Quadratic Equations Test

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Quadratic Equations Test - 34

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Question 1
1 / -0

What are the real factors of $$x^2 + 4$$?

A
$$(x^2+2)$$ and $$(x^2-2)$$

B
$$(x+2)$$ and $$(x-2)n$$

C
Does not exist

D
$$(x^2 + 2)$$ and $$(x + 2)$$

Solution

Real factors of the equation does not exist as it doesn't have real roots

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

$$\Rightarrow x=+2i$$ or $$- 2i$$
Question 2
1 / -0

Given that '$$x$$' is real then the solution set of the equation $$\sqrt { x-1 } +\sqrt { x+1 } =1$$.

A
No solution

B
Unique solution

C
$$2$$ solutions

D
None of these

Solution

$$\sqrt{x-1}+\sqrt{x+1}=1$$ As $$x > 1$$
$$\because x-1$$ can't be negative.
$$\Rightarrow \sqrt{x+1} > 1$$ $$0\sqrt{x-1}$$ is possitve
$$\therefore $$ there sum is $$ > 1$$ always . hence no solution.
Question 3
1 / -0

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

A
$$-3$$

B
$$13$$

C
$$11$$

D
$$12$$

Solution

Discrimant (D) is applied only to quadratic equations, but the equation $$ \sqrt{x^{2}+x+1} = 2 $$ is not an quadratic equation,
Hence, first convert the equation into quadratic by squaring it on both sides,

$$ \sqrt{x^{2}+x+1} = 2 $$

$$ \Rightarrow x^{2}+x+1 = 4 $$

$$ \Rightarrow x^{2}+x-3 = 0 $$

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

$$ D = 1-4\times 1\times (-3) $$

$$ \boxed{D = 13} $$
Question 4
1 / -0

If $$f(x)=x^2+4x+a$$ is a perfect square function then calculate the value of $$a$$.

A
$$3$$

B
$$4$$

C
$$2$$

D
$$6$$

Solution

Since f(x) is perfect square, then both the roots are equal.
$$\therefore$$ D=0
$$16-4a=0$$
$$a=4$$
Question 5
1 / -0

The roots of the quadratic equation $$(a+b-2c)x^2-(2a-b-c)x+(a-2b+c)=0$$ are-

A
$$(a+b+c)$$ and $$(a-b-c)$$

B
$$\dfrac 12$$ and $$a-2b+c$$

C
$$a-2b+c$$ and $$\dfrac 1{(a+b-2c)}$$

D
None of the above.

Solution

Clearly we see that x=1 satisfies the equation

$$\Rightarrow $$x=1 is a root of the that equation

Let $$\alpha $$ be other root

Product of roots =$$1×\alpha =\dfrac{a-2b+c}{a+b-2c}$$

other root $$=\alpha =\dfrac {a-2b+c}{a+b-2c}$$
Question 6
1 / -0

If $$\alpha,\beta$$ are roots of the equation $$2x^2-35x+2=0$$ then the value of $$(2\alpha-35)^3(2\beta-35)^3$$ is:

A
$$1$$

B
$$8$$

C
$$4$$

D
$$64$$

Solution

We will substitute $$\alpha, \beta $$ in the above equation

$$\Rightarrow 2\alpha - 35+\dfrac {2}{\alpha} =0$$

$$\Rightarrow 2\alpha - 35=\dfrac {-2}{\alpha} $$

Similarly for beta

$$\Rightarrow (2\alpha - 35)^{3}(2\beta - 35)^{3}=\dfrac {-8}{\alpha ^{3}}×\dfrac {-8}{\beta^{3}}=\dfrac {64}{1}=64$$
Question 7
1 / -0

Sum of the roots of the equation $$(x+3)^2-4|x+3|+3=0$$ is-

A
$$4$$

B
$$12$$

C
$$-12$$

D
$$-4$$

Solution

$${ (x+3) }^{ 2 }-4|x+3|+3=0\\ let\quad |x+3|=t\\ \therefore t\ge 0\\ { t }^{ 2 }-4t+3=0\\ (t-1)(t-3)=0\\ \therefore t=1\ or\ 3\\ when\quad t=1\\ |x+3|=1\\ x+3=1\quad or\quad x+3=-1\\ x=-2\quad or\quad x=-4\\ when\quad t=3\\ |x+3|=3\quad or\quad x+3=-3\\ x=0\quad or\quad x=-6\\ sum\quad of\quad all\quad roots\quad =\quad (-2)+(-4)+(0)+(-6)\\ =-12$$
Question 8
1 / -0

Solve the following quadratic equation by completing the square: $$x^2+6x-7=0$$

A
$${-7, 1}$$

B
$${-4, 1}$$

C
$${-7, 2}$$

D
None of these

Solution

$$\Rightarrow$$ The given quadratic equation is $$x^2+6x-7=0$$

$$\Rightarrow$$ $$x^2+6x=7$$

$$\Rightarrow$$ $$x^2+6x+9=7+9$$ [ Adding $$9$$ on both sides ]

$$\Rightarrow$$ $$(x+3)^2=16$$

$$\therefore$$ $$x+3=\pm 4$$ [ Taking square root on both sides ]

$$\Rightarrow$$ $$x+3=4$$ and $$x+3=-4$$

$$\therefore$$ $$x=1$$ and $$x=-7$$
Question 9
1 / -0

The minimum value of the expression $$4x^2+2x+1$$ is-

A
$$1$$

B
$$\dfrac{1}{3}$$

C
$$\dfrac{1}{2}$$

D
$$\dfrac{3}{4}$$

Solution

For a general quadratic equation $$ { ax }^{ 2 }+bx+c=0,\quad where\quad a>0\quad the\quad minimum\quad value\quad of\quad expression\quad occurs\quad at\quad \cfrac { -D }{ 4a } \\ \therefore { 4x }^{ 2 }+2x+1\\ minimum\quad value\quad =\quad \cfrac { -({ 2 }^{ 2 }-4.4.1) }{ 4.4 } \\ =\cfrac { 12 }{ 16 } =\cfrac { 3 }{ 4 } $$
Question 10
1 / -0

Solve the following quadratic equation by completing the square: $$x^2+3x+1=0$$

A
$$\left \{ \dfrac{-3+\sqrt{5}}{2}, \dfrac{-3-\sqrt{5}}{2} \right \}$$

B
$$\left \{ \dfrac{-3-\sqrt{5}}{2}, \dfrac{-3-\sqrt{5}}{2} \right \}$$

C
$$\left \{ \dfrac{-3+\sqrt{5}}{2}, \dfrac{-3+\sqrt{5}}{2} \right \}$$

D
None of these

Solution

$$x^2+3x+1=0$$
The 3 will have to be divided by $$2$$ then the result should be squared and the final result is $$\dfrac{9}{4}$$. This number will be added and subtracted in the equation on one side.
$$x^2+3x+\dfrac{9}{4}-\dfrac{9}{4}+1=0$$

$$\left(x^2+3x+\dfrac{9}{4}\right)-\dfrac{9}{4}+1=0$$

$$\left(x+\dfrac{3}{2}\right)^2-\dfrac{9}{4}+1=0$$

$$\left(x+\dfrac{3}{2}\right)^2-\dfrac{5}{4}=0$$

$$\left(x+\dfrac{3}{2}\right)^2=\dfrac{5}{4}$$

Now taking square root on both sides we get,
$$x+\dfrac{3}{2}=\pm\dfrac{\sqrt{5}}{2}$$

$$\therefore$$ $$x=-\dfrac{3}{2}\pm\dfrac{\sqrt{5}}{2}$$

$$\therefore$$ $$x=-\dfrac{3}{2}+\dfrac{\sqrt{5}}{2}$$ and  $$x=-\dfrac{3}{2}-\dfrac{\sqrt{5}}{2}$$

$$\Rightarrow$$  $$x=\left(\dfrac{-3+\sqrt{5}}{2},\dfrac{-3-\sqrt{5}}{2}\right)$$

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Quadratic Equations Test - 34

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Quadratic Equations Test - 34

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SHARING IS CARING

Question 1

$$(x^2+2)$$ and $$(x^2-2)$$

$$(x+2)$$ and $$(x-2)n$$

Does not exist

$$(x^2 + 2)$$ and $$(x + 2)$$

Solution

Question 2

No solution

Unique solution

$$2$$ solutions

None of these

Solution

Question 3

$$-3$$

$$13$$

$$11$$

$$12$$

Solution

Question 4

$$3$$

$$4$$

$$2$$

$$6$$

Solution

Question 5

$$(a+b+c)$$ and $$(a-b-c)$$

$$\dfrac 12$$ and $$a-2b+c$$

$$a-2b+c$$ and $$\dfrac 1{(a+b-2c)}$$

None of the above.

Solution

Question 6

$$1$$

$$8$$

$$4$$

$$64$$

Solution

Question 7

$$4$$

$$12$$

$$-12$$

$$-4$$

Solution

Question 8

$${-7, 1}$$

$${-4, 1}$$

$${-7, 2}$$

None of these

Solution

Question 9

$$1$$

$$\dfrac{1}{3}$$

$$\dfrac{1}{2}$$

$$\dfrac{3}{4}$$

Solution

Question 10

$$\left \{ \dfrac{-3+\sqrt{5}}{2}, \dfrac{-3-\sqrt{5}}{2} \right \}$$

$$\left \{ \dfrac{-3-\sqrt{5}}{2}, \dfrac{-3-\sqrt{5}}{2} \right \}$$

$$\left \{ \dfrac{-3+\sqrt{5}}{2}, \dfrac{-3+\sqrt{5}}{2} \right \}$$

None of these

Solution

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