Quadratic Equations Test

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Quadratic Equations Test - 17

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Weekly Quiz Competition

Question 1
1 / -0

The number of real roots of the equation $$(x - 1)^2 + (x - 2)^2 + (x- 3)^2 = 0$$ is :

A
$$1$$

B
$$2$$

C
$$3$$

D
None of these

Solution

Given,

$$(x - 1)^2 + (x - 2)^2 + (x - 3)^2 = 0$$

$$\Rightarrow x^2 - 2x + 1 + x^2 - 4x + 4 + x^2 - 6x + 9 = 0$$

$$\Rightarrow 3x^2 - 12x + 14 = 0$$

Let $$D$$ be the discriminant of the obtained quadratic equation.

$$D=(-12)^2-4\times 14\times 3 \\=144-168\\=-24$$

Since the value of discriminant is negative, the equation does not have any real roots
Question 2
1 / -0

If $$\alpha, \beta$$ are the roots of $$ax^2+bx+c=0$$ and $$\alpha +k, \beta +k$$ are the roots of $$px^2+qx+r=0$$, then $$\displaystyle\frac{b^2-4ac}{q^2-4pr}$$ is equal to:

A
$$\displaystyle\frac{a}{p}$$

B
$$1$$

C
$$\left(\displaystyle\frac{a}{p}\right)^2$$

D
$$0$$

Solution

Absolute difference of both the roots will be same,
$$\Rightarrow \left|\dfrac{\sqrt {D_1}}{a}\right|=\left|\dfrac{\sqrt{D_2}}{p} \right|$$
Where $$D_1,D_2$$ are the discriminant of the respective quadratic
$$\Rightarrow \dfrac{D_1}{D_2}=\dfrac{a^2}{p^2}$$, square above equation
$$\Rightarrow \dfrac{b^2-4ac}{q^2-4pr}=\left(\dfrac{a}{p}\right)^2$$
Question 3
1 / -0

The root of the equation $$2(1+i)x^2-4(2-i)x-5-3i=0$$, where $$i=\sqrt{-1}$$, which has eater modulus is

A
$$(3-5i)/2$$

B
$$(5-3i)/2$$

C
$$(3+i)/2$$

D
$$(1+3i)/2$$

Solution
Question 4
1 / -0

For the expression $$ax^2 + 7x + 2$$ to be quadratic, the necessary condition is

A
$$a=0$$

B
$$a \neq 0$$

C
$$ a> \cfrac{7}{2}$$

D
$$ a<-1 $$

Solution

For the expression $$a^2+7x+2$$ to be quadratic the possible values of $$a$$ must be non zero real numbers because we know that the expression $$ax^2+bx+c$$ where $$a, b, c$$ are real numbers is quadratic if $$a\neq 0$$
Question 5
1 / -0

Which of the following is a quadratic polynomial in one variable?

A
$$\sqrt{2x^{3}}+5$$

B
$$2x^2 + 2x^{-2}$$

C
$$x^2$$

D
$$2x^2+y^2$$

Solution

Polynomials in one variable are algebraic expressions that consist of terms in the form $$ax^n$$

Here $$n$$ is a non-negative (i.e. positive or zero) integer and $$a$$ is a real number and is called the coefficient of the term.

The degree of a polynomial in one variable is the largest exponent in the polynomial.

And for quadratic polynomial $$n=2.$$

Hence the only quadratic polynomial in one variable is $$x^2$$
Question 6
1 / -0

The roots of the equation $$\sqrt{3y + 1} = \sqrt{y - 1}$$ are?

A
$$0, - 1$$

B
$$2, 3$$

C
$$2, 1$$

D
None of these

Solution

Given: equation $$\sqrt {3y+1}=\sqrt {y-1}$$
To find the roots of the equation
Sol:
$$\sqrt {3y+1}=\sqrt {y-1}$$
Take square on both sides, we get
$$ {3y+1}= {y-1}\\\implies 3y-y=-1-1\\\implies 2y=-2$$
or, $$y=-1$$
But $$y\ne-1$$ as $$\sqrt{y-1} = \sqrt{-2}$$ which is not possible. Hence, none of the given options is the answer.
Question 7
1 / -0

If the discriminant of $$3x^{2}-14x+k=0$$ is $$100$$, then $$k=$$

A
$$8$$

B
$$32$$

C
$$16$$

D
$$24$$

Solution

Given that, the discriminant of $$3x^2-14x+k=0$$ is $$100$$.
To find out: The value of $$k$$.

We know that, the discriminant of a quadratic equation of the form $$ax^2+bx+c=0$$ is given by,
$$D=b^{2}-4ac$$

Here, $$a=3, b=-14, c=k$$

So, $$D=(-14)^{2}-4(3)k$$

$$\Rightarrow (-14)^{2}-4(3)k=100$$

$$\Rightarrow 196-12k=100$$

$$\Rightarrow 12k=96$$

$$\Rightarrow k=8$$

Hence, the value of $$k$$ is $$8$$.
Question 8
1 / -0

The discriminant (D) of $$\sqrt{x^{2}+x+1}=2$$ is:

A
-3

B
13

C
11

D
12

Solution

Squaring on the both sides of the given equation, we get
$$x^{2}+x+1=4$$
$$x^{2}+x-3=0$$
$$D=b^{2}-4ac$$
$$=1-4(1)(-3)$$
$$=1+12$$
$$=13$$
Question 9
1 / -0

A quadratic equation in $$x$$ is $$ax^2 + bx + c = 0$$, where $$a, b, c$$ are real numbers and the other condition is

A
$$a \neq 0$$

B
$$b \neq 0$$

C
$$c \neq 0$$

D
$$b = 0$$

Solution

$$a, b$$ and $$c$$ are constants in the equation.
As $$x$$ is raised to power $$2$$ it is a quadratic equation.
If $$a=0$$, then it will nullify $${ x }^{ 2 }$$, making it a linear equation.
So only if $$a\neq 0$$, the equation remains quadratic.
Option A is correct.
Question 10
1 / -0

The discriminant of $$ax^{2}-(a+b)x+b=0$$ is:

A
$$a-b$$

B
$$a+b$$

C
$$(a+b)^{2}$$

D
$$(a-b)^{2}$$

Solution

(D) $$b^{2}-4ac=\left [ (a+b) \right ]^{2}-4ab$$
$$=(a+b)^{2}-4ab$$
$$= a^2+b^2-2ab$$
$$=(a-b)^{2}$$

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Quadratic Equations Test - 17

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Quadratic Equations Test - 17

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Question 1

$$1$$

$$2$$

$$3$$

None of these

Solution

Question 2

$$\displaystyle\frac{a}{p}$$

$$1$$

$$\left(\displaystyle\frac{a}{p}\right)^2$$

$$0$$

Solution

Question 3

$$(3-5i)/2$$

$$(5-3i)/2$$

$$(3+i)/2$$

$$(1+3i)/2$$

Solution

Question 4

$$a=0$$

$$a \neq 0$$

$$ a> \cfrac{7}{2}$$

$$ a<-1 $$

Solution

Question 5

$$\sqrt{2x^{3}}+5$$

$$2x^2 + 2x^{-2}$$

$$x^2$$

$$2x^2+y^2$$

Solution

Question 6

$$0, - 1$$

$$2, 3$$

$$2, 1$$

None of these

Solution

Question 7

$$8$$

$$32$$

$$16$$

$$24$$

Solution

Question 8

-3

13

11

12

Solution

Question 9

$$a \neq 0$$

$$b \neq 0$$

$$c \neq 0$$

$$b = 0$$

Solution

Question 10

$$a-b$$

$$a+b$$

$$(a+b)^{2}$$

$$(a-b)^{2}$$

Solution

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