Quadratic Equations Test

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Quadratic Equations Test - 36

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Weekly Quiz Competition

Question 1
1 / -0

Which of the statements given below is correct, if discriminant for equation $$5x^2 - 4x +2 =0$$ is D ?

A
D > 0

B
D = 0

C
D < 0

D
D is not defined

Solution

The Discriminant($$D$$) of the quadratic equation $$ax^{2}+bx+c=0$$ is $$D=b^{2}-4ac$$.
Here, $$b=-4,a=5,c=2$$
Then,$$D=(-4)^{2}-4(5)(2)$$
$$D=16-40=-24$$
$$\Rightarrow$$ $$D<0$$
$$\therefore$$ Option C is correct.
Question 2
1 / -0

Which of the following equations are not quadratic?

A
$$x(2x+3)=x+2$$

B
$$(x-2)^2+1=2x-3$$

C
$$y(8y+5)=y^2+3$$

D
$$y(2y+15)=2(y^2+y+8)$$

Solution

As we know that in quadratic equation the highest degree of the polynomial is $$2$$. so all the option have degree of the polynomail $$2$$ except option D. because in option D the degree of the polynimail is $$1$$.so the correct option is D
Question 3
1 / -0

If $$b\in F'$$ then the roots of the equation $$\left( 2+b \right) { x }^{ 2 }+(3+b)x+(4+b)=0\quad $$ is

A
real and imaginary

B
real and equal

C
imagenary

D
cannot predicted

Solution

$$(2+b)x^{2}+(3+b)x+(4+b)=0$$
$$D=(3+b)^{2}-4(2+b)(4+b)=b^{2}+6{b}+9-4{b^{2}}-24{b}-32=-3{b^{2}}-18{b}-23=-3(b^{2}+6{b}+9)+4$$
$$=-3(b+3)^{2}+9<0$$
So roots are imaginary
Question 4
1 / -0

Directions For Questions

Consider the quadratic equation $$(1+k)x^2-2(1+2k)x+(3+k)=0$$, where k $$\epsilon$$ R$$-\{-1\}$$.

...view full instructions

The number of integral value of k such that the given quadratic equation has imaginary roots are?

A
$$0$$

B
$$\sqrt{2/3}$$

C
$$2$$

D
$$3$$

Solution
Question 5
1 / -0

If $$a, b,c$$ are distinct and the roots of $$(b-c)x^{2} +(c-a)x +(a-b)=0$$ are equal, then $$a,b,c$$ are in:

A
Arithmetic Progression

B
Geometric progression

C
Harmonic progression

D
Arithmetico-Geometric progression

Solution

$$f\left(x\right)=\left(b-c\right)x^{2}+\left(c-a\right)x+\left(a-b\right)=0$$
Given that the roots are equal.
$$\Rightarrow\Delta=0$$
Now,
$$\Delta=b^{2}-4ac=0$$
$$\Rightarrow\left(c-a\right)^{2}-4\left(b-c\right)\left(a-b\right)=0$$
$$\Rightarrow{c}^{2}-2ac+a^{2}+4b^{2}+4ac-4bc-4ab=0$$
$$\Rightarrow\left(a-2b+c\right)^{2}=0$$
$$\Rightarrow\left(a-2b+c\right)=0$$ $$\Rightarrow2b=a+c$$
$$\Rightarrow{a},b, c$$ are in A.P.
Question 6
1 / -0

Which of the following equations, is not a quadratic equation?

A
$$4x^{2} - 7x + 3 = 0$$

B
$$3x^{2} - 4x + 1 = 0$$

C
$$2x - 7 = 0$$

D
$$4x^{2} - 3 = 0$$

Solution

Alternate $$(A) : 4x^{2} - 7x + 3 = 0$$
The maximum exponent of variable $$x$$ is $$2$$. So it is a quadratic equation.
Alternate $$(B) : 3x^{2} - 4x + 1 = 0$$
The maximum exponent of variable $$x$$ is $$2$$. So it is a quadratic equation.
Alternate $$(C) : 2x - 7 = 0$$
The maximum exponent of variable $$x$$ is $$1$$. So it is not a quadratic equation.
Question 7
1 / -0

The discriminant of quadratic equation $$3x^{2} - 4x - 1 = 0$$ is _______.

A
$$0$$

B
$$4$$

C
$$12$$

D
$$28$$

Solution

Quadratic equation: $$3x^{2} - 4x - 1 = 0$$

Comparing the given equation with $$ax^{2} + bx + c = 0$$

we have $$a = 3, b = -4, c = -1$$

Now, discriminant

$$D = b^{2} - 4ac$$

$$= (-4)^{2} - 4(3) (-1)$$

$$= 16 + 12$$

$$= 28$$.
Question 8
1 / -0

Find the roots of the equation by the method of Completion of Squares:
$$x^2 - 6x + 9 = 0$$

A
$$3, 3$$

B
$$-3, -3$$

C
$$2, 2$$

D
$$-1, -1$$

Solution

$$x^2-6x+9=0$$
$$x^2-2*3*x+3^2=0$$
$$(x-3)^2=0$$
$$(x-3)(x-3)=0$$
Roots are $$3,3$$.
Question 9
1 / -0

What are the zeros of $$P(x) = 5 - x^{2}$$?

A
$$\sqrt {5}$$ and $$-5$$

B
$$\dfrac {1}{5}$$ and $$-\dfrac {1}{5}$$

C
$$5$$ and $$-5$$

D
$$\sqrt {5}$$ and $$-\sqrt {5}$$

Solution

For the zeros of $$P(x) = 5 - x^{2}$$,
$$P(x) = 0$$
$$\therefore 5 - x^{2} = 0$$
$$\therefore x^{2} = 5$$
$$\therefore x = \pm \sqrt {5}$$
$$\therefore x = \sqrt {5}$$ and $$-\sqrt {5}$$.
Question 10
1 / -0

If the roots of the quadratic equation $$5x^{2} - 2kx + 20 = 0$$ are real and equal then the value of $$k$$ is ________.

A
$$20$$

B
$$-10$$

C
$$10$$ or $$-10$$

D
$$10$$

Solution

Quadratic equation:
$$5x^{2} - 2kx + 20 = 0$$

It is given that the roots of the quadratic equation are real and equal, Then discriminant $$D = 0$$.

Comparing the given equation with $$ax^{2} + bx + c = 0$$

we have $$a = 5, b = -2k$$ and $$c = 20$$

Now, $$D = 0$$

$$\therefore b^{2} - 4ac = 0$$

$$\therefore (-2k)^{2} - 4(5)(20) = 0$$

$$\therefore 4k^{2} - 400 = 0$$

$$\therefore k^{2} = 100$$

$$\therefore k = \pm 10$$

Thus, $$k = 10$$ or $$-10$$.

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Quadratic Equations Test - 36

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Quadratic Equations Test - 36

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Question 1

D > 0

D = 0

D < 0

D is not defined

Solution

Question 2

$$x(2x+3)=x+2$$

$$(x-2)^2+1=2x-3$$

$$y(8y+5)=y^2+3$$

$$y(2y+15)=2(y^2+y+8)$$

Solution

Question 3

real and imaginary

real and equal

imagenary

cannot predicted

Solution

Question 4

$$0$$

$$\sqrt{2/3}$$

$$2$$

$$3$$

Solution

Question 5

Arithmetic Progression

Geometric progression

Harmonic progression

Arithmetico-Geometric progression

Solution

Question 6

$$4x^{2} - 7x + 3 = 0$$

$$3x^{2} - 4x + 1 = 0$$

$$2x - 7 = 0$$

$$4x^{2} - 3 = 0$$

Solution

Question 7

$$0$$

$$4$$

$$12$$

$$28$$

Solution

Question 8

$$3, 3$$

$$-3, -3$$

$$2, 2$$

$$-1, -1$$

Solution

Question 9

$$\sqrt {5}$$ and $$-5$$

$$\dfrac {1}{5}$$ and $$-\dfrac {1}{5}$$

$$5$$ and $$-5$$

$$\sqrt {5}$$ and $$-\sqrt {5}$$

Solution

Question 10

$$20$$

$$-10$$

$$10$$ or $$-10$$

$$10$$

Solution

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