Mister Exam
Other calculators
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How to use it?
- Equation:
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Equation (x-11)/(x-6)=11/16
-
Equation x^2-4x+6=0
-
Equation 5*x=12
-
Equation 2*sin(x)*cos(x)=0
- Express {x} in terms of y where:
- -20*x-7*y=-17
- -16*x-9*y=-5
- -8*x-11*y=1
- -20*x-18*y=20
- Graphing y =:
- sqrt(x^2-16)
- 8-x
- Derivative of:
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8-x
- Limit of the function:
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8-x
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Identical expressions
- sqrt(x^ two - sixteen)= eight -x
- square root of (x squared minus 16) equally 8 minus x
- square root of (x to the power of two minus sixteen) equally eight minus x
- √(x^2-16)=8-x
- sqrt(x2-16)=8-x
- sqrtx2-16=8-x
- sqrt(x²-16)=8-x
- sqrt(x to the power of 2-16)=8-x
- sqrtx^2-16=8-x
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Similar expressions
- sqrt(x^2-16)=8+x
- sqrt(x^2+16)=8-x
- log(7x^4-(x^4-9)^(1/2)+x^2+8)=log(x^8-(x^4-9)^(1/2)+x^2-10)
- (1/512)^(x+4)=8^-x
sqrt(x^2-16)=8-x equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given the equation
$$\sqrt{x^{2} - 16} = 8 - x$$
$$\sqrt{x^{2} - 16} = 8 - x$$
We raise the equation sides to 2-th degree
$$x^{2} - 16 = \left(8 - x\right)^{2}$$
$$x^{2} - 16 = x^{2} - 16 x + 64$$
Transfer the right side of the equation left part with negative sign
$$16 x - 80 = 0$$
Move free summands (without x)
from left part to right part, we given:
$$16 x = 80$$
Divide both parts of the equation by 16
We get the answer: x = 5
Because
$$\sqrt{x^{2} - 16} = 8 - x$$
and
$$\sqrt{x^{2} - 16} \geq 0$$
then
$$8 - x \geq 0$$
or
$$x \leq 8$$
$$-\infty < x$$
The final answer:
$$x_{1} = 5$$
$$\sqrt{x^{2} - 16} = 8 - x$$
$$\sqrt{x^{2} - 16} = 8 - x$$
We raise the equation sides to 2-th degree
$$x^{2} - 16 = \left(8 - x\right)^{2}$$
$$x^{2} - 16 = x^{2} - 16 x + 64$$
Transfer the right side of the equation left part with negative sign
$$16 x - 80 = 0$$
Move free summands (without x)
from left part to right part, we given:
$$16 x = 80$$
Divide both parts of the equation by 16
x = 80 / (16)
We get the answer: x = 5
Because
$$\sqrt{x^{2} - 16} = 8 - x$$
and
$$\sqrt{x^{2} - 16} \geq 0$$
then
$$8 - x \geq 0$$
or
$$x \leq 8$$
$$-\infty < x$$
The final answer:
$$x_{1} = 5$$