Mister Exam
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How to use it?
- Equation:
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Equation 8*x-3*(2*x+1)=2*x+4
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Equation 3*9^(x-1/2)-7*6^x+3*4^(x+1)=0
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Equation 9x-26=30-5x
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Equation x+19=12
- Express {x} in terms of y where:
- 12*x+14*y=-15
- -9*x-15*y=-4
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Identical expressions
- sqrt(x)^ three - three = zero
- square root of (x) cubed minus 3 equally 0
- square root of (x) to the power of three minus three equally zero
- √(x)^3-3=0
- sqrt(x)3-3=0
- sqrtx3-3=0
- sqrt(x)³-3=0
- sqrt(x) to the power of 3-3=0
- sqrtx^3-3=0
- sqrt(x)^3-3=O
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Similar expressions
- sqrt(x)^3+3=0
sqrt(x)^3-3=0 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given the equation
$$\left(\sqrt{x}\right)^{3} - 3 = 0$$
Because equation degree is equal to = 3/2 - does not contain even numbers in the numerator, then
the equation has single real root.
We raise the equation sides to 2/3-th degree:
We get:
$$\left(\left(1 x + 0\right)^{\frac{3}{2}}\right)^{\frac{2}{3}} = 3^{\frac{2}{3}}$$
or
$$x = 3^{\frac{2}{3}}$$
Expand brackets in the right part
We get the answer: x = 3^(2/3)
The final answer:
$$x_{1} = 3^{\frac{2}{3}}$$
$$\left(\sqrt{x}\right)^{3} - 3 = 0$$
Because equation degree is equal to = 3/2 - does not contain even numbers in the numerator, then
the equation has single real root.
We raise the equation sides to 2/3-th degree:
We get:
$$\left(\left(1 x + 0\right)^{\frac{3}{2}}\right)^{\frac{2}{3}} = 3^{\frac{2}{3}}$$
or
$$x = 3^{\frac{2}{3}}$$
Expand brackets in the right part
x = 3^2/3
We get the answer: x = 3^(2/3)
The final answer:
$$x_{1} = 3^{\frac{2}{3}}$$
Sum and product of roots
[src]
sum
2/3 0 + 3
$$0 + 3^{\frac{2}{3}}$$
=
2/3 3
$$3^{\frac{2}{3}}$$
product
2/3 1*3
$$1 \cdot 3^{\frac{2}{3}}$$
=
2/3 3
$$3^{\frac{2}{3}}$$
3^(2/3)
The graph