Mister Exam
Other calculators
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How to use it?
- Equation:
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Equation (x-2)/(x-7)=5/8
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Equation 6/(x+5)=-5
- Equation z^2+3+4*i=0
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Equation 3x-2(3x+4)=10
- Express {x} in terms of y where:
- -17*x+1*y=-11
- 1*x+4*y=20
- -18*x-2*y=9
- 7*x-1*y=-7
- Derivative of:
- sqrt(2*x+3)
- Integral of d{x}:
- sqrt(2*x+3)
-
Identical expressions
- sqrt(two *x+ three)=x
- square root of (2 multiply by x plus 3) equally x
- square root of (two multiply by x plus three) equally x
- √(2*x+3)=x
- sqrt(2x+3)=x
- sqrt2x+3=x
-
Similar expressions
- (3-2*sqrt(2))^x+(3+2*sqrt(2))^x=6
- sqrt(2x+3)+sqrt(x-2)=2*sqrt(x-1)
- sqrt(2x+3)+sqrt(x+5)=sqrt(6x+7)
- sqrt(2*x-3)=x
- sqrt(2*x+3-x^2)=0
sqrt(2*x+3)=x equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given the equation
$$\sqrt{2 x + 3} = x$$
$$\sqrt{2 x + 3} = x$$
We raise the equation sides to 2-th degree
$$2 x + 3 = x^{2}$$
$$2 x + 3 = x^{2}$$
Transfer the right side of the equation left part with negative sign
$$- x^{2} + 2 x + 3 = 0$$
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
$$a = -1$$
$$b = 2$$
$$c = 3$$
, then
Because D > 0, then the equation has two roots.
or
$$x_{1} = -1$$
$$x_{2} = 3$$
Because
$$\sqrt{2 x + 3} = x$$
and
$$\sqrt{2 x + 3} \geq 0$$
then
$$x \geq 0$$
or
$$0 \leq x$$
$$x < \infty$$
The final answer:
$$x_{2} = 3$$
$$\sqrt{2 x + 3} = x$$
$$\sqrt{2 x + 3} = x$$
We raise the equation sides to 2-th degree
$$2 x + 3 = x^{2}$$
$$2 x + 3 = x^{2}$$
Transfer the right side of the equation left part with negative sign
$$- x^{2} + 2 x + 3 = 0$$
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
$$a = -1$$
$$b = 2$$
$$c = 3$$
, then
D = b^2 - 4 * a * c =
(2)^2 - 4 * (-1) * (3) = 16
Because D > 0, then the equation has two roots.
x1 = (-b + sqrt(D)) / (2*a)
x2 = (-b - sqrt(D)) / (2*a)
or
$$x_{1} = -1$$
$$x_{2} = 3$$
Because
$$\sqrt{2 x + 3} = x$$
and
$$\sqrt{2 x + 3} \geq 0$$
then
$$x \geq 0$$
or
$$0 \leq x$$
$$x < \infty$$
The final answer:
$$x_{2} = 3$$
The graph