Mister Exam
Other calculators
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How to use it?
- Equation:
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Equation x^2+5*x-14=0
-
Equation x²-4x+3=0
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Equation x=x^9
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Equation 12-y+2=0
- Express {x} in terms of y where:
- 16*x+5*y=-14
- 12*x-2*y=-18
- -19*x-12*y=13
- -16*x-19*y=-4
- Derivative of:
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(x-1)^2
- Integral of d{x}:
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(x-1)^2
- Canonical form:
- (x-1)^2
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Identical expressions
- (x- one)^ two = one
- (x minus 1) squared equally 1
- (x minus one) to the power of two equally one
- (x-1)2=1
- x-12=1
- (x-1)²=1
- (x-1) to the power of 2=1
- x-1^2=1
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Similar expressions
- (x+1)^2=1
(x-1)^2=1 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Move right part of the equation to
left part with negative sign.
The equation is transformed from
$$\left(x - 1\right)^{2} = 1$$
to
$$\left(x - 1\right)^{2} - 1 = 0$$
Expand the expression in the equation
$$\left(x - 1\right)^{2} - 1 = 0$$
We get the quadratic equation
$$x^{2} - 2 x = 0$$
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
$$a = 1$$
$$b = -2$$
$$c = 0$$
, then
Because D > 0, then the equation has two roots.
or
$$x_{1} = 2$$
$$x_{2} = 0$$
left part with negative sign.
The equation is transformed from
$$\left(x - 1\right)^{2} = 1$$
to
$$\left(x - 1\right)^{2} - 1 = 0$$
Expand the expression in the equation
$$\left(x - 1\right)^{2} - 1 = 0$$
We get the quadratic equation
$$x^{2} - 2 x = 0$$
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
$$a = 1$$
$$b = -2$$
$$c = 0$$
, then
D = b^2 - 4 * a * c =
(-2)^2 - 4 * (1) * (0) = 4
Because D > 0, then the equation has two roots.
x1 = (-b + sqrt(D)) / (2*a)
x2 = (-b - sqrt(D)) / (2*a)
or
$$x_{1} = 2$$
$$x_{2} = 0$$
The graph