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# The real roots of the equation ${x^{2/3}} + {x^{1/3}} - 2 = 0$ areA. 1, 8B. $- 1$, $- 8$C. $- 1$, 8D. 1, $- 8$

Last updated date: 12th Sep 2024
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Hint: Here, we will first take ${x^{1/3}} = y$ in the given equation and then find the real roots of the obtained equation by factorization. Then we will put the value of $y$ back to find the real roots of the given equation.

It is given that the equation is ${x^{2/3}} + {x^{1/3}} - 2 = 0$.

Taking ${x^{1/3}} = y$ in the above equation, we get

${y^2} + y - 2 = 0$

We will now factor the above equation to find the root of the equation.

Factoring the above equation to find the value of $y$, we get

$\Rightarrow {y^2} - y + 2y - 2 = 0 \\ \Rightarrow y\left( {y - 1} \right) + 2\left( {y - 1} \right) = 0 \\ \Rightarrow \left( {y + 2} \right)\left( {y - 1} \right) = 0 \\$

Taking $y + 2 = 0$ and $y - 1 = 0$ in the above equation, we get

$\Rightarrow y + 2 = 0$ or $y - 1 = 0$
$\Rightarrow y = - 2$ or $y = 1$

Replacing ${x^{1/3}}$ for $y$ in these above equations, we get

$\Rightarrow {x^{1/3}} = 1$ or ${x^{1/3}} = - 2$

Taking the square in the above equations, we get

$\Rightarrow x = {\left( 1 \right)^3} \\ \Rightarrow x = 1 \\$ or $x = {\left( { - 2} \right)^3} \\ x = - 8 \\$

Thus, we have found that the real roots of the given equation are 1 and $- 8$.

Hence, the option D is correct.

Note: In this question, the equation can also be solved using the quadratic formula. Students should also know the concept of real roots before solving this question. Also, we are supposed to write the values properly to avoid any miscalculation.