Mister Exam
Other calculators
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How to use it?
- Equation:
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Equation x^2+x=12
-
Equation 3x-2(3x-2)=19
-
Equation x^4=(x-56)^2
-
Equation (-2x+1)*(-2x-7)=0
- Express {x} in terms of y where:
- -7*x-13*y=-12
- 8*x+1*y=4
- 20*x-16*y=7
- -3*x+19*y=16
- Derivative of:
- 1-9/x^2
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Identical expressions
- one - nine /x^ two = zero
- 1 minus 9 divide by x squared equally 0
- one minus nine divide by x to the power of two equally zero
- 1-9/x2=0
- 1-9/x²=0
- 1-9/x to the power of 2=0
- 1-9/x^2=O
- 1-9 divide by x^2=0
-
Similar expressions
- 1+9/x^2=0
1-9/x^2=0 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given the equation
$$1 - \frac{9}{x^{2}} = 0$$
Because equation degree is equal to = -2 - contains the even number -2 in the numerator, then
the equation has two real roots.
Get the root -2-th degree of the equation sides:
We get:
$$\frac{1}{3 \sqrt{\frac{1}{x^{2}}}} = \frac{1}{\sqrt{1}}$$
$$\frac{1}{3 \sqrt{\frac{1}{x^{2}}}} = \left(-1\right) \frac{1}{\sqrt{1}}$$
or
$$\frac{x}{3} = 1$$
$$\frac{x}{3} = -1$$
Divide both parts of the equation by 1/3
We get the answer: x = 3
Divide both parts of the equation by 1/3
We get the answer: x = -3
or
$$x_{1} = -3$$
$$x_{2} = 3$$
The final answer:
$$x_{1} = -3$$
$$x_{2} = 3$$
$$1 - \frac{9}{x^{2}} = 0$$
Because equation degree is equal to = -2 - contains the even number -2 in the numerator, then
the equation has two real roots.
Get the root -2-th degree of the equation sides:
We get:
$$\frac{1}{3 \sqrt{\frac{1}{x^{2}}}} = \frac{1}{\sqrt{1}}$$
$$\frac{1}{3 \sqrt{\frac{1}{x^{2}}}} = \left(-1\right) \frac{1}{\sqrt{1}}$$
or
$$\frac{x}{3} = 1$$
$$\frac{x}{3} = -1$$
Divide both parts of the equation by 1/3
x = 1 / (1/3)
We get the answer: x = 3
Divide both parts of the equation by 1/3
x = -1 / (1/3)
We get the answer: x = -3
or
$$x_{1} = -3$$
$$x_{2} = 3$$
The final answer:
$$x_{1} = -3$$
$$x_{2} = 3$$
The graph