Mister Exam
Other calculators
-
How to use it?
- Equation:
-
Equation 2x-3=10
-
Equation 4*x^4-5*x^2+1=0
-
Equation 5*x^2+7*x+2=0
-
Equation 3/8*x+15=1/6*x+10
- Express {x} in terms of y where:
- -2*x+4*y=3
- -14*x-13*y=11
- -1*x-5*y=-1
- -12*x-6*y=-3
-
Identical expressions
- (2x- three)(four -3x)= zero
- (2x minus 3)(4 minus 3x) equally 0
- (2x minus three)(four minus 3x) equally zero
- 2x-34-3x=0
- (2x-3)(4-3x)=O
-
Similar expressions
- (2x-3)(4+3x)=0
- (2x-3)*(4-3x)=0
- (2x+3)(4-3x)=0
(2x-3)(4-3x)=0 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Expand the expression in the equation
$$\left(4 - 3 x\right) \left(2 x - 3\right) = 0$$
We get the quadratic equation
$$- 6 x^{2} + 17 x - 12 = 0$$
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
$$a = -6$$
$$b = 17$$
$$c = -12$$
, then
Because D > 0, then the equation has two roots.
or
$$x_{1} = \frac{4}{3}$$
$$x_{2} = \frac{3}{2}$$
$$\left(4 - 3 x\right) \left(2 x - 3\right) = 0$$
We get the quadratic equation
$$- 6 x^{2} + 17 x - 12 = 0$$
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
$$a = -6$$
$$b = 17$$
$$c = -12$$
, then
D = b^2 - 4 * a * c =
(17)^2 - 4 * (-6) * (-12) = 1
Because D > 0, then the equation has two roots.
x1 = (-b + sqrt(D)) / (2*a)
x2 = (-b - sqrt(D)) / (2*a)
or
$$x_{1} = \frac{4}{3}$$
$$x_{2} = \frac{3}{2}$$
Sum and product of roots
[src]
sum
4/3 + 3/2
$$\frac{4}{3} + \frac{3}{2}$$
=
17/6
$$\frac{17}{6}$$
product
4*3 --- 3*2
$$\frac{3 \cdot 4}{2 \cdot 3}$$
=
2
$$2$$
2