Mister Exam
Other calculators
-
How to use it?
- Equation:
-
Equation cos(x)=2/3
-
Equation sin(5*x)=cos(4*x)
-
Equation sqrt(x+1)=x-1
-
Equation sin(x)^2=1/2
- Express {x} in terms of y where:
- 12*x+14*y=-15
- -4*x+19*y=-13
- -9*x-4*y=-20
- 12*x-13*y=-14
- Integral of d{x}:
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sqrt(x+1)
- Derivative of:
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sqrt(x+1)
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x-1
- Graphing y =:
- sqrt(x+1)
- x-1
- Canonical form:
- x-1
-
Identical expressions
- sqrt(x+ one)=x- one
- square root of (x plus 1) equally x minus 1
- square root of (x plus one) equally x minus one
- √(x+1)=x-1
- sqrtx+1=x-1
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Similar expressions
- sqrt(x-1)=x-1
- sin(2*x)-12*(sin(x)-cos(x))+12=0
- (arctan(sqrt(2)tan(x)+1)+arctan(sqrt(2)tan(x)-1))/sqrt(2)-1/(sqrt(2))arctan(tan(2x)/sqrt(2))=0
- 2^sqrt(x)+1-2^(sqrt(x)+1)-2^(sqrt(x)-1)=12
- f*x=-2*sqrt(x)+12*x^4-3*sin(x)+9*acos(x)-5
- sqrt(x+1)=x+1
sqrt(x+1)=x-1 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given the equation
$$\sqrt{x + 1} = x - 1$$
$$\sqrt{x + 1} = x - 1$$
We raise the equation sides to 2-th degree
$$x + 1 = \left(x - 1\right)^{2}$$
$$x + 1 = x^{2} - 2 x + 1$$
Transfer the right side of the equation left part with negative sign
$$- x^{2} + 3 x = 0$$
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
$$a = -1$$
$$b = 3$$
$$c = 0$$
, then
Because D > 0, then the equation has two roots.
or
$$x_{1} = 0$$
$$x_{2} = 3$$
Because
$$\sqrt{x + 1} = x - 1$$
and
$$\sqrt{x + 1} \geq 0$$
then
$$x - 1 \geq 0$$
or
$$1 \leq x$$
$$x < \infty$$
The final answer:
$$x_{2} = 3$$
$$\sqrt{x + 1} = x - 1$$
$$\sqrt{x + 1} = x - 1$$
We raise the equation sides to 2-th degree
$$x + 1 = \left(x - 1\right)^{2}$$
$$x + 1 = x^{2} - 2 x + 1$$
Transfer the right side of the equation left part with negative sign
$$- x^{2} + 3 x = 0$$
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
$$a = -1$$
$$b = 3$$
$$c = 0$$
, then
D = b^2 - 4 * a * c =
(3)^2 - 4 * (-1) * (0) = 9
Because D > 0, then the equation has two roots.
x1 = (-b + sqrt(D)) / (2*a)
x2 = (-b - sqrt(D)) / (2*a)
or
$$x_{1} = 0$$
$$x_{2} = 3$$
Because
$$\sqrt{x + 1} = x - 1$$
and
$$\sqrt{x + 1} \geq 0$$
then
$$x - 1 \geq 0$$
or
$$1 \leq x$$
$$x < \infty$$
The final answer:
$$x_{2} = 3$$
The graph