Mister Exam
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How to use it?
- Equation:
-
Equation sin(x)=cos(x)
-
Equation cos(x/2)=1/2
-
Equation sin(x)=√3/2
-
Equation sin(x)=1,5
- Express {x} in terms of y where:
- -2*x+1*y=-11
- 8*x+10*y=5
- -8*x-5*y=-1
- -3*x-1*y=9
-
Identical expressions
- -(y- seven)²-12y+ thirty-six
- minus (y minus 7)² minus 12y plus 36
- minus (y minus seven)² minus 12y plus thirty minus six
- -y-7²-12y+36
-
Similar expressions
- -(y+7)²-12y+36
- (y-7)²-12y+36
- -(y-7)²+12y+36
- -(y-7)²-12y-36
-(y-7)²-12y+36 equation
The teacher will be very surprised to see your correct solution 😉
The solution
You have entered
[src]
2 - (y - 7) - 12*y + 36 = 0
$$\left(- 12 y - \left(y - 7\right)^{2}\right) + 36 = 0$$
Detail solution
Expand the expression in the equation
$$\left(- 12 y - \left(y - 7\right)^{2}\right) + 36 = 0$$
We get the quadratic equation
$$- y^{2} + 2 y - 13 = 0$$
This equation is of the form
A quadratic equation can be solved
using the discriminant.
The roots of the quadratic equation:
$$y_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$y_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = -1$$
$$b = 2$$
$$c = -13$$
, then
Because D<0, then the equation
has no real roots,
but complex roots is exists.
or
$$y_{1} = 1 - 2 \sqrt{3} i$$
$$y_{2} = 1 + 2 \sqrt{3} i$$
$$\left(- 12 y - \left(y - 7\right)^{2}\right) + 36 = 0$$
We get the quadratic equation
$$- y^{2} + 2 y - 13 = 0$$
This equation is of the form
a*y^2 + b*y + c = 0
A quadratic equation can be solved
using the discriminant.
The roots of the quadratic equation:
$$y_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$y_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = -1$$
$$b = 2$$
$$c = -13$$
, then
D = b^2 - 4 * a * c =
(2)^2 - 4 * (-1) * (-13) = -48
Because D<0, then the equation
has no real roots,
but complex roots is exists.
y1 = (-b + sqrt(D)) / (2*a)
y2 = (-b - sqrt(D)) / (2*a)
or
$$y_{1} = 1 - 2 \sqrt{3} i$$
$$y_{2} = 1 + 2 \sqrt{3} i$$
Rapid solution
[src]
___ y1 = 1 - 2*I*\/ 3
$$y_{1} = 1 - 2 \sqrt{3} i$$
___ y2 = 1 + 2*I*\/ 3
$$y_{2} = 1 + 2 \sqrt{3} i$$
y2 = 1 + 2*sqrt(3)*i
Sum and product of roots
[src]
sum
___ ___ 1 - 2*I*\/ 3 + 1 + 2*I*\/ 3
$$\left(1 - 2 \sqrt{3} i\right) + \left(1 + 2 \sqrt{3} i\right)$$
=
2
$$2$$
product
/ ___\ / ___\ \1 - 2*I*\/ 3 /*\1 + 2*I*\/ 3 /
$$\left(1 - 2 \sqrt{3} i\right) \left(1 + 2 \sqrt{3} i\right)$$
=
13
$$13$$
13
Numerical answer
[src]
y1 = 1.0 - 3.46410161513775*i
y2 = 1.0 + 3.46410161513775*i
y2 = 1.0 + 3.46410161513775*i