Quadratic Equations Test

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Quadratic Equations Test - 22

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Weekly Quiz Competition

Question 1
1 / -0

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

A
$$a - b$$

B
$$a + b$$

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

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

Solution

$$ D = (b^{1})^{2}-4a^{1}c^{1} $$
$$ = (a+b)^{2}-4\times (a)\times (b) $$
$$ = a^{2}+b^{2}+2ab-4ab = a^{2}+b^{2}-2ab $$
$$ \Rightarrow \boxed{D = (a-b)^{2}} $$
Question 2
1 / -0

Which of the following is a quadratic equation?

A
$$x^{1/2} + 2x + 3 = 0$$

B
$$(x - 1)(x + 4) = x^{2} + 1$$

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

D
$$(2x + 1)(3x - 4) = 6x^{2} + 3$$

Solution

i) As highest Power is '1' $$ \therefore $$ not quadratic
ii) $$ (x-1)(x+4) = x^{2}+1\Rightarrow x^{2}+3x-4 = x^{2}+1\Rightarrow \boxed{3x-5 = 0} $$
$$ \therefore $$ not quadratic (higher degree 1)
iii) Highest delrle is '2' $$ \therefore $$ quadratic
iv) $$ (2x+1)(3x-4) = 6x^{2}+3\Rightarrow 6x^{2}-5x-4 = 6x^{2}+3 $$
$$ \Rightarrow 5x+7 = 0 $$ ($$ \therefore $$ not quadratic)
Question 3
1 / -0

If $$x^{2} + 10 x = 24,$$ where $$x>0$$, then the value of $$x + 5$$ is

A
$$7$$

B
$$5$$

C
$$-12$$

D
$$2$$

Solution

$$x^2+10x-24 = 0$$
$$x^2+12x-2x-24=0$$

On factoring we get,

$$(x – 2)(x + 12) = 0$$

$$x = 2, x = -12$$

Condition: $$x > 0$$, then the value of $$x + 5$$ is $$2 + 5 = 7$$.
Question 4
1 / -0

The roots of the quadratic equations
$$(x-1)^2 = \frac{9}{4}$$, are

A
$$x=-\frac{5}{3}, x=\frac{5}{3}$$

B
$$x=-\frac{1}{2}, x=\frac{5}{2}$$

C
$$x=\frac{5}{9}, x=\frac{13}{9}$$

D
$$x=1\pm \sqrt{\frac{2}{3}}$$

Solution

Given, $$(x-1)^2=\dfrac{9}{4}=\left(\dfrac{3}{2}\right)^2$$
$$\Rightarrow x-1=\pm \dfrac{3}{2}$$
$$\Rightarrow x=\pm \dfrac{3}{2}+1$$
$$\Rightarrow x=-\dfrac{1}{2}, \dfrac{5}{2}$$
Question 5
1 / -0

Solve the given equation by the method of Completion of Squares: $${x}^{2}+4x+4=0$$?

A
$$2$$

B
$$-2$$

C
$$4$$

D
$$-4$$

Solution

Given that:
$$p(x)=x^2+4x+4=0$$

Solution:
Zeroes of the polynomial $$p(x)=x^2+4x+4=0$$

$$\Rightarrow x^2 + 2 \times 2 \times x + (2)^2 = (x + 2)^2$$

or, $$(x+2)(x+2)=0$$

or, $$x=-2$$ or $$x=-2$$

Therefore, the correct answer is $$-2$$
Question 6
1 / -0

The discriminant ($$D$$) of the equation $$5x-6+\cfrac { 1 }{ x } =0$$ is ............

A
$$4$$

B
$$\sqrt { 56 } $$

C
$$16$$

D
$$56$$

Solution

Given: $$5x-6+\cfrac { 1 }{ x } =0$$

$$\Rightarrow 5{ x }^{ 2 }-6x+1=0$$

Comparing above equation with standard quadratic equation $$ax^2+bx+c=0$$, we get $$a=5,b=-6,c=1$$

Now, Discriminant $$(D) ={ b }^{ 2 }-4ac$$

$$=36-20$$

$$=16$$
Question 7
1 / -0

The roots of quadratic equation $$\dfrac{x}{k} = \dfrac{k}{x} $$ are ...........

A
$$k, -k$$

B
$$-k, -k$$

C
$$k, k$$

D
$$k^2, - k^2$$

Solution

Given: $$\dfrac{x}{k} = \dfrac{k}{x}$$

$$\Rightarrow x^2 = k^2$$

$$\Rightarrow x = \sqrt{k^2}$$

$$\Rightarrow x = \pm k$$

$$\Rightarrow x = k, -k$$

Hence, the roots are $$k,-k$$
Question 8
1 / -0

The discriminant of $$x^2 - 3x + k = 0$$ is $$1$$ then the value of $$k = .............$$

A
$$-2$$

B
$$4$$

C
$$-4$$

D
$$2$$

Solution

Comparing the equation $$x^2 - 3x + k = 0$$ with $$ax^2 + bx + c = 0$$, we get $$a = 1, b = - 3, c = k$$

Here, discriminant $$= 1$$

$$\therefore b^2 - 4ac = 1$$ ...... $$[\because D=b^2 - 4ac]$$

$$\therefore (-3)^2 - 4(1)(k) = 1$$

$$\therefore 9 - 4k = 1$$

$$\therefore -4k = - 8$$

$$\therefore k = 2$$
Question 9
1 / -0

__________ is true for the discriminant of a quadratic equation $$x^2+x+1=0$$.

A
$$D=0$$

B
$$D<0$$

C
$$D>0$$

D
$$D$$ is a perfect square

Solution

For $$x^2+x+1=0$$
$$a=b=c=1$$ { by comparing with the general equation $$ax^2+bx+c$$ }
$$D=b^{2}-4ac$$
$$D=1^{2}-4\times 1\times 1$$
$$D=1-4=-3$$
Therefore D is negative
Question 10
1 / -0

The value of the product $$\left (3 + \dfrac {5}{x}\right )$$ and $$\left (9 - \dfrac {15}{x} + \dfrac {25}{x^{2}}\right )$$ at $$x = 1$$ is _______.

A
$$150$$

B
$$148$$

C
$$152$$

D
$$140$$

Solution

The value of product $$(3+\dfrac{5}{x})$$ and $$\left (9-\dfrac{15}{x}+\dfrac{25}{x^2}\right)$$ at $$x=1$$ be $$f$$
$$\left (3+\dfrac{5}{x}\right)$$ at $$x=1$$ is equal to $$3+\dfrac{5}{1}=3+5=8$$
$$\left (9-\dfrac{15}{x}+\dfrac{25}{x^2}\right)$$ at $$x=1$$ is equal to $$9-\dfrac{15}{1}+\dfrac{25}{1}=9-15+25=19$$
$$y=8\times 19=152$$ at $$x=1$$
Therefore, required product is $$152$$.

Option (C) is correct.

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Quadratic Equations Test - 22

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Quadratic Equations Test - 22

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SHARING IS CARING

Question 1

$$a - b$$

$$a + b$$

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

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

Solution

Question 2

$$x^{1/2} + 2x + 3 = 0$$

$$(x - 1)(x + 4) = x^{2} + 1$$

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

$$(2x + 1)(3x - 4) = 6x^{2} + 3$$

Solution

Question 3

$$7$$

$$5$$

$$-12$$

$$2$$

Solution

Question 4

$$x=-\frac{5}{3}, x=\frac{5}{3}$$

$$x=-\frac{1}{2}, x=\frac{5}{2}$$

$$x=\frac{5}{9}, x=\frac{13}{9}$$

$$x=1\pm \sqrt{\frac{2}{3}}$$

Solution

Question 5

$$2$$

$$-2$$

$$4$$

$$-4$$

Solution

Question 6

$$4$$

$$\sqrt { 56 } $$

$$16$$

$$56$$

Solution

Question 7

$$k, -k$$

$$-k, -k$$

$$k, k$$

$$k^2, - k^2$$

Solution

Question 8

$$-2$$

$$4$$

$$-4$$

$$2$$

Solution

Question 9

$$D=0$$

$$D<0$$

$$D>0$$

$$D$$ is a perfect square

Solution

Question 10

$$150$$

$$148$$

$$152$$

$$140$$

Solution

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