Algebra Expressions MCQ Questions

Solution: 3x – $\frac{1}{4\mathrm{y}}$ = 6

⇒ 4x – $\frac{1}{3\mathrm{y}}$ = 8 (multiplying 4/3 both side)

Solution: $\mathrm{x}-\frac{1}{\mathrm{x}}$ = 5

⇒ $\mathrm{x}^2+\frac{1}{\mathrm{x}^2}\,\,-2.\mathrm{x}.\frac{1}{\mathrm{x}}=25$ (squaring both side)

⇒ $\mathrm{x}^2+\frac{1}{\mathrm{x}^2}\,\,=\,\,27$

Using Trick ⇒$\mathrm{x}^2+\frac{1}{\mathrm{x}^2}\,\,=\,\,5^2+2=27$

Short Trick: If $\mathrm{x}+\frac{1}{\mathrm{x}}=\mathrm{n}$ then $\mathrm{x}^2+\frac{1}{\mathrm{x}^2}$ = n² – 2 and If $\mathrm{x}-\frac{1}{\mathrm{x}}=\mathrm{n}$ then $\mathrm{x}^2+\frac{1}{\mathrm{x}^2}$ = n² + 2

Solution: $\mathrm{x}+\frac{1}{\mathrm{x}}$ = 7

⇒ $\mathrm{x}^2+\frac{1}{\mathrm{x}^2}\,\,+2.\mathrm{x}.\frac{1}{\mathrm{x}}=49$ (squaring both side)

⇒ $\mathrm{x}^2+\frac{1}{\mathrm{x}^2}\,\,=\,\,47$

Using Trick ⇒$\mathrm{x}^2+\frac{1}{\mathrm{x}^2}\,\,=\,\,7^2-2=47$

Short Trick: If $\mathrm{x}+\frac{1}{\mathrm{x}}=\mathrm{n}$ then $\mathrm{x}^2+\frac{1}{\mathrm{x}^2}$ = n² – 2 and If $\mathrm{x}-\frac{1}{\mathrm{x}}=\mathrm{n}$ then $\mathrm{x}^2+\frac{1}{\mathrm{x}^2}$ = n² + 2

Solution: $\mathrm{x}+\frac{1}{\mathrm{x}}\,\,=2$

⇒ x² +1 = 2x

⇒ x² – 2x + 1 = 0

⇒ (x – 1)² = 0

⇒ x = 1

∴ $\mathrm{x}^{2020}+\frac{1}{\mathrm{x}^{2020}}\,\,=\,\,1^2+\frac{1}{1^2}=2$

Short Tricks: If   $\mathrm{x}+\frac{1}{\mathrm{x}}\,\,=2$ then the value of$\mathrm{x}^{\mathrm{n}}+\frac{1}{\mathrm{x}^{\mathrm{n}}}\,\,=\,\,2$, where n = integer

Solution: $\mathrm{x}+\frac{1}{\mathrm{x}}$ = 5

⇒ $\mathrm{x}^2+\frac{1}{\mathrm{x}^2}\,\,+2.\mathrm{x}.\frac{1}{\mathrm{x}}=25$ (squaring both side)

⇒ $\mathrm{x}^2+\frac{1}{\mathrm{x}^2}\,\,=\,\,23$

⇒ $\mathrm{x}^4+\frac{1}{\mathrm{x}^4}\,\,+2.\mathrm{x}^2.\frac{1}{\mathrm{x}^2}=529$ (squaring both side)

⇒ $\mathrm{x}^4+\frac{1}{\mathrm{x}^4}\,\,=\,\,527$

Using Trick ⇒$\mathrm{x}^4+\frac{1}{\mathrm{x}^4}\,\,=\,\,\left( 5^2-2 \right) ^2-2=527$

Short Trick: If $\mathrm{x}+\frac{1}{\mathrm{x}}=\mathrm{n}$ then $\mathrm{x}^4+\frac{1}{\mathrm{x}^4}$ = (n² – 2)² – 2 and

If $\mathrm{x}-\frac{1}{\mathrm{x}}=\mathrm{n}$ then $\mathrm{x}^4+\frac{1}{\mathrm{x}^4}$ = (n² + 2)² – 2

Solution: $\mathrm{x}+\frac{1}{\mathrm{x}}$ = 9

⇒ $\mathrm{x}^2+\frac{1}{\mathrm{x}^2}\,\,-2.\mathrm{x}.\frac{1}{\mathrm{x}}=9$ (squaring both side)

⇒ $\mathrm{x}^2+\frac{1}{\mathrm{x}^2}\,\,=\,\,11$

⇒ $\mathrm{x}^4+\frac{1}{\mathrm{x}^4}\,\,+2.\mathrm{x}^2.\frac{1}{\mathrm{x}^2}=121$ (squaring both side)

⇒ $\mathrm{x}^4+\frac{1}{\mathrm{x}^4}\,\,=\,\,119$

Using Trick ⇒$\mathrm{x}^4+\frac{1}{\mathrm{x}^4}\,\,=\,\,\left( 3^2+2 \right) ^2-2=119$

Short Trick: If $\mathrm{x}+\frac{1}{\mathrm{x}}=\mathrm{n}$ then $\mathrm{x}^4+\frac{1}{\mathrm{x}^4}$ = (n² – 2)² – 2 and

If $\mathrm{x}-\frac{1}{\mathrm{x}}=\mathrm{n}$ then $\mathrm{x}^4+\frac{1}{\mathrm{x}^4}$ = (n² + 2)² – 2

Solution: $\mathrm{x}+\frac{1}{\mathrm{x}}$ = 3

⇒ $\mathrm{x}^3+\frac{1}{\mathrm{x}^3}\,\,+3.\mathrm{x}.\frac{1}{\mathrm{x}}\left( \mathrm{x}+\frac{1}{\mathrm{x}} \right) =27$ (cubing both side)

⇒ $\mathrm{x}^3+\frac{1}{\mathrm{x}^3}\,\,=\,\,27 -9=18$

Using Trick ⇒$\mathrm{x}^3+\frac{1}{\mathrm{x}^3}\,\,=\,\,3^3-3\times 3=18$

Short Trick: If $\mathrm{x}+\frac{1}{\mathrm{x}}=\mathrm{n}$ then $\mathrm{x}^3+\frac{1}{\mathrm{x}^3}$ = n³ – 3.n and If $\mathrm{x}-\frac{1}{\mathrm{x}}=\mathrm{n}$ then $\mathrm{x}^3-\frac{1}{\mathrm{x}^3}$ = n³ + 3.n

Solution: $\mathrm{x}+\frac{1}{\mathrm{x}}$ = 5

⇒ $\mathrm{x}^3-\frac{1}{\mathrm{x}^3}\,\,-3.\mathrm{x}.\frac{1}{\mathrm{x}}\left( \mathrm{x}-\frac{1}{\mathrm{x}} \right) =125$ (cubing both side)

⇒ $\mathrm{x}^3-\frac{1}{\mathrm{x}^3}\,\,=\,\,125 +15=140$

Using Trick ⇒$\mathrm{x}^3-\frac{1}{\mathrm{x}^3}\,\,=\,\,5^3+3\times 5=140$

Short Trick: If $\mathrm{x}+\frac{1}{\mathrm{x}}=\mathrm{n}$ then $\mathrm{x}^3+\frac{1}{\mathrm{x}^3}$ = n³ – 3.n and If $\mathrm{x}-\frac{1}{\mathrm{x}}=\mathrm{n}$ then $\mathrm{x}^3-\frac{1}{\mathrm{x}^3}$ = n³ + 3.n

Solution: $\mathrm{x}+\frac{1}{\mathrm{x}}$ = 3

⇒ $\mathrm{x}^3+\frac{1}{\mathrm{x}^3}\,\,+3.\mathrm{x}.\frac{1}{\mathrm{x}}\left( \mathrm{x}+\frac{1}{\mathrm{x}} \right) =27$ (cubing both side)

⇒ $\mathrm{x}^3+\frac{1}{\mathrm{x}^3}\,\,=\,\,27 -9=18$

⇒ $\mathrm{x}^6+\frac{1}{\mathrm{x}^6}\,\,+2.\mathrm{x}^3.\frac{1}{\mathrm{x}^3}=\,\,324$(squaring both side)

⇒ $\mathrm{x}^6+\frac{1}{\mathrm{x}^6}\,\,=\,\,324-2=322$

Using Trick ⇒$\mathrm{x}^6+\frac{1}{\mathrm{x}^6}\,\,=\,\,\left( 3^3-3\times 3 \right) ^2-2=322$

Short Trick: If $\mathrm{x}+\frac{1}{\mathrm{x}}=\mathrm{n}$ then $\mathrm{x}^6+\frac{1}{\mathrm{x}^6}$ = (n³ – 3.n)² – 2 and

If $\mathrm{x}-\frac{1}{\mathrm{x}}=\mathrm{n}$ then $\mathrm{x}^6+\frac{1}{\mathrm{x}^6}$ = (n³ + 3.n)² + 2

Solution: $\mathrm{x}-\frac{1}{\mathrm{x}}$ = 2

⇒ $\mathrm{x}^3-\frac{1}{\mathrm{x}^3}\,\,-3.\mathrm{x}.\frac{1}{\mathrm{x}}\left( \mathrm{x}-\frac{1}{\mathrm{x}} \right) =8$ (cubing both side)

⇒ $\mathrm{x}^3-\frac{1}{\mathrm{x}^3}\,\,=\,\,8 +6=14$

⇒ $\mathrm{x}^6+\frac{1}{\mathrm{x}^6}\,\,-2.\mathrm{x}^3.\frac{1}{\mathrm{x}^3}=\,\,196$ (squaring both side)

⇒ $\mathrm{x}^6+\frac{1}{\mathrm{x}^6}\,\,=\,\,196+2=198$

Using Trick ⇒$\mathrm{x}^6+\frac{1}{\mathrm{x}^6}\,\,=\,\,\left( 2^3+3\times 2 \right) ^2+2=198$

Short Trick: If $\mathrm{x}+\frac{1}{\mathrm{x}}=\mathrm{n}$ then $\mathrm{x}^6+\frac{1}{\mathrm{x}^6}$ = (n³ – 3.n)² – 2 and

If $\mathrm{x}-\frac{1}{\mathrm{x}}=\mathrm{n}$ then $\mathrm{x}^6+\frac{1}{\mathrm{x}^6}$ = (n³ + 3.n)² + 2