## Are you ready to take your first step toward achieving your goal?

- 10 diagnosis quiz questions
- 10 minutes in duration
- Required to attempt all questions
- A mixture of short answer and multiple-choice questions
- Instant feedback straight after the test

## OK, let’s get started now!

#### Quiz Summary

0 of 10 Questions completed

Questions:

#### Information

You have already completed the quiz before. Hence you can not start it again.

Quiz is loading…

You must sign in or sign up to start the quiz.

You must first complete the following:

#### Results

#### Results

0 of 10 Questions answered correctly

Your time:

Time has elapsed

You have reached 0 of 0 point(s), (0)

Earned Point(s): 0 of 0, (0)

0 Essay(s) Pending (Possible Point(s): 0)

#### Categories

- Not categorized 0%

### The result is not promising.

### You will need to push yourself a bit further.

### You are almost getting there.

### Stay focused to obtain better results!

### Excellent!

### We are so impressed with your result!

### Fantastic Result!

### We hope you’ll keep continued this serial of good marks.

- 1
- 2
- 3
- 4
- 5
- 6
- 7
- 8
- 9
- 10

- Current
- Review
- Answered
- Correct
- Incorrect

- Question 1 of 10
##### 1. Question

Find \( x \) and \( y \) if \({12^3} = {2^x} \times {3^y}\).

\( x = \) , \( y = \)

CorrectIncorrect##### Hint

\({\left( {{a^x}} \right)^y} = {a^{xy}}\)

- Question 2 of 10
##### 2. Question

Write \(\dfrac{1}{{{3^{2 – x}}}}\) in non-fractional form.

CorrectIncorrect##### Hint

\({a^{ – 1}} = \dfrac{1}{a}\)

- Question 3 of 10
##### 3. Question

Find \(a \), \(b \) and \(c \) if $\dfrac{{\sqrt {3x{y^5}} }}{{\sqrt {2{x^6}{y^2}} }} \times \sqrt {\dfrac{{6{x^7}{y^3}}}{{2{x^5}y}}} = \dfrac{{a{y^b}\sqrt y }}{{x\sqrt {cx} }}$.

\(a = \) , \(b = \) , \(c = \)

CorrectIncorrect##### Hint

$\dfrac{{\sqrt a }}{{\sqrt b }} = \sqrt {\dfrac{a}{b}} $

- Question 4 of 10
##### 4. Question

Find \(a \) and \(b \) , if $\dfrac{1}{{\sqrt 8 – 2}} + \dfrac{1}{{2\sqrt 8 – 2}} = \dfrac{{a + b\sqrt 2 }}{{14}}$.

\(a = \) , \(b = \)

CorrectIncorrect##### Hint

\(\dfrac{1}{{\sqrt a + \sqrt b }} = \dfrac{1}{{\sqrt a + \sqrt b }} \times \dfrac{{\sqrt a – \sqrt b }}{{\sqrt a – \sqrt b }} = \dfrac{{\sqrt a – \sqrt b }}{{a – b}}\)

- Question 5 of 10
##### 5. Question

Solve $\sqrt {x + 1} + \sqrt {2x + 3} = 5$.

CorrectIncorrect - Question 6 of 10
##### 6. Question

Solve \(\left( {2x + 1} \right)\left( {3x + 2} \right) = 0\).

CorrectIncorrect - Question 7 of 10
##### 7. Question

State whether \(y = 4{x^2} + 2x – 5\) is satisfied by \(\left( {2, 12} \right)\).

CorrectIncorrect - Question 8 of 10
##### 8. Question

Find any values of \(x \) for \(y = {x^2} + 2x – 5\) if \(y = – 6\).

\(x = \)

CorrectIncorrect - Question 9 of 10
##### 9. Question

Find \(A \), if \(\dfrac{1}{{\left( {x + 2} \right)\left( {2x – 5} \right)}} \div \dfrac{{x – 9}}{{2x – 5}} = \dfrac{A}{{\left( {x + 2} \right)\left( {x – 9} \right)}}\).

\(A = \)

CorrectIncorrect - Question 10 of 10
##### 10. Question

Find all values of \(k\) for which \(2{x^2} + \left( {k – 2} \right)x + 2 = 0\) has two real roots.

CorrectIncorrect##### Hint

\(\begin{align} \displaystyle

{b^2} – 4ac &\gt 0 \text{ two real solutions }\\

{b^2} – 4ac &= 0 \text{ one real solution }\\

{b^2} – 4ac &\lt 0 \text{ no real solutions }

\end{align}\)