Mister Exam
Other calculators
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How to use it?
- Equation:
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Equation x^2+3*x+9=0
-
Equation x^2+6x-2=0
-
Equation 3*x^2-8*x-3=0
-
Equation -2x-7=-4x
- Express {x} in terms of y where:
- -13*x-14*y=0
- -13*x+15*y=1
- 11*x-6*y=-1
- -12*x-4*y=-17
- Derivative of:
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(x-3)^2
- Integral of d{x}:
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(x-3)^2
- Graphing y =:
- (x-3)^2
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Identical expressions
- (x- three)^ two = four
- (x minus 3) squared equally 4
- (x minus three) to the power of two equally four
- (x-3)2=4
- x-32=4
- (x-3)²=4
- (x-3) to the power of 2=4
- x-3^2=4
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Similar expressions
- (x+3)^2=4
(x-3)^2=4 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
$$\left(x - 3\right)^{2} = 4$$
to
$$\left(x - 3\right)^{2} - 4 = 0$$
Expand the expression in the equation
$$\left(x - 3\right)^{2} - 4 = 0$$
We get the quadratic equation
$$x^{2} - 6 x + 5 = 0$$
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
$$a = 1$$
$$b = -6$$
$$c = 5$$
, then
Because D > 0, then the equation has two roots.
or
$$x_{1} = 5$$
$$x_{2} = 1$$
left part with negative sign.
The equation is transformed from
$$\left(x - 3\right)^{2} = 4$$
to
$$\left(x - 3\right)^{2} - 4 = 0$$
Expand the expression in the equation
$$\left(x - 3\right)^{2} - 4 = 0$$
We get the quadratic equation
$$x^{2} - 6 x + 5 = 0$$
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
$$a = 1$$
$$b = -6$$
$$c = 5$$
, then
D = b^2 - 4 * a * c =
(-6)^2 - 4 * (1) * (5) = 16
Because D > 0, then the equation has two roots.
x1 = (-b + sqrt(D)) / (2*a)
x2 = (-b - sqrt(D)) / (2*a)
or
$$x_{1} = 5$$
$$x_{2} = 1$$
The graph