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Solve ${{x}^{\dfrac{2}{3}}}+{{x}^{\dfrac{1}{3}}}-2=0$ .

Last updated date: 12th Sep 2024
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Hint:
In this question we need to solve ${{x}^{\dfrac{2}{3}}}+{{x}^{\dfrac{1}{3}}}-2=0$ which means we need to find the value of x. For this we will consider ${{x}^{\dfrac{1}{3}}}$ as y. And then substitute it in the given equation to get a quadratic equation in terms of y. After that, we will use splitting the middle term method to find the value of y. At last, using the relation ${{x}^{\dfrac{1}{3}}}=y$ we will find the value of x.

Here we are given the equation as ${{x}^{\dfrac{2}{3}}}+{{x}^{\dfrac{1}{3}}}-2=0$ .
As we can see that we cannot solve it directly, so let us suppose that ${{x}^{\dfrac{1}{3}}}=y\cdots \cdots \cdots \left( 1 \right)$ .
Squaring both sides we have, ${{\left( {{x}^{\dfrac{1}{3}}} \right)}^{2}}={{y}^{2}}$ .
Using ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}$ we get ${{x}^{\dfrac{2}{3}}}={{y}^{2}}\cdots \cdots \cdots \left( 2 \right)$ .
Substituting values from (1) and (2) in the given equation we have ${{y}^{2}}+y-2=0$ .
Now let us solve this equation using splitting the middle term method. We know, 2-1 = 1 which is coefficient of y and also 2(-1) = -2 which is the constant term. So we have, ${{y}^{2}}+\left( 2-1 \right)y-2=0\Rightarrow {{y}^{2}}+2y-y-2=0$ .
Taking y common from the first two terms and -1 common from the third and the fourth term we get, $y\left( y+2 \right)-1\left( y+2 \right)=0$ .
Taking (y+2) common we get $\left( y+2 \right)\left( y-1 \right)=0$ .
We can say that y+2 = 0 and y-1 = 0.
Solving y+2 = 0 we get y = -2.
Also solving y-1 = 0 we get y = 1.
Hence we have values of y as -2, 1.
Now let us find the value of x.
We suppose ${{x}^{\dfrac{1}{3}}}$ as y, i.e. ${{x}^{\dfrac{1}{3}}}=y$ .
Taking cube on both sides we get, ${{\left( {{x}^{\dfrac{1}{3}}} \right)}^{3}}={{y}^{3}}$ .
Using ${{\left( {{a}^{m}} \right)}^{n}}$ we get $x={{y}^{3}}$ .
Putting y = -2 we get,
$x={{\left( -2 \right)}^{3}}\Rightarrow x=\left( -2 \right)\times \left( -2 \right)\times \left( -2 \right)\Rightarrow x=-8$ .
Putting y = 1 we get, $x={{\left( 1 \right)}^{3}}\Rightarrow x=1$ .
Hence the value of x is -8 and 1 which is our required answer.

Note:
Students should take care of the signs while splitting the middle term. Make proper factors to put them equal to zero. Students can also use the quadratic formula to solve the quadratic equation. For an equation of the form $a{{x}^{2}}+bx+c$ quadratic formula is given by $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ .