Quadratic Equations Test

Result

Quadratic Equations Test - 34

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

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

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

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Quadratic Equations Test - 34

Result

Quadratic Equations Test - 34

Score

-

Rank

-

SHARING IS CARING

Question 1

(x2+2)(x^2+2)(x2+2) and (x2−2)(x^2-2)(x2−2)

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

Does not exist

(x2+2)(x^2 + 2)(x2+2) and (x+2)(x + 2)(x+2)

Solution

Question 2

No solution

Unique solution

222 solutions

None of these

Solution

Question 3

−3-3−3

131313

111111

121212

Solution

Question 4

333

444

222

666

Solution

Question 5

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

12\dfrac 1221​ and a−2b+ca-2b+ca−2b+c

a−2b+ca-2b+ca−2b+c and 1(a+b−2c)\dfrac 1{(a+b-2c)}(a+b−2c)1​

None of the above.

Solution

Question 6

111

888

444

646464

Solution

Question 7

444

121212

−12-12−12

−4-4−4

Solution

Question 8

−7,1{-7, 1}−7,1

−4,1{-4, 1}−4,1

−7,2{-7, 2}−7,2

None of these

Solution

Question 9

111

13\dfrac{1}{3}31​

12\dfrac{1}{2}21​

34\dfrac{3}{4}43​

Solution

Question 10

{−3+52,−3−52}\left \{ \dfrac{-3+\sqrt{5}}{2}, \dfrac{-3-\sqrt{5}}{2} \right \}{2−3+5​​,2−3−5​​}

{−3−52,−3−52}\left \{ \dfrac{-3-\sqrt{5}}{2}, \dfrac{-3-\sqrt{5}}{2} \right \}{2−3−5​​,2−3−5​​}

{−3+52,−3+52}\left \{ \dfrac{-3+\sqrt{5}}{2}, \dfrac{-3+\sqrt{5}}{2} \right \}{2−3+5​​,2−3+5​​}

None of these

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.

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

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

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

$2$ solutions

$-3$

$13$

$11$

$12$

$3$

$4$

$2$

$6$

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

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

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

$1$

$8$

$4$

$64$

$4$

$12$

$-12$

$-4$

${-7, 1}$

${-4, 1}$

${-7, 2}$

$1$

$\dfrac{1}{3}$

$\dfrac{1}{2}$

$\dfrac{3}{4}$

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

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

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