Quadratic Equations Test

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Quadratic Equations Test - 25

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Weekly Quiz Competition

Question 1
1 / -0

Which constant must be added and subtracted to solve the quadratic equation $$\displaystyle 9x^2 +\frac{3}{4}x+ 2 = 0$$ by the method of completing the square?

A
$$\displaystyle \frac{1}{64}$$

B
$$\displaystyle \frac{1}{576}$$

C
$$\displaystyle \frac{1}{144}$$

D
$$\displaystyle \frac{127}{64}$$

Solution

$$9x^2 + \dfrac 34x +2 = 0$$
Dividing by $$9$$ on both sides we get,
$$x^2 + \dfrac{x}{12} + \dfrac 29 = 0$$

$$x^2 + \dfrac{x}{12} = -\dfrac 29$$

Third term $$= \left(\dfrac 12 \times\text{coefficient of }x\right)^2$$

$$= \left(\dfrac 12 \times\dfrac {1}{12}\right)^2$$ = $$\left(\dfrac {1}{24}\right)^2 =\dfrac{1}{576}$$

Adding $$\dfrac {1}{576}$$ on both sides we get,

$$ \therefore x^2 + \dfrac{x}{12} + \dfrac {1}{576}$$ = $$\dfrac{1}{576} - \dfrac 29$$

$$\left(x + \dfrac{1}{24}\right)^2 =$$ $$\dfrac{1}{576} - \dfrac 29$$

Hence, option $$B$$.
Question 2
1 / -0

Assertion (A): The equation $$x-\displaystyle \cfrac{2}{x-1}=1-\cfrac{2}{x-1}$$ has no root.
Reason (R): $$x-1\neq 0$$, then only above equation is defined.

A
Both A and R are true and R is the correct explanation of A.

B
Both A and R are true and R is not the correct explanation of A.

C
A is true but R is false.

D
A is false but R is true

Solution

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

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

$$x=1$$
However at $$x=1$$ the expression $$\dfrac{2}{x-1}$$ is not defined.
Hence the above equation has no real roots.
Question 3
1 / -0

lf the roots of $${p}{x}^{2}+2{q}{x}+{r}=0$$ and $$qx^{2}-2\sqrt{pr}x+q=0$$ are simultaneously real, then

A
$$ p=q$$ ; $$r\neq 0$$

B
$$ 2q=\sqrt{pr}$$

C
$$ pr=q^{2}$$

D
$$ {p}{r}={q}$$

Solution

Since roots are real,
$$\Rightarrow (2q)^{2}-4pr\geq 0$$ and $$(2\sqrt{pr})^{2}-4q^{2}\geq 0$$
$$4q^{2}\geq 4pr$$ and $$4pr\geq 4q^{2}$$
To hold above two equations simultenously
$$\therefore q^{2}=pr$$
Question 4
1 / -0

If $$a > 0$$, then the expression $$ax^{2}+bx+c$$ is positive for all values of $$x$$ provided

A
$$b^{2}-4ac> 0$$

B
$$b^{2}-4ac< 0$$

C
$$b^{2}-4ac= 0$$

D
$$b^{2}-ac< 0$$

Solution

Consider the equation,
$$ax^{2}+bx+c$$
Now if $$b^{2}-4ac<0$$, then it does not have any real roots and hence it will not intersect the $$x$$ axis at any point.
If $$a>0$$ then $$y=ax^{2}+bx+c$$ will lie completely above $$x$$ axis and
If $$a<0$$ then $$y=ax^{2}+bx+c$$ will lie completely below $$x$$ axis.
It is given that the above equation is always greater than zero,
Or
$$ax^{2}+bx+c>0$$ for $$\epsilon R$$.
This can only occur is
$$b^{2}-4ac<0$$ and $$a>0$$.
Question 5
1 / -0

If the expression $$(a-2)x^{2}+2(2a-3)x+(5a-6)$$ is positive for all real values of $$x$$, then

A
$$a$$ can be any real number

B
$$a>1$$

C
$$a>3$$

D
$$a=3$$

Solution

The expression is always positive that means the discriminant of the equation is always less than zero .
So, $$4(2a-3)^2-4(a-2)(5a-6)<0$$
$$(16a^2+36-48a)-(20a^2-64a+48)<0$$
$$(-4a^2+16a-12)<0$$
$$(-4)(a^2-4a+3)<0$$
$$(a-3)(a-1)>0$$
So, the expression will be positive for $$a>3$$ and $$a<1$$
Option C is correct .
Question 6
1 / -0

Which of the following is not a quadratic equation :

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

B
$$x(x+1)+8=(x+2)(x-2)$$

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

D
$$(x+2)^{3}=x^{3}-4$$
Question 7
1 / -0

If $$a=0$$, then the equation $$\displaystyle \frac{x-a-1}{x-a}=a +1-\displaystyle \frac{1}{x-a}$$ has

A
one root.

B
two roots.

C
many roots.

D
no roots.

Solution

$$\dfrac{x-a-1}{x-a}=a+1-\dfrac{1}{x-a}$$

$$1-\dfrac{1}{x-a}=a+1-\dfrac{1}{x-a}$$
Since, given $$a=0$$, equation holds for all $$x$$ except $$x=a$$
$$\therefore $$ The equation has many roots.
Question 8
1 / -0

Which of the following is not a quadratic equation?

A
$$2(x + 1)^2 = 4x^2 + 2x + 1$$

B
$$2x + x^2 = 2x^2 + 5$$

C
$$( \sqrt{2} \times \sqrt{3}x)^2+ x^2 = 3x^2 + 5x$$

D
$$(x^2 + 2x)^2 = x^5 + 3 + 4x^3$$

Solution

A. $$2(x+1)^{2}=4x^{2}+2x+1$$
$$2(x^{2}+2x+1)=4x^{2}+2x+1$$
$$2x^{2}+4x+2=4x^{2}+2x+1$$
$$2x^{2}+4x+2-4x^{2}-2x-1=0$$
$$-2x^{2}+2x+1=0$$
is a quadratic equation
B. $$2x+x^2=2x^2+5$$
$$-x^2+2x=5$$
is a quadratic equation
C. $$[\sqrt{2}x\sqrt{3}]^2+x^2=3x^2+5x$$
$$(6x^2)+x^2-3x^2-5x=0$$
$$4x^2-5x=0$$
is a quadratic equation
D. $$(x^2+2x)^2=x^5+4x^3+3$$
As degree of $$x$$ is $$5$$, it is not quadratic.
Question 9
1 / -0

Values of $$k$$ for which the quadratic equation $$2x^2+ kx + k = 0$$ has equal roots.

A
$$4, 8$$

B
$$0, 4$$

C
$$4,-8$$

D
$$0,8$$

Solution

For equal roots, discriminant D of equation $$2x^2+kx+k=0$$, should be zero.
$$D =k^2-4(2)(k)=0$$
$$k^2-8k=0$$
$$k(k-8)=0$$
$$k=0, 8$$
Question 10
1 / -0

If $$14$$ is the maximum of $$-\lambda x^{2} + \lambda x + 8$$, then the value of $$\lambda$$ is

A
$$24$$

B
$$\sqrt[6]{3}$$

C
$$\sqrt[-6]{3}$$

D
$$- 12$$

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Quadratic Equations Test - 25

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Quadratic Equations Test - 25

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SHARING IS CARING

Question 1

$$\displaystyle \frac{1}{64}$$

$$\displaystyle \frac{1}{576}$$

$$\displaystyle \frac{1}{144}$$

$$\displaystyle \frac{127}{64}$$

Solution

Question 2

Both A and R are true and R is the correct explanation of A.

Both A and R are true and R is not the correct explanation of A.

A is true but R is false.

A is false but R is true

Solution

Question 3

$$ p=q$$ ; $$r\neq 0$$

$$ 2q=\sqrt{pr}$$

$$ pr=q^{2}$$

$$ {p}{r}={q}$$

Solution

Question 4

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

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

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

$$b^{2}-ac< 0$$

Solution

Question 5

$$a$$ can be any real number

$$a>1$$

$$a>3$$

$$a=3$$

Solution

Question 6

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

$$x(x+1)+8=(x+2)(x-2)$$

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

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

Question 7

one root.

two roots.

many roots.

no roots.

Solution

Question 8

$$2(x + 1)^2 = 4x^2 + 2x + 1$$

$$2x + x^2 = 2x^2 + 5$$

$$( \sqrt{2} \times \sqrt{3}x)^2+ x^2 = 3x^2 + 5x$$

$$(x^2 + 2x)^2 = x^5 + 3 + 4x^3$$

Solution

Question 9

$$4, 8$$

$$0, 4$$

$$4,-8$$

$$0,8$$

Solution

Question 10

$$24$$

$$\sqrt[6]{3}$$

$$\sqrt[-6]{3}$$

$$- 12$$

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