# Algebra Questions For All Exams With Complete Solution

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In this article, I’ll discuss some of the most asked questions in the exams related to algebra. These types of algebra questions are very important to know.

If you miss these algebra questions, definitely you will regret for that at last.

Of course, the data or values of the questions can vary, but the type of the questions will be the same as what I’m going to discuss below.

Before telling these algebra questions, you must ensure that you know the basic formulas related to algebra otherwise at the time of solving you may face hassles.

If you are interested to solve trigonometry questions, check the article by clicking the provided link.

Firstly, I’ll discuss some of the important formulas which should be stored into your mind. After that I’ll discuss the most important types of the algebra questions.

Basic Algebraic formulas –

• ( a + b )² = a²+b²+2ab
• ( a – b )² = a²+b²-2ab
• ( a + b )² = ( a – b )²+4ab
• ( a – b )² = ( a + b )²-4ab
• ( a²-b² ) = ( a + b ) ( a – b )
• ( a + b )³ = a³+b³+3ab ( a + b )
• ( a – b )³ = a³-b³-3ab ( a – b )
• ( a³ – b³ ) = ( a – b ) ( a²+b²+ab )
• ( a³ + b³ ) = ( a + b ) ( a²+b²-ab )
• ( a + 1/a )² = a²+1/a²+2
• ( a – 1/a )² = a²+1/a²-2
• ( a – 1/a )² = ( a + 1/a )² – 4
• ( a + 1/a )² = ( a – 1/a )² + 4
• ( a – 1/a )³ = a³-1/a³-3(a-1/a)
• ( a + 1/a )³ = a³+1/a³+3(a+1/a)
• ( a + b + c ) ² = a²+b²+c²+2ab+2bc+2ca = a²+b²+c²+2( ab + bc + ca )
• a³+b³+c³-3abc = ( a + b + c ) ( a²+b²+c²-ab-bc-ca ) = 1/2×(a + b + c ) [ ( a – b )²+( b – c )²+( c-a )² ]

Now, we focus on the most important types of the algebra questions, which can definitely be asked in upcoming exams.

Let’s start the topic.

## Algebra Questions For All Exams With Complete Solution –

### Q1. If x = 7-4√3, then Find the value of x+\frac{1}{x}

Solution – 1/x = \frac{1}{7-4\sqrt{3}}

= \frac{1}{7-4\sqrt{3}}\times\frac{7+4\sqrt{3}}{7+4\sqrt{3}}

= \frac{7+4\sqrt{3}}{49-48}

1/x = 7+4√3

\therefore x+\frac{1}{x} = 7-4√3+7+4√3 = 14 ( Ans )

### Q2. If \sqrt{x}+\frac{1}{\sqrt{x}}=\sqrt{6}, then find the value of x^{2}+\frac{1}{x^{2}}.

Solution\sqrt{x}+\frac{1}{\sqrt{x}}=\sqrt{6}

squaring both the sides

\Rightarrow (\sqrt{x}+\frac{1}{\sqrt{x}})^{2}=(\sqrt{6})^{2}

\Rightarrow x+\frac{1}{x}+2=6

\Rightarrow x+\frac{1}{x}=4

again squaring both the sides

\Rightarrow x^{2}+\frac{1}{x^{2}}+2=16

\Rightarrow x^{2}+\frac{1}{x^{2}}=14 ( Ans )

### Q3. If 5^{5x+5} = 1, then find the value of x.

Solution 5^{5x+5} = 1

\Rightarrow 5^{5x+5} = 5^{0}

comparing both the sides

\Rightarrow 5x+5 = 0

\Rightarrow 5x = -5

\Rightarrow x = -1 ( Ans )

### Q4. If x = \frac{\sqrt{3}+1}{\sqrt{3}-1} & y = \frac{\sqrt{3}-1}{\sqrt{3}+1}, then find the value of x²+y².

Solution – x = \frac{\sqrt{3}+1}{\sqrt{3}-1} & y = \frac{\sqrt{3}-1}{\sqrt{3}+1} ( given )

x² = (\frac{\sqrt{3}+1}{\sqrt{3}-1})^{2}

\Rightarrow \frac{3+1+2\sqrt{3}}{3+1-2\sqrt{3}}

\Rightarrow \frac{4+2\sqrt{3}}{4-2\sqrt{3}}

y² = (\frac{\sqrt{3}-1}{\sqrt{3}+1})^{2}

\Rightarrow \frac{3+1-2\sqrt{3}}{3+1+2\sqrt{3}}

\Rightarrow \frac{4-2\sqrt{3}}{4+2\sqrt{3}}

now, x²+y²

\frac{4+2\sqrt{3}}{4-2\sqrt{3}}+\frac{4-2\sqrt{3}}{4+2\sqrt{3}}

= \frac{(4+2\sqrt{3})^{2}+(4-2\sqrt{3})^{2}}{(4+2\sqrt{3})(4-2\sqrt{3})}

= \frac{16+12+16√3+16+12-16√3}{16-12}

= \frac{16+12+16√3+16+12-16√3}{16-12}

= \frac{56}{4}

= 14 ( Ans )

### Q5. If x : y = 7 : 3, then find the value of (\frac{xy+y^{2}}{x^{2}-y^{2}}).

Solution (\frac{xy+y^{2}}{x^{2}-y^{2}})

= \frac{21+9}{49-9}

= \frac{30}{40}

= \frac{3}{4} ( Ans )

### Q6. If a = 7, b = 5 and c = 3, then find the value of a²+b²+c²-ab-bc-ca.

Solution

\because a²+b²+c²-ab-bc-ca = \frac{1}{2} [ ( a-b )²+( b-c )²+(c-a)² ] ( formula )

= \frac{1}{2} [ ( 7-5 )²+( 5-3 )²+(3-7)² ]

= \frac{1}{2} [ 4+4+16 ]

= \frac{1}{2} [ 24 ]

= 12 ( Ans )

### Q7. If \frac{x}{y} = \frac{3}{4}, find the value of \frac{6}{7}+\frac{y-x}{y+x}.

Solution \frac{6}{7}+\frac{y-x}{y+x}

= \frac{6}{7}+\frac{y(1-\frac{x}{y})}{y(1+\frac{x}{y})}

= \frac{6}{7}+\frac{(1-\frac{3}{4})}{(1+\frac{3}{4})}

= \frac{6}{7}+\frac{4-3}{4}×\frac{4}{4+3}

= \frac{6}{7}+ \frac{1}{7}

= \frac{7}{7}

= 1 ( Ans )

### Q8. If for non-zero x, x²-4x-1 = 0, the value of x^{2}+\frac{1}{x^{2}} ? –

Solution x^{2}+\frac{1}{x^{2}} = ?

\because x²-4x-1 = 0 ( given )

\Rightarrow x²-1 = 4x

divide by x both the sides

\Rightarrow x-\frac{1}{x} = 4

squaring both the sides

\Rightarrow x^{2}+\frac{1}{x^{2}}-2 = 16

\Rightarrow x^{2}+\frac{1}{x^{2}} = 18

### Q9. (x+\frac{1}{x})(x-\frac{1}{x})(x^{2}+\frac{1}{x^{2}}-1)(x^{2}+\frac{1}{x^{2}}+1) is equal to ? –

Solution(x+\frac{1}{x})(x-\frac{1}{x})(x^{2}+\frac{1}{x^{2}}-1)(x^{2}+\frac{1}{x^{2}}+1)

(x+\frac{1}{x})(x^{2}+\frac{1}{x^{2}}-1)(x-\frac{1}{x})(x^{2}+\frac{1}{x^{2}}+1)

\because ( a³ – b³ ) = ( a – b ) ( a²+b²+ab ) and ( a³ + b³ ) = ( a + b ) ( a²+b²-ab )

so (x^{3}+\frac{1}{x^{3}})(x^{3}-\frac{1}{x^{3}})

= (x^{6}-\frac{1}{x^{6}}) ( Ans )

### Q10. If p = 999, then find the value of \sqrt[3]{p(p^{2}+3p+3)+1}.

Solution\sqrt[3]{p(p^{2}+3p+3)+1}

= \sqrt[3]{p^{3}+3p^{2}+3p+1}

= \sqrt[3]{(p+1)^{3}}

put p = 999

= \sqrt[3]{(999+1)^{3}}

= \sqrt[3]{(1000)^{3}}

= 1000 ( Ans )

### Q11. If (\frac{3}{5})^{3}(\frac{3}{5})^{-6} = (\frac{3}{5})^{2x-1}, find the value of x.

Solution – (\frac{3}{5})^{3}(\frac{3}{5})^{-6} = (\frac{3}{5})^{2x-1}

\Rightarrow (\frac{3}{5})^{3-6} = (\frac{3}{5})^{2x-1}

\Rightarrow -3 = 2x-1

\Rightarrow x = -1 ( Ans )

### Q12. If 4b^{2}+\frac{1}{b^{2}} = 2, find the value of 8b^{3}+\frac{1}{b^{3}}.

Solution4b^{2}+\frac{1}{b^{2}} = 2 ( given )

add 4 & subtract 4 in L.H.S

4b^{2}+\frac{1}{b^{2}} + 4 – 4 = 2

(2b)^{2}+(\frac{1}{b})^{2} + 4 – 4 = 2

(2b+\frac{1}{b})^{2} – 4 = 2

(2b+\frac{1}{b})^{2} = 6

(2b+\frac{1}{b}) = √6

cube both the sides

(2b+\frac{1}{b})^{3} = (√6)³

8b^{3}+\frac{1}{b^{3}} + 3×2b×\frac{1}{b}(2b+\frac{1}{b})= 6√6

8b^{3}+\frac{1}{b^{3}} + 6(2b+\frac{1}{b})= 6√6

since (2b+\frac{1}{b}) = √6, so

8b^{3}+\frac{1}{b^{3}} + 6√6 = 6√6

8b^{3}+\frac{1}{b^{3}} = 0

### Q13. If x = (\sqrt{2}+1)^{1/3}, find the value of x^{3}-\frac{1}{x^{3}}.

Solution – x = (\sqrt{2}+1)^{1/3}

cube both the sides

x^{3} = (\sqrt{2}+1)

\frac{1}{x^{3}} = \frac{1}{\sqrt{2}+1} × \frac{\sqrt{2}-1}{\sqrt{2}-1} = \frac{\sqrt{2}-1}{1} = √2-1

now, x³-\frac{1}{x^{3}} = √2+1-√2+1 = 2 ( Ans )

### Q14. If 4x²-6x+1 = 0, then find the value of 8x³+( 8x )^{-1}.

Solution – 4x²-6x+1 = 0 ( given )

\Rightarrow 2x ( 2x-3+\frac{1}{2x} ) = 0

\Rightarrow 2x+\frac{1}{2x} = 3

cube both the sides

\Rightarrow 8x³+\frac{1}{8x³} = 27-3×3 = 18 ( Ans )

### Q15. If x = 2-p, then find the value of x³+6xp+p³.

Solution – x = 2-p ( given )

\Rightarrow x+p = 2

cube both the sides

( x+p )³ = ( 2 )³

\Rightarrow x³+p³+3xp(x+p) = 8

since x+p = 2, so x³+p³+3xp×2 = 8

\Rightarrow x³+p³+6xp = 8 ( Ans )

### Q16. If a³+b³ = 5824 and a+b = 28, find the value of ( a-b )²+ab.

Solution – we know that a³+b³ = ( a+b ) ( a²+b²-ab )

now, 5824 = 28 ( a²+b²-ab )

\Rightarrow ( a²+b²-ab ) = 5824/28

\Rightarrow ( a²+b²-ab ) = 208

add ab & subtract ab in L.H.S

\Rightarrow ( a²+b²-ab+ab-ab ) = 208

\Rightarrow ( a²+b²-2ab+ab ) = 208

\Rightarrow ( a-b )²+ab = 208 ( Ans )

### Q17. 5^{x}=30^{-y}=6^{z} what is the value of \frac{xy+yz+zx}{xyz} ?

Solution – Let 5^{x}=30^{-y}=6^{z} = k

now, 5=k^{1/x}, 30=k^{-1/y}, 6=k^{1/z}

30=6×5

k^{-1/y} = k^{1/z}×k^{1/x}

k^{\frac{1}{x}+\frac{1}{z}} =k^{-1/y}

\frac{1}{x}+\frac{1}{z} = \frac{-1}{y}

\frac{1}{x}+\frac{1}{z}+\frac{1}{y} = 0

\frac{xy+yz+zx}{xyz} = 0 ( Ans )

### Questions for practice –

( A ) If √x-\frac{1}{√x} = √5, find the value of x²+\frac{1}{x²}Ans is 47

( B ) If a + b + c =8 & ab + bc + ca = 20, then find the value of a³+b³+c³-3abc – Ans is 32

( C ) If x+\frac{1}{16x} = 3, find the value of 16x³+\frac{1}{256x³}Ans is 423

( D ) If √x-\frac{1}{√x} = 2√2, find the value of x²+\frac{1}{x²}Ans is 98

( E ) If \frac{3(x²+1)-7x}{3x} = 6 & x\neq0, then find the value of √x+\frac{1}{√x}Ans is √(\frac{31}{3})

( F ) If 3x+4y-11 = 18 & 8x-6y+12= 6, then find the value of 5x-3y-9 – Ans is -9

( G ) If a=1+√3 & b=1-√3, then find the value of a²+b² – Ans is 8

Eventually, I hope that you’ve got an important article about algebra questions. In this article, I covered all types of algebra questions.

Moreover, I have provided a list of unsolved questions related to algebra. For more practice, you can definitely solve them.

If you really liked the article, please share it to those who really need it.

Thank you.

## Vijay Gupta

Hello everyone, My name is Vijay Gupta and I belong to a very small town that is situated in district Hardoi, which is in Uttar Pradesh. 1. Education – I’ve completed my primary education from a private school that is situated in my hometown and upper primary, matric and higher secondary education have been completed from a government college. Well, I was an average student till class 5th, but I accelerated my preference towards studies from class six. Consequently, I passed out many classes with good positions. Even I passed out 12th with good marks ( 405/500 ) and topped my college. Due to getting good marks, I got a cheque of 500 rupees and was rewarded by the Principal of my college. After completing my 12th, I prepared twice for IIT ( Indian Institute of Technology ) from Aakash institute, but unfortunately, I failed to get selected into the best IIT colleges. But during the preparation, I was being graduated from CSJMU Kanpur. I completed my graduation in 2016 and now I’m pursuing an educational degree ( B.Ed. ). 2. Profession – Although I love teaching, but I also do blogging. Both are my favorite jobs.