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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Question 1

164\displaystyle \frac{1}{64}641​

1576\displaystyle \frac{1}{576}5761​

1144\displaystyle \frac{1}{144}1441​

12764\displaystyle \frac{127}{64}64127​

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 p=qp=q ; r≠0r\neq 0r=0

2q=pr 2q=\sqrt{pr}2q=pr​

pr=q2 pr=q^{2}pr=q2

pr=q {p}{r}={q}pr=q

Solution

Question 4

b2−4ac>0b^{2}-4ac> 0b2−4ac>0

b2−4ac<0b^{2}-4ac< 0b2−4ac<0

b2−4ac=0b^{2}-4ac= 0b2−4ac=0

b2−ac<0b^{2}-ac< 0b2−ac<0

Solution

Question 5

aaa can be any real number

a>1a>1a>1

a>3a>3a>3

a=3a=3a=3

Solution

Question 6

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

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

x(2x+3)=x2+1x(2x+3)=x^{2}+1x(2x+3)=x2+1

(x+2)3=x3−4(x+2)^{3}=x^{3}-4(x+2)3=x3−4

Question 7

one root.

two roots.

many roots.

no roots.

Solution

Question 8

2(x+1)2=4x2+2x+12(x + 1)^2 = 4x^2 + 2x + 12(x+1)2=4x2+2x+1

2x+x2=2x2+52x + x^2 = 2x^2 + 52x+x2=2x2+5

(2×3x)2+x2=3x2+5x( \sqrt{2} \times \sqrt{3}x)^2+ x^2 = 3x^2 + 5x(2​×3​x)2+x2=3x2+5x

(x2+2x)2=x5+3+4x3(x^2 + 2x)^2 = x^5 + 3 + 4x^3(x2+2x)2=x5+3+4x3

Solution

Question 9

4,84, 84,8

0,40, 40,4

4,−84,-84,−8

0,80,80,8

Solution

Question 10

242424

36\sqrt[6]{3}63​

3−6\sqrt[-6]{3}−63​

−12- 12−12

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

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

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

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

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

$2q=\sqrt{pr}$

$pr=q^{2}$

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

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

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

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

$b^{2}-ac< 0$

$a$ can be any real number

$a>1$

$a>3$

$a=3$

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

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

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

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

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

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

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

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

$4, 8$

$0, 4$

$4,-8$

$0,8$

$24$

$\sqrt[6]{3}$

$\sqrt[-6]{3}$

$- 12$