Quadratic Equations Test

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Quadratic Equations Test - 24

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Weekly Quiz Competition

Question 1
1 / -0

If the roots of the equation $$ax^2+bx+c=0$$ are all real equal then which one of the following is true?

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

B
$$b^2-4ac\ge0$$

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

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

Solution

Given, the roots of a quadratic equation $$ax^2+bx+c=0$$ are all real and equal.
Then the discriminant of this equation is $$= 0$$.
The discriminant of this equation is $$b^2-4ac$$.
So $$b^2-4ac= 0$$.
Question 2
1 / -0

If the roots of a quadratic equation are equal, than discriminate is

A
$$1$$

B
$$0$$

C
greater than $$0$$

D
less than $$0$$

Solution

We know that to find the roots we have

$$\dfrac{-b\pm \sqrt{b^{2}-4ac}}{2a}$$

If roots are equal $${b^{2}-4ac=0}$$ $$(Discriminate )$$
Question 3
1 / -0

Which is a quadratic equation?

A
$$x+\dfrac{1}{x}=2$$

B
$$x^{2}+1=(x+3)^{2}$$

C
$$x(x+2)$$

D
$$x+\dfrac{1}{x}$$
Question 4
1 / -0

The number of integral values of $$'x'$$ for which $$x^{2}+19x+92$$ is a perfect square is:

A
$$0$$

B
$$1$$

C
$$2$$

D
$$3$$

Solution

The above quadratic equation can be expressed as : $$x^2 + 19x + 92 = \left(x + \dfrac{19}{2} \right)^2 + 92 - \dfrac{361}{4}$$
$$= \left(x + \dfrac{19}{2} \right)^2 + \dfrac{7}{4}$$

From the above expression, we can infer that no integer value of x will make it a perfect square because we have a fraction in the expression.
Question 5
1 / -0

The solution of the equation $$x+\cfrac{1}{x}=2$$ will be

A
$$2,-1$$

B
$$0,-1,-\cfrac{1}{5}$$

C
$$-1,-\cfrac{1}{5}$$

D
None of these

Solution

$$x+\dfrac{1}{x}=2$$
$$x^{2}-2x+1=0$$
$$(x-1)^{2}=0$$
$$x=1$$
Question 6
1 / -0

If $$81$$ is discriminant of $$2{ x }^{ 2 }+5x-k=0$$ then the value of $$k$$ is

A
$$5$$

B
$$7$$

C
$$-7$$

D
$$2$$

Solution

the given equation $$2x^2+5x-k=0$$ is in the form of $$ax^2+bx+c$$

By comparing $$ a=2, b=5, c=-k$$

Given, Descriminant $$D=81$$ $$(D=b^2-4ac)$$

By equating,
$$\Rightarrow b^2-4ac=81$$

$$\Rightarrow 25+8k=81$$

$$\Rightarrow 8k=81-25$$

$$\Rightarrow 8k=56$$

$$\Rightarrow k=7$$
Question 7
1 / -0

Which of the following is non-quadratic polynomial.

A
$$x ^ { 2 } + 2$$

B
$$x ^ { 2 } + 3 x$$

C
$$4 x ^ { 4 } + 3 y$$

D
$$x ^ { 2 }$$

Solution

$$x^2+2, x^2+3x$$ and $$x^2$$ are all quadratic polynomials. But $$4x^4+3y$$ is not a quadratic polynomial because this polynomial has a degree $$4$$.
Question 8
1 / -0

Which of the following must be added to $$x^2-6x+5$$ to make it a perfect square ?

A
$$3$$

B
$$4$$

C
$$5$$

D
$$6$$

Solution

We've,
$$x^2-6x+5$$
$$=x^2-2.x.3+3^2-4$$
$$=(x-3)^2-4$$.
So to make it perfect square i.e. $$(x-3)^2$$, we are to add $$4$$ to add.
Question 9
1 / -0

The discriminant of the quadratic equation $$ ax^2+bx+c= 0$$ is

A
$$\Delta = b^2 + 4ac$$

B
$$\Delta = b^2 - 4ac$$

C
$$\Delta =4abc$$

D
$$\Delta = b^2 \times 4ac$$

Solution

The discriminant of the quadratic equation $$ax^2 + bx + c = 0$$ is $$\Delta = b^2 - 4ac$$
Question 10
1 / -0

Which of the following is a quadratic equation?

A
$$x^3 + 2x + 1 = x^2 + 3$$

B
$$-2x^2=(5-x)\left ( 2x-\cfrac{2}{5}\right )$$

C
$$(k+1)x^2+\cfrac{3}{2}x=7,$$where $$k=-1$$

D
$$x^3 x^2 = (x)^3$$

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Re-Attempt Weekly Quiz Competition

Quadratic Equations Test - 24

Result

Quadratic Equations Test - 24

Score

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SHARING IS CARING

Question 1

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

$$b^2-4ac\ge0$$

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

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

Solution

Question 2

$$1$$

$$0$$

greater than $$0$$

less than $$0$$

Solution

Question 3

$$x+\dfrac{1}{x}=2$$

$$x^{2}+1=(x+3)^{2}$$

$$x(x+2)$$

$$x+\dfrac{1}{x}$$

Question 4

$$0$$

$$1$$

$$2$$

$$3$$

Solution

Question 5

$$2,-1$$

$$0,-1,-\cfrac{1}{5}$$

$$-1,-\cfrac{1}{5}$$

None of these

Solution

Question 6

$$5$$

$$7$$

$$-7$$

$$2$$

Solution

Question 7

$$x ^ { 2 } + 2$$

$$x ^ { 2 } + 3 x$$

$$4 x ^ { 4 } + 3 y$$

$$x ^ { 2 }$$

Solution

Question 8

$$3$$

$$4$$

$$5$$

$$6$$

Solution

Question 9

$$\Delta = b^2 + 4ac$$

$$\Delta = b^2 - 4ac$$

$$\Delta =4abc$$

$$\Delta = b^2 \times 4ac$$

Solution

Question 10

$$x^3 + 2x + 1 = x^2 + 3$$

$$-2x^2=(5-x)\left ( 2x-\cfrac{2}{5}\right )$$

$$(k+1)x^2+\cfrac{3}{2}x=7,$$where $$k=-1$$

$$x^3 x^2 = (x)^3$$

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