Mister Exam
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How to use it?
- Equation:
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Equation 7*x^3-28*x=0
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Equation (x^2-9)^2+(x^2+x-6)^2=0
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Equation x^3+x^2-x-2=0
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Equation 60/x-60/(x+10)=1/5
- Express {x} in terms of y where:
- -5*x+14*y=15
- 15*x+14*y=-16
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Identical expressions
- (x+ zero , six)(x- nine)= zero
- (x plus 0,6)(x minus 9) equally 0
- (x plus zero , six)(x minus nine) equally zero
- x+0,6x-9=0
- (x+0,6)(x-9)=O
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Similar expressions
- (x-0,6)(x-9)=0
- (x+0,6)(x+9)=0
(x+0,6)(x-9)=0 equation
The teacher will be very surprised to see your correct solution 😉
The solution
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[src]
(x + 3/5)*(x - 9) = 0
$$\left(x - 9\right) \left(x + \frac{3}{5}\right) = 0$$
Detail solution
Expand the expression in the equation
$$\left(x - 9\right) \left(x + \frac{3}{5}\right) = 0$$
We get the quadratic equation
$$x^{2} - \frac{42 x}{5} - \frac{27}{5} = 0$$
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
$$a = 1$$
$$b = - \frac{42}{5}$$
$$c = - \frac{27}{5}$$
, then
Because D > 0, then the equation has two roots.
or
$$x_{1} = 9$$
$$x_{2} = - \frac{3}{5}$$
$$\left(x - 9\right) \left(x + \frac{3}{5}\right) = 0$$
We get the quadratic equation
$$x^{2} - \frac{42 x}{5} - \frac{27}{5} = 0$$
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
$$a = 1$$
$$b = - \frac{42}{5}$$
$$c = - \frac{27}{5}$$
, then
D = b^2 - 4 * a * c =
(-42/5)^2 - 4 * (1) * (-27/5) = 2304/25
Because D > 0, then the equation has two roots.
x1 = (-b + sqrt(D)) / (2*a)
x2 = (-b - sqrt(D)) / (2*a)
or
$$x_{1} = 9$$
$$x_{2} = - \frac{3}{5}$$
Sum and product of roots
[src]
sum
9 - 3/5
$$- \frac{3}{5} + 9$$
=
42/5
$$\frac{42}{5}$$
product
9*(-3) ------ 5
$$\frac{\left(-3\right) 9}{5}$$
=
-27/5
$$- \frac{27}{5}$$
-27/5