Download App

Maths 3 Marks Question

Operations on Algebraic Expressions · Jharkhand Board (JAC) STD 8 (Jharkhand - English Medium). 21 questions.

Questions

3 Marks Question

Take a timed test

21 questions · self-marked practice — reveal the answer and mark yourself.

Question 13 Marks
If $\Big(\text{x}+\frac{1}{\text{x}}\Big)=4$ Find the values of: $\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)$
Answer
$\Big(\text{x}+\frac{1}{\text{x}}\Big)=4$ Squaring on both sides: $\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=(4)^2$ $\Rightarrow​​\text{x}^2+\frac{1}{\text{x}^2}+2\times\text{x}+\frac{1}{\text{x}}=16$ $\Rightarrow\text{x}^2+\frac{1}{\text{x}^2}+2=16$ $\Rightarrow\text{x}^2+\frac{1}{\text{x}^2}=16-2=14$ $\therefore\text{x}^2+\frac{1}{\text{x}^2}=14$
View full question & answer→
Question 23 Marks
Write the quotient and remainder when we divide: $\left(2 x^3-5 x^2+8 x-5\right)$ by $\left(2 x^2-3 x+5\right)$
View full question & answer→
Question 43 Marks
Find the following products: $\Big(\text{x}^4+\frac{1}{\text{x}^4})×\Big(\text{x}+\frac{1}{\text{x}}\Big)$
Answer
$\Big(\text{x}^4+\frac{1}{\text{x}^4})×\Big(\text{x}+\frac{1}{\text{x}}\Big)$$=\text{x}^4\Big(\text{x}+\frac{1}{\text{x}}\Big)+\frac{1}{\text{x}^4}\Big(\text{x}+\frac{1}{\text{x}}\Big)$
$=\text{x}^4\times\text{x}+\frac{1}{\text{x}}+\frac{1}{\text{x}^4}\times\text{x}+\frac{1}{\text{x}^4}\times\frac{1}{\text{x}}$
$=\text{x}^5+\text{x}^3+\frac{1}{\text{x}^3}+\frac{1}{\text{x}^5}$
View full question & answer→
Question 53 Marks
Write the quotient and remainder when we divide: $\left(x^3-6 x^2+11 x-6\right)$ by $\left(x^2-5 x+6\right)$
View full question & answer→
Question 63 Marks
Write the quotient and remainder when we divide: $\left(x^4-2 x^3+2 x^2+x+4\right)$ by $\left(x^2+x+1\right)$
View full question & answer→
Question 73 Marks
If $x+y=12$ and $x y=14$, find the value of $\left(x^2+y^2\right)$.
Answer
$x+y=12$
On squaring both the sides:
$(x+y)^2=(12)^2$
$\Rightarrow x^2+y^2+2 x y=144$
$\Rightarrow x^2+y^2+2 \times 14=144$
$\Rightarrow x^2+y^2+28=144$
$\Rightarrow x^2+y^2=144-28$
$x^2+y^2=116$
View full question & answer→
Question 83 Marks
Find following products: $\Big(\frac{1}{3}\text{x}^2-9\Big)\Big(\frac{1}{3}\text{x}^2-9\Big)$
Answer
$\Big(\frac{1}{3}\text{x}^2-9\Big)\Big(\frac{1}{3}\text{x}^2-9\Big)$ $=\Big(\frac{1}{3}\text{x}^2-9\Big)^2$ $=\Big(\frac{1}{3}\text{x}^2\Big)^2-2\times\frac{1}{3}\text{x}^2\times9+(9)^2 $ $=\frac{1}{9}\text{x}^4-6\text{x}^2+81$
View full question & answer→
Question 93 Marks
If $x - y = 7$ and $xy = 9,$ find the value of $\left(x^2+y^2\right)$.
Answer
$x+y=7$
Squaring both sides:
$(x-y)^2=(7)^2$
$\Rightarrow x^2+y^2-2 xy=49$
$\Rightarrow x^2+y^2-2 \times 9=49$
$\Rightarrow x^2+y^2-18=49$
$\Rightarrow x^2+y^2=49+18=67$
$x^2+y^2=67$
View full question & answer→
Question 103 Marks
Write the quotient and remainder when we divide: $\left(5 x^3-12 x^2+12 x+13\right)$ by $\left(x^2-3 x+4\right)$
View full question & answer→
Question 113 Marks
If $\Big(\text{x}+\frac{1}{\text{x}}\Big)=4$ Find the values of: $\Big(\text{x}^4+\frac{1}{\text{x}^4}\Big)$
Answer
$\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)^2=(14)^2$ Squaring on both sides: $\Rightarrow(\text{x}^2)^2+\Big(\frac{1}{\text{x}^2}\Big)^2+2\times\text{x}^2\times\frac{1}{\text{x}^2}=196$ $\Rightarrow\text{x}^4+\frac{1}{\text{x}^4}+2=196$ $\Rightarrow\text{x}^4+\frac{1}{\text{x}^4}=196-2=194$ $\therefore\text{x}^4+\frac{1}{\text{x}^4}=194$
View full question & answer→
Question 123 Marks
Write the quotient and remainder when we divide: $\left(8 x^4+10 x^3-5 x^2-4 x+1\right)$ by $\left(2 x^2+x-1\right)$
View full question & answer→
Question 143 Marks
If $\Big(\text{x}-\frac{1}{\text{x}}\Big)=5$ Find the values of: $\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)$
Answer
$\Big(\text{x}-\frac{1}{\text{x}}\Big)=5$ Squaring on both sides: $\Big(\text{x}-\frac{1}{\text{x}}\Big)^2=(5)^2$ $\Rightarrow\Big(\text{x}^2+\frac{1}{\text{x}^2}-2(\text{x})\times\Big(\frac{1}{\text{x}}\Big)\Big)=25$ $\Rightarrow\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)-2=25$ $\Rightarrow\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)=25+2=27$ $\therefore\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)=27$ Therefore, the value of $\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)\text{ is }27$
View full question & answer→
Question 153 Marks
Find following products: $\Big(\frac{3}{4}\text{x}-\frac{5}{6}\text{y}\Big)\Big(\frac{3}{4}\text{x}-\frac{5}{6}\text{y}\Big)$
Answer
$\Big(\frac{3}{4}\text{x}-\frac{5}{6}\text{y}\Big)\Big(\frac{3}{4}\text{x}-\frac{5}{6}\text{y}\Big)$ $=\Big(\frac{3}{4}\text{x}-\frac{5}{6}\text{y}\Big)^2​​​​​​​$ $=\Big(\frac{3}{4}\text{x}\Big)^2-2×\frac{3}{4}\text{x}×\frac{5}{6}\text{y}+\Big(\frac{5}{6}\text{y}\Big)^2 $ $=\frac{9}{16}\text{x}^2-\frac{5}{4}\text{xy}+\frac{25}{36}\text{y}^2$
View full question & answer→
Question 163 Marks
If $\Big(\text{x}-\frac{1}{\text{x}}\Big)=5$ Find the values of: $\Big(\text{x}^4+\frac{1}{\text{x}^4}\Big)$
Answer
$\Big(\text{x}^2+\frac{1}{\text{x}^2}=27\Big)$Squaring on both sides:
$\Rightarrow\Big(\text{x}^4-\frac{1}{\text{x}^4}-2(\text{x}^2\Big(\frac{1}{\text{x}^2}\Big)\Big)=(27)^2$ $\Rightarrow\Big(\text{x}^4+\frac{1}{\text{x}^4}\Big)-2=729$ $\Rightarrow\Big(\text{x}^4+\frac{1}{\text{x}^4}\Big)=729+2$ $\Rightarrow\Big(\text{x}^4+\frac{1}{\text{x}^4}\Big)=731$ Therefore, the value of $\Big(\text{x}^4+\frac{1}{\text{x}^4}\Big)\text { is }731.$
View full question & answer→
Question 183 Marks
Find following products: $\Big(\text{x}-\frac{3}{\text{x}}\Big)\Big(\text{x}-\frac{3}{\text{x}}\Big)$
Answer
$\Big(\text{x}-\frac{3}{\text{x}}\Big)\Big(\text{x}-\frac{3}{\text{x}}\Big)$ $=\Big(\text{x}-\frac{3}{\text{x}}\Big)^2$ $=(\text{x})^2-2\times\text{x}\times\frac{3}{\text{x}}+\Big(\frac{3}{\text{x}}\Big)^2 $ $=\text{x}^2-6+\frac{9}{\text{x}^2}$
View full question & answer→
Question 193 Marks
Find following products: $\Big(\frac{1}{2}\text{y}^2-\frac{1}{3}\text{y}\Big)\Big(\frac{1}{2}\text{y}^2-\frac{1}{3}\text{y}\Big)$
Answer
$\Big(\frac{1}{2}\text{y}^2-\frac{1}{3}\text{y}\Big)\Big(\frac{1}{2}\text{y}^2-\frac{1}{3}\text{y}\Big)$ $=\Big(\frac{1}{2}\text{y}^2-\frac{1}{3}\text{y}\Big)^2$ $=\Big(\frac{1}{2}\text{y}^2\Big)2-2\times\frac{1}{2}\text{y}^2\times\frac{1}{3}\text{y}+\Big(\frac{1}{3}\text{y}\Big)^2$ $=\frac{1}{4}\text{y}^4-\frac{1}{3}\text{y}^3+\frac{1}{9}\text{y}^2$
View full question & answer→
Generate Paper FreeAll Operations on Algebraic Expressions topics