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Question: The real roots of the equation \[{x^{2/3}} + {x^{1/3}} - 2 = 0\] are A. 1, 8 B. \[ - 1\], \[ - 8...

The real roots of the equation x2/3+x1/32=0{x^{2/3}} + {x^{1/3}} - 2 = 0 are
A. 1, 8
B. 1 - 1, 8 - 8
C. 1 - 1, 8
D. 1, 8 - 8

Explanation

Solution

Hint: Here, we will first take x1/3=y{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 yy back to find the real roots of the given equation.

Complete step by step answer:
It is given that the equation is x2/3+x1/32=0{x^{2/3}} + {x^{1/3}} - 2 = 0.

Taking x1/3=y{x^{1/3}} = y in the above equation, we get

y2+y2=0{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 yy, we get

y2y+2y2=0 y(y1)+2(y1)=0 (y+2)(y1)=0  \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=0y + 2 = 0 and y1=0y - 1 = 0 in the above equation, we get

y+2=0 \Rightarrow y + 2 = 0 or y1=0y - 1 = 0
y=2\Rightarrow y = - 2 or y=1y = 1

Replacing x1/3{x^{1/3}} for yy in these above equations, we get

x1/3=1 \Rightarrow {x^{1/3}} = 1 or x1/3=2{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.