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This is one of our many quizzes on Solving cube root equations. Click on this to access more such competitive quizzes on Cubes and cube roots.

Solving cube root equations Quiz 2

mathematics

 Evaluate: $\sqrt[3]{{{x^2} - 20x + 78}} = 3$

 Cubing on both sides of the equation and then solving the quadratic equation to get your answer.

We have: $\sqrt[3]{{{x^2} - 20x + 78}} = 3$ $ \\ $ On cubing both sides: $ \\ {x^2} - 20x + 78 = 27 \\   \Rightarrow {x^2} - 20x + 51 \\   \Rightarrow \left( {x - 17} \right)\left( {x - 3} \right) \\   \Rightarrow x = 17,3 \\  $

 Find the factors of: $\sqrt[3]{{{x^2} - 2x + 666}} = 9$

We have: $\sqrt[3]{{{x^2} - 2x + 666}} = 9$ $ \\ $ Take cube on both the sides of the equation $ \\   {x^2} - 2x + 666 = 729 \\   \Rightarrow {x^2} - 2x - 63 = 0 \\   \Rightarrow \left( {x + 7} \right)\left( {x - 9} \right) \\   \Rightarrow x =  - 7,9 \\  $

 Solve for $x: \dfrac{{\sqrt[3]{{2x - 7}}}}{{\sqrt[3]{{x - 1}}}} = 1$

 Cubing on both sides of the equation and then carrying all the terms on one side of the equation. Take L.C.M and solve for x.

We have: $\dfrac{{\sqrt[3]{{2x - 7}}}}{{\sqrt[3]{{x - 1}}}} = 1$ $ \\ $ Take cube on both the sides. $ \\   \dfrac{{2x - 7}}{{x - 1}} = 1 \\   \Rightarrow \dfrac{{2x - 7}}{{x - 1}} - 1 = 0 \\   \Rightarrow \dfrac{{2x - 7 - x + 1}}{{x - 1}} = 0 \\   \Rightarrow \dfrac{{x - 6}}{{x - 1}} = 0 \\   \Rightarrow x - 6 = 0 \\<br>&nbsp;&nbsp;&nbsp;\Rightarrow x = 6 \\ $

 Find the values of x in $5 + \sqrt[3]{{{x^2} - 23x + 51}} = 7$

 To solve the question, carry the constants on one side and then take a cube on both the sides of the equation. Solve the quadratic equation using $\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$

 We have: $5 + \sqrt[3]{{{x^2} - 23x + 50}} = 7$ $ \\ $ This can be written as:  $ \\   \sqrt[3]{{{x^2} - 23x + 50}} = 7 - 5 \\   \Rightarrow \sqrt[3]{{{x^2} - 23x + 50}} = 2 \\  $ Take cube on both the sides $ \\ {x^2} - 23x + 50 = 8 \\   \Rightarrow {x^2} - 23x + 42 = 0 \\  $ Use $\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ to solve the quadratic equation. $ \\  \dfrac{{23 \pm \sqrt {{{\left( { - 23} \right)}^2} - 4\left( 1 \right)\left( {42} \right)} }}{2} \\   \Rightarrow 21,2 \\   \Rightarrow x = 21,2 \\  $

 Simplify: $ - \dfrac{1}{5} + \dfrac{{\sqrt[3]{{{x^3} + 2{x^2} - 49x - 97}}}}{5} = 0$

 To solve the question, first take the constant terms on one side of the equation and then take the cube on both the sides. To solve the cubic equation, find one of the factors by hit and trial method and then use it to find the quadratic equation. Solve it to get the factors.

We have: $ - \dfrac{1}{5} + \dfrac{{\sqrt[3]{{{x^3} + 2{x^2} - 49x - 97}}}}{5} = 0$ $ \\ $ Take the constant on other side of the equation: $ \\ \dfrac{{\sqrt[3]{{{x^3} + 2{x^2} - 49x - 97}}}}{5} = \dfrac{1}{5} \\ $ This can be written as:  $ \\  \sqrt[3]{{{x^3} + 2{x^2} - 49x - 97}} = \dfrac{5}{5} \\   \Rightarrow \sqrt[3]{{{x^3} + 2{x^2} - 49x - 97}} = 1 \\  $ Take cube on both the sides: $ \\  {x^3} + 2{x^2} - 49x - 97 = 1 \\   \Rightarrow {x^3} + 2{x^2} - 49x - 98 = 0 \\  $ By hit and trial method find one of the factors of the equation: $ \\ $ $\left( {x + 2} \right)$ is one of the factors of the above equation. Use this to divide the equation. We get: $ \\  \left( {x + 2} \right)\left( {{x^2} - 5x - 14} \right) = 0 \\   \Rightarrow \left( {x + 2} \right)\left( {x - 7} \right)\left( {x + 7} \right) = 0 \\  $ So, $x = 7, - 2, - 7$