Mister Exam
Other calculators
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How to use it?
- Equation:
-
Equation 9-8*x=-9*x+15
-
Equation 4^x-1/2-6*2^x-1+3=0
-
Equation 3x-5(2x+1)=3(3-2x)
-
Equation 2*6-y+2=0
- Express {x} in terms of y where:
- -16*x-9*y=-1
- -15*x+7*y=1
- -19*x+16*y=-9
- -5*x+1*y=19
- Graphing y =:
- sqrt(x+2)
- Integral of d{x}:
-
sqrt(x+2)
- Sum of series:
- sqrt(x+2)
-
Identical expressions
- sqrt(x+ two)=x
- square root of (x plus 2) equally x
- square root of (x plus two) equally x
- √(x+2)=x
- sqrtx+2=x
-
Similar expressions
- sqrt(x-2)=x
- sqrt4x+8-sqrt3x-2=sqrtx+2.
- x-3sqrt(x)+2=0
sqrt(x+2)=x equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given the equation
$$\sqrt{x + 2} = x$$
$$\sqrt{x + 2} = x$$
We raise the equation sides to 2-th degree
$$x + 2 = x^{2}$$
$$x + 2 = x^{2}$$
Transfer the right side of the equation left part with negative sign
$$- x^{2} + x + 2 = 0$$
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
$$a = -1$$
$$b = 1$$
$$c = 2$$
, then
Because D > 0, then the equation has two roots.
or
$$x_{1} = -1$$
$$x_{2} = 2$$
Because
$$\sqrt{x + 2} = x$$
and
$$\sqrt{x + 2} \geq 0$$
then
$$x \geq 0$$
or
$$0 \leq x$$
$$x < \infty$$
The final answer:
$$x_{2} = 2$$
$$\sqrt{x + 2} = x$$
$$\sqrt{x + 2} = x$$
We raise the equation sides to 2-th degree
$$x + 2 = x^{2}$$
$$x + 2 = x^{2}$$
Transfer the right side of the equation left part with negative sign
$$- x^{2} + x + 2 = 0$$
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
$$a = -1$$
$$b = 1$$
$$c = 2$$
, then
D = b^2 - 4 * a * c =
(1)^2 - 4 * (-1) * (2) = 9
Because D > 0, then the equation has two roots.
x1 = (-b + sqrt(D)) / (2*a)
x2 = (-b - sqrt(D)) / (2*a)
or
$$x_{1} = -1$$
$$x_{2} = 2$$
Because
$$\sqrt{x + 2} = x$$
and
$$\sqrt{x + 2} \geq 0$$
then
$$x \geq 0$$
or
$$0 \leq x$$
$$x < \infty$$
The final answer:
$$x_{2} = 2$$
The graph