Mister Exam
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Equation sqrt(2x+5)+sqrt(5x+6)=sqrt(12x+25)
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Identical expressions
- sqrt(2x+ five)+sqrt(5x+ six)=sqrt(12x+ twenty-five)
- square root of (2x plus 5) plus square root of (5x plus 6) equally square root of (12x plus 25)
- square root of (2x plus five) plus square root of (5x plus six) equally square root of (12x plus twenty minus five)
- √(2x+5)+√(5x+6)=√(12x+25)
- sqrt2x+5+sqrt5x+6=sqrt12x+25
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Similar expressions
- sqrt(2x+5)+sqrt(5x+6)=sqrt(12x-25)
- sqrt(2x+5)+sqrt(5x-6)=sqrt(12x+25)
- sqrt(2x+5)-sqrt(5x+6)=sqrt(12x+25)
- sqrt(2x-5)+sqrt(5x+6)=sqrt(12x+25)
sqrt(2x+5)+sqrt(5x+6)=sqrt(12x+25) equation
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The solution
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_________ _________ ___________ \/ 2*x + 5 + \/ 5*x + 6 = \/ 12*x + 25
$$\sqrt{2 x + 5} + \sqrt{5 x + 6} = \sqrt{12 x + 25}$$
Detail solution
Given the equation
$$\sqrt{2 x + 5} + \sqrt{5 x + 6} = \sqrt{12 x + 25}$$
We raise the equation sides to 2-th degree
$$\left(\sqrt{2 x + 5} + \sqrt{5 x + 6}\right)^{2} = 12 x + 25$$
or
$$1^{2} \cdot \left(5 x + 6\right) + \left(1 \cdot 2 \cdot 1 \sqrt{\left(2 x + 5\right) \left(5 x + 6\right)} + 1^{2} \cdot \left(2 x + 5\right)\right) = 12 x + 25$$
or
$$7 x + 2 \sqrt{10 x^{2} + 37 x + 30} + 11 = 12 x + 25$$
transform:
$$2 \sqrt{10 x^{2} + 37 x + 30} = 5 x + 14$$
We raise the equation sides to 2-th degree
$$40 x^{2} + 148 x + 120 = \left(5 x + 14\right)^{2}$$
$$40 x^{2} + 148 x + 120 = 25 x^{2} + 140 x + 196$$
Transfer the right side of the equation left part with negative sign
$$15 x^{2} + 8 x - 76 = 0$$
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
$$a = 15$$
$$b = 8$$
$$c = -76$$
, then
Because D > 0, then the equation has two roots.
or
$$x_{1} = 2$$
Simplify
$$x_{2} = - \frac{38}{15}$$
Simplify
Because
$$\sqrt{10 x^{2} + 37 x + 30} = \frac{5 x}{2} + 7$$
and
$$\sqrt{10 x^{2} + 37 x + 30} \geq 0$$
then
$$\frac{5 x}{2} + 7 \geq 0$$
or
$$- \frac{14}{5} \leq x$$
$$x < \infty$$
$$x_{1} = 2$$
$$x_{2} = - \frac{38}{15}$$
check:
$$x_{1} = 2$$
$$\sqrt{2 x_{1} + 5} + \sqrt{5 x_{1} + 6} - \sqrt{12 x_{1} + 25} = 0$$
=
$$- \sqrt{12 \cdot 2 + 25} + \left(\sqrt{2 \cdot 2 + 5} + \sqrt{6 + 5 \cdot 2}\right) = 0$$
=
- the identity
$$x_{2} = - \frac{38}{15}$$
$$\sqrt{2 x_{2} + 5} + \sqrt{5 x_{2} + 6} - \sqrt{12 x_{2} + 25} = 0$$
=
$$- \sqrt{12 \left(- \frac{38}{15}\right) + 25} + \left(\sqrt{2 \left(- \frac{38}{15}\right) + 5} + \sqrt{5 \left(- \frac{38}{15}\right) + 6}\right) = 0$$
=
- No
The final answer:
$$x_{1} = 2$$
$$\sqrt{2 x + 5} + \sqrt{5 x + 6} = \sqrt{12 x + 25}$$
We raise the equation sides to 2-th degree
$$\left(\sqrt{2 x + 5} + \sqrt{5 x + 6}\right)^{2} = 12 x + 25$$
or
$$1^{2} \cdot \left(5 x + 6\right) + \left(1 \cdot 2 \cdot 1 \sqrt{\left(2 x + 5\right) \left(5 x + 6\right)} + 1^{2} \cdot \left(2 x + 5\right)\right) = 12 x + 25$$
or
$$7 x + 2 \sqrt{10 x^{2} + 37 x + 30} + 11 = 12 x + 25$$
transform:
$$2 \sqrt{10 x^{2} + 37 x + 30} = 5 x + 14$$
We raise the equation sides to 2-th degree
$$40 x^{2} + 148 x + 120 = \left(5 x + 14\right)^{2}$$
$$40 x^{2} + 148 x + 120 = 25 x^{2} + 140 x + 196$$
Transfer the right side of the equation left part with negative sign
$$15 x^{2} + 8 x - 76 = 0$$
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
$$a = 15$$
$$b = 8$$
$$c = -76$$
, then
D = b^2 - 4 * a * c =
(8)^2 - 4 * (15) * (-76) = 4624
Because D > 0, then the equation has two roots.
x1 = (-b + sqrt(D)) / (2*a)
x2 = (-b - sqrt(D)) / (2*a)
or
$$x_{1} = 2$$
Simplify
$$x_{2} = - \frac{38}{15}$$
Simplify
Because
$$\sqrt{10 x^{2} + 37 x + 30} = \frac{5 x}{2} + 7$$
and
$$\sqrt{10 x^{2} + 37 x + 30} \geq 0$$
then
$$\frac{5 x}{2} + 7 \geq 0$$
or
$$- \frac{14}{5} \leq x$$
$$x < \infty$$
$$x_{1} = 2$$
$$x_{2} = - \frac{38}{15}$$
check:
$$x_{1} = 2$$
$$\sqrt{2 x_{1} + 5} + \sqrt{5 x_{1} + 6} - \sqrt{12 x_{1} + 25} = 0$$
=
$$- \sqrt{12 \cdot 2 + 25} + \left(\sqrt{2 \cdot 2 + 5} + \sqrt{6 + 5 \cdot 2}\right) = 0$$
=
0 = 0
- the identity
$$x_{2} = - \frac{38}{15}$$
$$\sqrt{2 x_{2} + 5} + \sqrt{5 x_{2} + 6} - \sqrt{12 x_{2} + 25} = 0$$
=
$$- \sqrt{12 \left(- \frac{38}{15}\right) + 25} + \left(\sqrt{2 \left(- \frac{38}{15}\right) + 5} + \sqrt{5 \left(- \frac{38}{15}\right) + 6}\right) = 0$$
=
2*i*sqrt(15)/15 = 0
- No
The final answer:
$$x_{1} = 2$$
The graph