
Therefore, $$(-2p)^{2}-4(1)(8p-15)=0$$
$$\Rightarrow 4p^{2}-32p+60=0$$
$$\Rightarrow p^{2}-8p+15=0$$
$$\Rightarrow (p-5)(p-3)=0$$
i.e., $$p=5 $$ or $$ 3$$
Question 9
1 / -0

Find the argument of $$-1 - i\sqrt{3}$$

A
$$\theta= -2\pi/3$$

B
$$\theta= 2\pi/3$$

C
$$\theta= -4\pi/3$$

D
$$\theta= 4\pi/3$$

Solution

Let $$ z = -1 -i\sqrt{3}$$
Then $$\alpha = \tan^{-1}\left|b/a\right| = \tan^{-1} \left|\sqrt{3}/1\right| = \pi/3$$
Here, $$z$$ is in the third quadrant.
Therefore, argument is $$\theta= -(\pi - \alpha) = -(\pi - \pi/3) = -2\pi/3$$

Ans: A
Question 10
1 / -0

The roots of $$x^{2}-x+1=0$$ are:

A
Real and equal

B
Real and not equal

C
Imaginary

D
Reciprocals

Solution

Given equation is $$x^2-x+1=0$$
We know $$D=b^{2}-4ac$$
Here $$a=1, b=-1, c=1$$
Therefore, $$D=(-1)^{2}-4(1)(1)$$
$$=1-4$$
$$=-3< 0$$
Thus roots are imaginary.

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Complex Numbers and Quadratic Equations Test -6

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Question 1

$$2x^2-3\sqrt 2 x+\dfrac 94=0$$

$$x^{2}+x-5=0$$

$$x^{2}+3x+2\sqrt{2}=0$$

$$5x^{2}-3x+1=0$$

Solution

Question 2

$$a=0$$

$$b^2-4ac = 0$$

$$b^2-4ac < 0$$

$$b^2-4ac > 0$$

Solution

Question 3

real and unequal

real and equal

imaginary

none of these

Solution

Question 4

Real and distinct

Non-real and distinct

Real and equal

Non-real and equal

Solution

Question 5

$$ b^{2}=4a$$

$$ b^{2}<4a$$

$$ b^{2}>4a$$

$$ab=\dfrac{1}{4}$$

Solution

Question 6

$$\dfrac{\pi }{2}$$

$$\dfrac{-\pi }{2}$$

$$\pi $$

$$-\pi $$

Solution

Question 7

$$0$$

$$1$$

$$-1$$

$$2$$

Solution

Question 8

$$3$$ or $$-5$$

$$3$$ or $$5$$

$$-3$$ or $$5$$

$$-3$$ or $$-5$$

Solution

Question 9

$$\theta= -2\pi/3$$

$$\theta= 2\pi/3$$

$$\theta= -4\pi/3$$

$$\theta= 4\pi/3$$

Solution

Question 10

Real and equal

Real and not equal

Imaginary

Reciprocals

Solution

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