Mister Exam
Other calculators
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How to use it?
- Equation:
-
Equation x+12=5
-
Equation x^3=4
-
Equation 0,144:(3,4-x)=2,4
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Equation x^4=(x-56)^2
- Express {x} in terms of y where:
- 15*x-5*y=-5
- 18*x+20*y=-2
- -5*x-6*y=-1
- -17*x-15*y=-17
-
Identical expressions
- x^ two -(ten / three)*x- one = zero
- x squared minus (10 divide by 3) multiply by x minus 1 equally 0
- x to the power of two minus (ten divide by three) multiply by x minus one equally zero
- x2-(10/3)*x-1=0
- x2-10/3*x-1=0
- x²-(10/3)*x-1=0
- x to the power of 2-(10/3)*x-1=0
- x^2-(10/3)x-1=0
- x2-(10/3)x-1=0
- x2-10/3x-1=0
- x^2-10/3x-1=0
- x^2-(10/3)*x-1=O
- x^2-(10 divide by 3)*x-1=0
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Similar expressions
- x^2+(10/3)*x-1=0
- x^2-(10/3)*x+1=0
x^2-(10/3)*x-1=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
$$a = 1$$
$$b = - \frac{10}{3}$$
$$c = -1$$
, then
Because D > 0, then the equation has two roots.
or
$$x_{1} = \frac{5}{3} + \frac{\sqrt{34}}{3}$$
$$x_{2} = \frac{5}{3} - \frac{\sqrt{34}}{3}$$
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
$$a = 1$$
$$b = - \frac{10}{3}$$
$$c = -1$$
, then
D = b^2 - 4 * a * c =
(-10/3)^2 - 4 * (1) * (-1) = 136/9
Because D > 0, then the equation has two roots.
x1 = (-b + sqrt(D)) / (2*a)
x2 = (-b - sqrt(D)) / (2*a)
or
$$x_{1} = \frac{5}{3} + \frac{\sqrt{34}}{3}$$
$$x_{2} = \frac{5}{3} - \frac{\sqrt{34}}{3}$$
Vieta's Theorem
it is reduced quadratic equation
$$p x + q + x^{2} = 0$$
where
$$p = \frac{b}{a}$$
$$p = - \frac{10}{3}$$
$$q = \frac{c}{a}$$
$$q = -1$$
Vieta Formulas
$$x_{1} + x_{2} = - p$$
$$x_{1} x_{2} = q$$
$$x_{1} + x_{2} = \frac{10}{3}$$
$$x_{1} x_{2} = -1$$
$$p x + q + x^{2} = 0$$
where
$$p = \frac{b}{a}$$
$$p = - \frac{10}{3}$$
$$q = \frac{c}{a}$$
$$q = -1$$
Vieta Formulas
$$x_{1} + x_{2} = - p$$
$$x_{1} x_{2} = q$$
$$x_{1} + x_{2} = \frac{10}{3}$$
$$x_{1} x_{2} = -1$$
Rapid solution
[src]
____
5 \/ 34
x1 = - - ------
3 3
$$x_{1} = \frac{5}{3} - \frac{\sqrt{34}}{3}$$
____
5 \/ 34
x2 = - + ------
3 3
$$x_{2} = \frac{5}{3} + \frac{\sqrt{34}}{3}$$
x2 = 5/3 + sqrt(34)/3
Sum and product of roots
[src]
sum
____ ____ 5 \/ 34 5 \/ 34 - - ------ + - + ------ 3 3 3 3
$$\left(\frac{5}{3} - \frac{\sqrt{34}}{3}\right) + \left(\frac{5}{3} + \frac{\sqrt{34}}{3}\right)$$
=
10/3
$$\frac{10}{3}$$
product
/ ____\ / ____\ |5 \/ 34 | |5 \/ 34 | |- - ------|*|- + ------| \3 3 / \3 3 /
$$\left(\frac{5}{3} - \frac{\sqrt{34}}{3}\right) \left(\frac{5}{3} + \frac{\sqrt{34}}{3}\right)$$
=
-1
$$-1$$
-1