Mister Exam
Other calculators
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How to use it?
- Equation:
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Equation 7(4x-1)=6-2(3-14x)
- Equation 836-16x=420
- Equation 6p²-p-2=0
- Equation 2x^3-50x=0
- Express {x} in terms of y where:
- (x^2+y^2-1)^3+x^2*y^3=0
- -1*x-19*y=-10
- 4*x-16*y=6
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Identical expressions
- 6p²-p- two = zero
- 6p² minus p minus 2 equally 0
- 6p² minus p minus two equally zero
- 6p²-p-2=O
-
Similar expressions
- 6p²+p-2=0
- 6p²-p+2=0
6p²-p-2=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:
$$p_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$p_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = 6$$
$$b = -1$$
$$c = -2$$
, then
Because D > 0, then the equation has two roots.
or
$$p_{1} = \frac{2}{3}$$
$$p_{2} = - \frac{1}{2}$$
a*p^2 + b*p + c = 0
A quadratic equation can be solved
using the discriminant.
The roots of the quadratic equation:
$$p_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$p_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = 6$$
$$b = -1$$
$$c = -2$$
, then
D = b^2 - 4 * a * c =
(-1)^2 - 4 * (6) * (-2) = 49
Because D > 0, then the equation has two roots.
p1 = (-b + sqrt(D)) / (2*a)
p2 = (-b - sqrt(D)) / (2*a)
or
$$p_{1} = \frac{2}{3}$$
$$p_{2} = - \frac{1}{2}$$
Vieta's Theorem
rewrite the equation
$$\left(6 p^{2} - p\right) - 2 = 0$$
of
$$a p^{2} + b p + c = 0$$
as reduced quadratic equation
$$p^{2} + \frac{b p}{a} + \frac{c}{a} = 0$$
$$p^{2} - \frac{p}{6} - \frac{1}{3} = 0$$
$$2 p^{2} + q = 0$$
where
$$p = \frac{b}{a}$$
$$p = - \frac{1}{6}$$
$$q = \frac{c}{a}$$
$$q = - \frac{1}{3}$$
Vieta Formulas
$$p_{1} + p_{2} = - p$$
$$p_{1} p_{2} = q$$
$$p_{1} + p_{2} = \frac{1}{6}$$
$$p_{1} p_{2} = - \frac{1}{3}$$
$$\left(6 p^{2} - p\right) - 2 = 0$$
of
$$a p^{2} + b p + c = 0$$
as reduced quadratic equation
$$p^{2} + \frac{b p}{a} + \frac{c}{a} = 0$$
$$p^{2} - \frac{p}{6} - \frac{1}{3} = 0$$
$$2 p^{2} + q = 0$$
where
$$p = \frac{b}{a}$$
$$p = - \frac{1}{6}$$
$$q = \frac{c}{a}$$
$$q = - \frac{1}{3}$$
Vieta Formulas
$$p_{1} + p_{2} = - p$$
$$p_{1} p_{2} = q$$
$$p_{1} + p_{2} = \frac{1}{6}$$
$$p_{1} p_{2} = - \frac{1}{3}$$
Sum and product of roots
[src]
sum
-1/2 + 2/3
$$- \frac{1}{2} + \frac{2}{3}$$
=
1/6
$$\frac{1}{6}$$
product
-2 --- 2*3
$$- \frac{1}{3}$$
=
-1/3
$$- \frac{1}{3}$$
-1/3