Mister Exam
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How to use it?
- Equation:
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Equation 5x(x-4)-x(3+5x)=4
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Equation sinx=1/3
- Equation √7x-5-√3x-2=√4x-3
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Identical expressions
- √7x- five -√ three x- two =√4x-3
- √7x minus 5 minus √3x minus 2 equally √4x minus 3
- √7x minus five minus √ three x minus two equally √4x minus 3
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Similar expressions
- √7x+5-√3x-2=√4x-3
- √7x-5-√3x+2=√4x-3
- √(зx+1)/x+4(√(3x+1)/3-√(4x-3))=13/3-4x-1/x
- x=√(4x-3)
- √(4x-3)=3
- √7x-5-√3x-2=√4x+3
- √7x-5+√3x-2=√4x-3
√7x-5-√3x-2=√4x-3 equation
The teacher will be very surprised to see your correct solution 😉
The solution
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_____ _____ _____ \/ 7*x - 5 - \/ 3*x - 2 = \/ 4*x - 3
$$\left(- \sqrt{3 x} + \left(\sqrt{7 x} - 5\right)\right) - 2 = \sqrt{4 x} - 3$$
Detail solution
Given the equation
$$\left(- \sqrt{3 x} + \left(\sqrt{7 x} - 5\right)\right) - 2 = \sqrt{4 x} - 3$$
Transfer the right side of the equation left part with negative sign
$$\sqrt{x} \left(-2 - \sqrt{3} + \sqrt{7}\right) = 4$$
We raise the equation sides to 2-th degree
$$x \left(-2 - \sqrt{3} + \sqrt{7}\right)^{2} = 16$$
$$x \left(-2 - \sqrt{3} + \sqrt{7}\right)^{2} = 16$$
Transfer the right side of the equation left part with negative sign
$$x \left(-2 - \sqrt{3} + \sqrt{7}\right)^{2} - 16 = 0$$
Expand brackets in the left part
Looking for similar summands in the left part:
Move free summands (without x)
from left part to right part, we given:
$$x \left(-2 - \sqrt{3} + \sqrt{7}\right)^{2} = 16$$
Divide both parts of the equation by (-2 + sqrt(7) - sqrt(3))^2
We get the answer: x = 16/(2 + sqrt(3) - sqrt(7))^2
Because
$$\sqrt{x} = \frac{4}{-2 - \sqrt{3} + \sqrt{7}}$$
and
$$\sqrt{x} \geq 0$$
then
$$\frac{4}{-2 - \sqrt{3} + \sqrt{7}} \geq 0$$
The final answer:
This equation has no roots
$$\left(- \sqrt{3 x} + \left(\sqrt{7 x} - 5\right)\right) - 2 = \sqrt{4 x} - 3$$
Transfer the right side of the equation left part with negative sign
$$\sqrt{x} \left(-2 - \sqrt{3} + \sqrt{7}\right) = 4$$
We raise the equation sides to 2-th degree
$$x \left(-2 - \sqrt{3} + \sqrt{7}\right)^{2} = 16$$
$$x \left(-2 - \sqrt{3} + \sqrt{7}\right)^{2} = 16$$
Transfer the right side of the equation left part with negative sign
$$x \left(-2 - \sqrt{3} + \sqrt{7}\right)^{2} - 16 = 0$$
Expand brackets in the left part
-16 + x-2+sqrt+7 - sqrt3)^2 = 0
Looking for similar summands in the left part:
-16 + x*(-2 + sqrt(7) - sqrt(3))^2 = 0
Move free summands (without x)
from left part to right part, we given:
$$x \left(-2 - \sqrt{3} + \sqrt{7}\right)^{2} = 16$$
Divide both parts of the equation by (-2 + sqrt(7) - sqrt(3))^2
x = 16 / ((-2 + sqrt(7) - sqrt(3))^2)
We get the answer: x = 16/(2 + sqrt(3) - sqrt(7))^2
Because
$$\sqrt{x} = \frac{4}{-2 - \sqrt{3} + \sqrt{7}}$$
and
$$\sqrt{x} \geq 0$$
then
$$\frac{4}{-2 - \sqrt{3} + \sqrt{7}} \geq 0$$
The final answer:
This equation has no roots