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This is one of our many quizzes on Variable expressions. Click on this to access more such competitive quizzes on Algebra.

Variable expressions Quiz 2

mathematics

 Solve for $x$, if ${{x}^{2}}-\sqrt{3}x+2=0$.

  Apply quadratic formula $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ and further solve.

$ -{{x}^{2}}-\sqrt{3}x+2=0 \\ \text{On the comparing with }a{{x}^{2}}+bx+c=0 \\ a=-1,b=-\sqrt{3},c=2 \\ x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} \\ \Rightarrow x=\dfrac{\sqrt{3}\pm \sqrt{11}}{-2} \\ \Rightarrow x=\dfrac{\sqrt{3}+\sqrt{11}}{-2}\text{ and }x=\dfrac{\sqrt{3}-\sqrt{11}}{-2} $

 Solve for $x$, if  $3x+7=2\left( x+5 \right)$.

 It is a linear equation so we will transfer all $x$ left hand side and other constants will transfer right hand side.

 $ 3x+7=2\left( x+5 \right) \\ \Rightarrow 3x+7=2x+10 \\ \Rightarrow 3x-2x=10-7 \\ \Rightarrow x=3 \\ \text{Hence 3 is correct}\text{.} $

 Find the volume of a cube whose side is $2\left( x-1 \right)$.

 We know that if the side of a cube is a then the volume will be ${{\text{a}}^{3}}$.

 $\text{We know that if side of a cube is a then the volume will be }{{\text{a}}^{3}} \\ \text{So,} \\ \Rightarrow volume={{\left( 2\left( x-1 \right) \right)}^{3}} \\ \Rightarrow volume=8{{\left( x-1 \right)}^{3}} $ 

 Simplify the given expression $\dfrac{2x+4}{{{x}^{2}}+5x+6}$ .

 Firstly try to take common from all the things, and if possible then try to factorise the quadratic equation so that we can take common them.

 $  \dfrac{2x+4}{{{x}^{2}}+5x+6} \\ \Rightarrow \dfrac{2\left( x+2 \right)}{{{x}^{2}}+2x+3x+6} \\ \Rightarrow \dfrac{2\left( x+2 \right)}{x\left( x+2 \right)+3\left( x+2 \right)} \\ \Rightarrow \dfrac{2\left( x+2 \right)}{\left( x+2 \right)\left( x+3 \right)} \\ \Rightarrow \dfrac{2}{x+3} $

 If  $\dfrac{{{x}^{3}}-8}{x-2}=2$ , then find the value of  $\dfrac{{{x}^{2}}+2x+6}{{{\left( {{x}^{2}}+2x+1 \right)}^{2}}+5}.$

  Firstly we will solve the given expression and then try to get a expression value in terms of ${{x}^{2}}$, for this use the formula ${{x}^{3}}-{{a}^{3}}=\left( x-a \right)\left( {{x}^{2}}+ax+{{a}^{2}} \right)$. And then put this value in the expression which we need to find. 

 $ \text{We have the data }\dfrac{{{x}^{3}}-8}{x-2},\text{ So we will first simplify it,} \\ \Rightarrow \dfrac{{{x}^{3}}-8}{x-2}=2 \\ \Rightarrow \dfrac{\left( x-2 \right)\left( {{x}^{2}}+4+2x \right)}{x-2}=2 \\ \Rightarrow {{x}^{2}}+2x=-2. \\ \text{Now put this value directly in the expression }\dfrac{{{x}^{2}}+2x+6}{{{\left( {{x}^{2}}+2x+1 \right)}^{2}}+5} \\ \text{So, }\\ \Rightarrow \dfrac{{{x}^{2}}+2x+6}{{{\left( {{x}^{2}}+2x+1 \right)}^{2}}+5}\Rightarrow \dfrac{-2+6}{{{\left( -2+1 \right)}^{2}}+5}\Rightarrow \dfrac{4}{1+5}\Rightarrow \dfrac{4}{6} \\ \Rightarrow \dfrac{2}{3}  $