Mister Exam
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How to use it?
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Identical expressions
- 3x²-2x- forty = zero
- 3x² minus 2x minus 40 equally 0
- 3x² minus 2x minus forty equally zero
- 3x²-2x-40=O
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Similar expressions
- 3x²-2x+40=0
- 3x²+2x-40=0
3x²-2x-40=0 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
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
$$a = 3$$
$$b = -2$$
$$c = -40$$
, then
Because D > 0, then the equation has two roots.
or
$$x_{1} = 4$$
$$x_{2} = - \frac{10}{3}$$
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
$$a = 3$$
$$b = -2$$
$$c = -40$$
, then
D = b^2 - 4 * a * c =
(-2)^2 - 4 * (3) * (-40) = 484
Because D > 0, then the equation has two roots.
x1 = (-b + sqrt(D)) / (2*a)
x2 = (-b - sqrt(D)) / (2*a)
or
$$x_{1} = 4$$
$$x_{2} = - \frac{10}{3}$$
Vieta's Theorem
rewrite the equation
$$\left(3 x^{2} - 2 x\right) - 40 = 0$$
of
$$a x^{2} + b x + c = 0$$
as reduced quadratic equation
$$x^{2} + \frac{b x}{a} + \frac{c}{a} = 0$$
$$x^{2} - \frac{2 x}{3} - \frac{40}{3} = 0$$
$$p x + q + x^{2} = 0$$
where
$$p = \frac{b}{a}$$
$$p = - \frac{2}{3}$$
$$q = \frac{c}{a}$$
$$q = - \frac{40}{3}$$
Vieta Formulas
$$x_{1} + x_{2} = - p$$
$$x_{1} x_{2} = q$$
$$x_{1} + x_{2} = \frac{2}{3}$$
$$x_{1} x_{2} = - \frac{40}{3}$$
$$\left(3 x^{2} - 2 x\right) - 40 = 0$$
of
$$a x^{2} + b x + c = 0$$
as reduced quadratic equation
$$x^{2} + \frac{b x}{a} + \frac{c}{a} = 0$$
$$x^{2} - \frac{2 x}{3} - \frac{40}{3} = 0$$
$$p x + q + x^{2} = 0$$
where
$$p = \frac{b}{a}$$
$$p = - \frac{2}{3}$$
$$q = \frac{c}{a}$$
$$q = - \frac{40}{3}$$
Vieta Formulas
$$x_{1} + x_{2} = - p$$
$$x_{1} x_{2} = q$$
$$x_{1} + x_{2} = \frac{2}{3}$$
$$x_{1} x_{2} = - \frac{40}{3}$$
Sum and product of roots
[src]
sum
4 - 10/3
$$- \frac{10}{3} + 4$$
=
2/3
$$\frac{2}{3}$$
product
4*(-10) ------- 3
$$\frac{\left(-10\right) 4}{3}$$
=
-40/3
$$- \frac{40}{3}$$
-40/3