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Equation sin^3x+cos^3x=1
- Equation sqrt(3x+1)-sqrt(16-3x)=5
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Identical expressions
- sqrt(3x+ one)-sqrt(sixteen -3x)= five
- square root of (3x plus 1) minus square root of (16 minus 3x) equally 5
- square root of (3x plus one) minus square root of (sixteen minus 3x) equally five
- √(3x+1)-√(16-3x)=5
- sqrt3x+1-sqrt16-3x=5
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Similar expressions
- sqrt(3x-1)-sqrt(16-3x)=5
- sqrt(3x+1)-sqrt(16+3x)=5
- sqrt(3x+1)+sqrt(16-3x)=5
sqrt(3x+1)-sqrt(16-3x)=5 equation
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The solution
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_________ __________ \/ 3*x + 1 - \/ 16 - 3*x = 5
$$- \sqrt{16 - 3 x} + \sqrt{3 x + 1} = 5$$
Detail solution
Given the equation
$$- \sqrt{16 - 3 x} + \sqrt{3 x + 1} = 5$$
We raise the equation sides to 2-th degree
$$\left(- \sqrt{16 - 3 x} + \sqrt{3 x + 1}\right)^{2} = 25$$
or
$$\left(-1\right)^{2} \left(16 - 3 x\right) + \left(\left(-1\right) 2 \sqrt{\left(16 - 3 x\right) \left(3 x + 1\right)} + 1^{2} \left(3 x + 1\right)\right) = 25$$
or
$$17 - 2 \sqrt{- 9 x^{2} + 45 x + 16} = 25$$
transform:
$$- 2 \sqrt{- 9 x^{2} + 45 x + 16} = 8$$
We raise the equation sides to 2-th degree
$$- 36 x^{2} + 180 x + 64 = 64$$
$$- 36 x^{2} + 180 x + 64 = 64$$
Transfer the right side of the equation left part with negative sign
$$- 36 x^{2} + 180 x = 0$$
This equation is of the form
A quadratic equation can be solved
using the discriminant.
The roots of the quadratic equation:
$$x_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$x_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = -36$$
$$b = 180$$
$$c = 0$$
, then
Because D > 0, then the equation has two roots.
or
$$x_{1} = 0$$
$$x_{2} = 5$$
Because
$$\sqrt{- 9 x^{2} + 45 x + 16} = -4$$
and
$$\sqrt{- 9 x^{2} + 45 x + 16} \geq 0$$
then
$$-4 \geq 0$$
The final answer:
This equation has no roots
$$- \sqrt{16 - 3 x} + \sqrt{3 x + 1} = 5$$
We raise the equation sides to 2-th degree
$$\left(- \sqrt{16 - 3 x} + \sqrt{3 x + 1}\right)^{2} = 25$$
or
$$\left(-1\right)^{2} \left(16 - 3 x\right) + \left(\left(-1\right) 2 \sqrt{\left(16 - 3 x\right) \left(3 x + 1\right)} + 1^{2} \left(3 x + 1\right)\right) = 25$$
or
$$17 - 2 \sqrt{- 9 x^{2} + 45 x + 16} = 25$$
transform:
$$- 2 \sqrt{- 9 x^{2} + 45 x + 16} = 8$$
We raise the equation sides to 2-th degree
$$- 36 x^{2} + 180 x + 64 = 64$$
$$- 36 x^{2} + 180 x + 64 = 64$$
Transfer the right side of the equation left part with negative sign
$$- 36 x^{2} + 180 x = 0$$
This equation is of the form
a*x^2 + b*x + c = 0
A quadratic equation can be solved
using the discriminant.
The roots of the quadratic equation:
$$x_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$x_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = -36$$
$$b = 180$$
$$c = 0$$
, then
D = b^2 - 4 * a * c =
(180)^2 - 4 * (-36) * (0) = 32400
Because D > 0, then the equation has two roots.
x1 = (-b + sqrt(D)) / (2*a)
x2 = (-b - sqrt(D)) / (2*a)
or
$$x_{1} = 0$$
$$x_{2} = 5$$
Because
$$\sqrt{- 9 x^{2} + 45 x + 16} = -4$$
and
$$\sqrt{- 9 x^{2} + 45 x + 16} \geq 0$$
then
$$-4 \geq 0$$
The final answer:
This equation has no roots