Mister Exam
Other calculators
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How to use it?
- Equation:
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Equation x^6=(3x-2)^3
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Equation 2x^2+5x-2=0
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Equation 1/4*x^2-4=0
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Equation x^2-10x=0
- Express {x} in terms of y where:
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- -16*x+1*y=-14
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Identical expressions
- (3x- one)(four -5x)= zero
- (3x minus 1)(4 minus 5x) equally 0
- (3x minus one)(four minus 5x) equally zero
- 3x-14-5x=0
- (3x-1)(4-5x)=O
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Similar expressions
- (3x+1)(4-5x)=0
- (3x-1)(4+5x)=0
(3x-1)(4-5x)=0 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Expand the expression in the equation
$$\left(4 - 5 x\right) \left(3 x - 1\right) = 0$$
We get the quadratic equation
$$- 15 x^{2} + 17 x - 4 = 0$$
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
$$a = -15$$
$$b = 17$$
$$c = -4$$
, then
Because D > 0, then the equation has two roots.
or
$$x_{1} = \frac{1}{3}$$
$$x_{2} = \frac{4}{5}$$
$$\left(4 - 5 x\right) \left(3 x - 1\right) = 0$$
We get the quadratic equation
$$- 15 x^{2} + 17 x - 4 = 0$$
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
$$a = -15$$
$$b = 17$$
$$c = -4$$
, then
D = b^2 - 4 * a * c =
(17)^2 - 4 * (-15) * (-4) = 49
Because D > 0, then the equation has two roots.
x1 = (-b + sqrt(D)) / (2*a)
x2 = (-b - sqrt(D)) / (2*a)
or
$$x_{1} = \frac{1}{3}$$
$$x_{2} = \frac{4}{5}$$
Sum and product of roots
[src]
sum
1/3 + 4/5
$$\frac{1}{3} + \frac{4}{5}$$
=
17 -- 15
$$\frac{17}{15}$$
product
4 --- 3*5
$$\frac{4}{3 \cdot 5}$$
=
4/15
$$\frac{4}{15}$$
4/15