Quadratic Equations Test

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Quadratic Equations Test - 32

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Weekly Quiz Competition

Question 1
1 / -0

Find the discriminant of $$\displaystyle pqr{ x }^{ 2 }-8pqx+pr=0$$

A
$$\displaystyle { p }^{ 2 }{ q }^{ 2 }-4{ p }^{ 2 }{ qr }^{ 2 }$$

B
$$\displaystyle 64{ p }^{ 2 }{ q }^{ 2 }-{ p }^{ 2 }{ r }^{ 2 }$$

C
$$\displaystyle 64{ p }^{ 2 }{ q }^{ 2 }-4{ p }^{ 2 }{ qr }^{ 2 }$$

D
$$\displaystyle 64{ p }^{ 2 }{ q }^{ 2 }-4{ p }^{ 2 }{ r }^{ 2 }$$

Solution

$$\displaystyle D={ b }^{ 2 }-4ac$$
Here, $$\displaystyle b=-8pq,a=pqr,c=pr$$
$$\displaystyle D={ \left( -8pq \right) }^{ 2 }-4\left( pq \right) \left( pr \right) $$
$$\displaystyle =64{ p }^{ 2 }{ q }^{ 2 }-4{ p }^{ 2 }{ qr }^{ 2 }$$
Question 2
1 / -0

Which of the following is correct if one root of quadratic equation is real?

A
$$\displaystyle D\le 0$$

B
$$\displaystyle D=0$$

C
$$\displaystyle D\ge 0$$

D
None

Solution

$$\displaystyle D\ge 0\quad \because $$ both are real
$$\displaystyle D=0\quad \because $$ both are equal
$$\displaystyle D\le 0\quad \because $$ both are imaginary
Question 3
1 / -0

Which of the following is correct, if the roots of quadratic equation are equal?

A
$$\displaystyle D>0$$

B
$$\displaystyle D<0$$

C
$$\displaystyle D=0$$

D
None

Solution

If both roots of a quadratic equation are equal and real, then
$$\displaystyle D=0$$, $$\displaystyle \propto =\beta $$
Question 4
1 / -0

Find the roots of equation by completing the square method:
$$\displaystyle 3{ x }^{ 2 }-12qx+12{ q }^{ 2 }=0$$

A
$$2q, 2q$$

B
$$-2q, -2q$$

C
$$-2q, 2q$$

D
None

Solution

$$\displaystyle 3{ x }^{ 2 }-12qx+12{ q }^{ 2 }=0$$
$$\displaystyle { x }^{ 2 }-4qx+4{ q }^{ 2 }=0$$
$$\displaystyle { x }^{ 2 }-2*2qx+4{ q }^{ 2 }=0$$
$$\displaystyle \left( x-2q \right) ^2 =0$$
$$\displaystyle \left( x-2q \right) \left( x-2q \right) =0$$
$$\displaystyle x=2q\quad or\quad x=2q$$
Question 5
1 / -0

The solution of which of the following equations is same as that of $$40-6x=x^2$$?

A
$$y=(x-6)^2-40$$

B
$$y=(x-6)^2+40$$

C
$$y=(x+3)^2-49$$

D
$$y=(x+3)^2+49$$

Solution

Given, $$40-6x=x^2$$
$$\Rightarrow x^{2}+6x-40=0$$
$$\Rightarrow (x+3)^{2}-9-40=0$$
$$\Rightarrow (x+3)^{2}-49=0$$
Hence, the correct option is C.
Question 6
1 / -0

Value(s) of $$x$$ which satisfies the equation $$x^{2} + 2kx = \dfrac {j}{3}$$, where $$j$$ and $$k$$ are constants, is/are

A
$$x = -k\pm \dfrac {\sqrt {3(3k^{2} + j)}}{3}$$

B
$$x = -6k\pm \dfrac {\sqrt {3(3k^{2} + j)}}{3}$$

C
$$x = -k\pm \dfrac {\sqrt {3(3k^{2} + j)}}{6}$$

D
$$x = -6k\pm \left (k + \dfrac {\sqrt {3j}}{3}\right )$$

Solution

We know the quadratic equation formula,
$$\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$$
$$x^2+2kx=\dfrac{j}{3}$$
On cross multiplying, we get
$$3x^2+6kx-j=0$$
Here $$a = 3, b = 6k, c = -j$$
On substituting the values, we get
$$=$$ $$\dfrac{-6k\pm\sqrt{(6k)^2-4\times 3\times -j}}{2\times 3}$$
$$=$$ $$\dfrac{-6k\pm\sqrt{36k^2+12 j}}{6}$$
$$=$$ $$\dfrac{-6k\pm\sqrt{12(3k^2+j)}}{6}$$
$$=$$ $$\dfrac{-6k\pm4\sqrt{3(3k^2+j)}}{6}$$
Take $$2$$ as common, we get
$$=$$ $$-3k\pm\dfrac{\sqrt{3(3k^2+j)}}{3}$$
Therefore, $$x =-k\pm \dfrac{\sqrt{3(3k^2+j)}}{3}$$
Question 7
1 / -0

Among the following, find the values of $$x$$ that are real and satisfy the equation $$2x^2-5x-2=0$$

A
$$1$$ and $$4$$

B
$$-\dfrac{5}{4} + \dfrac{\sqrt{41}}{4}$$ and $$-\dfrac{5}{4}-\dfrac{\sqrt{41}}{4}$$

C
$$\dfrac{5}{4}+\dfrac{\sqrt{41}}{4}$$ and $$\dfrac{5}{4}-\dfrac{\sqrt{41}}{4}$$

D
No real solutions

Solution

For finding the values of $$x$$ which satisfy the above equation, we solve the equation for $$x$$.
$$2x^2 - 5x - 2 = 0$$
$$\Rightarrow x = \cfrac{5 \pm \sqrt{25 + 16}}{4}$$
$$\Rightarrow x = \cfrac{5 \pm \sqrt{41}}{4}$$
$$\Rightarrow x = \cfrac{5 + \sqrt{41}}{4}, \cfrac{5 - \sqrt{41}}{4}$$
Question 8
1 / -0

Solve: $$p^2-12p+32=0$$

A
$$p=-8, 4$$

B
$$p=8, 4$$

C
$$p=8, -4$$

D
$$p=-8, -4$$

Solution
Question 9
1 / -0

If a, b and c are real, then both the roots of the equation $$\left( x-b \right) \left( x-c \right) +\left( x-c \right) \left( x-a \right) +\left( x-a \right) \left( x-b \right) =0$$ are always

A
Positive

B
Negative

C
Real

D
Imaginary

Solution

Given equation is $$(x-b)(x-c)+(x-c)(x-a)+(x-a)(x-b)=0$$
We can further simplify this equation by multiplying terms and taking common powers of $$x$$.
$$(x^2-bx-cx+bc)+(x^2-cx-ax+ac)+(x^2-ax-bx+ab)=0$$
$$\Rightarrow 3x^2-2ax-2bx-2cx+ab+bc+ca$$
$$\Rightarrow 3{ x }^{ 2 }-2x\left( a+b+c \right) +ab+bc+ca=0$$
We know that, discriminant of a quadratic equation of the form $$ax^2+bx+c=0$$ is given by:
$$D=b^2-4ac$$

$$\therefore \ D=[-2(a+b+c)]^2-4\times 3\times (ab+bc+ca)$$
$$\Rightarrow D=4{ \left( a+b+c \right) }^{ 2 }-4\times 3\left( ab+bc+ca \right) $$
$$\Rightarrow 4\left( { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }-ab-bc-ca \right) $$
$$\Rightarrow 2\left[ { \left( a-b \right) }^{ 2 }+{ \left( b-c \right) }^{ 2 }+{ \left( c-a \right) }^{ 2 } \right] $$
Given that, $$a,\ b\text { and }c$$ are real.
Hence, $$D\ge 0$$
Therefore, roots are real.

Hence, option C is correct.
Question 10
1 / -0

Let $$a$$ be the solution of the equation $$4x^2-12x+9=16$$, then the value of $$10a-15$$ is

A
20

B
25

C
30

D
35

Solution

$$4x^2 - 12x + 9 = 16$$
$$\Rightarrow 4x^2 - 12x - 7 = 0$$
$$\Rightarrow x = \cfrac{12 \pm \sqrt{144 + 112}}{8}$$
$$\Rightarrow x = \cfrac{12 \pm 16}{8}$$
$$\Rightarrow x = -0.5, 3.5$$
Substituting the value of a in $$10a - 15$$, we get $$35 - 15 = 20$$ and $$5 - 15 = -10$$

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Quadratic Equations Test - 32

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Quadratic Equations Test - 32

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Question 1

$$\displaystyle { p }^{ 2 }{ q }^{ 2 }-4{ p }^{ 2 }{ qr }^{ 2 }$$

$$\displaystyle 64{ p }^{ 2 }{ q }^{ 2 }-{ p }^{ 2 }{ r }^{ 2 }$$

$$\displaystyle 64{ p }^{ 2 }{ q }^{ 2 }-4{ p }^{ 2 }{ qr }^{ 2 }$$

$$\displaystyle 64{ p }^{ 2 }{ q }^{ 2 }-4{ p }^{ 2 }{ r }^{ 2 }$$

Solution

Question 2

$$\displaystyle D\le 0$$

$$\displaystyle D=0$$

$$\displaystyle D\ge 0$$

None

Solution

Question 3

$$\displaystyle D>0$$

$$\displaystyle D<0$$

$$\displaystyle D=0$$

None

Solution

Question 4

$$2q, 2q$$

$$-2q, -2q$$

$$-2q, 2q$$

None

Solution

Question 5

$$y=(x-6)^2-40$$

$$y=(x-6)^2+40$$

$$y=(x+3)^2-49$$

$$y=(x+3)^2+49$$

Solution

Question 6

$$x = -k\pm \dfrac {\sqrt {3(3k^{2} + j)}}{3}$$

$$x = -6k\pm \dfrac {\sqrt {3(3k^{2} + j)}}{3}$$

$$x = -k\pm \dfrac {\sqrt {3(3k^{2} + j)}}{6}$$

$$x = -6k\pm \left (k + \dfrac {\sqrt {3j}}{3}\right )$$

Solution

Question 7

$$1$$ and $$4$$

$$-\dfrac{5}{4} + \dfrac{\sqrt{41}}{4}$$ and $$-\dfrac{5}{4}-\dfrac{\sqrt{41}}{4}$$

$$\dfrac{5}{4}+\dfrac{\sqrt{41}}{4}$$ and $$\dfrac{5}{4}-\dfrac{\sqrt{41}}{4}$$

No real solutions

Solution

Question 8

$$p=-8, 4$$

$$p=8, 4$$

$$p=8, -4$$

$$p=-8, -4$$

Solution

Question 9

Positive

Negative

Real

Imaginary

Solution

Question 10

20

25

30

35

Solution

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