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}}.
Solution – 4b^{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.
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