Mister Exam
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How to use it?
- Equation:
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Equation e^x=4
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Equation x^3=x^2-7x+7
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Identical expressions
- two x^2- nine *x- eight = zero
- 2x squared minus 9 multiply by x minus 8 equally 0
- two x squared minus nine multiply by x minus eight equally zero
- 2x2-9*x-8=0
- 2x²-9*x-8=0
- 2x to the power of 2-9*x-8=0
- 2x^2-9x-8=0
- 2x2-9x-8=0
- 2x^2-9*x-8=O
-
Similar expressions
- 2x^2-9*x+8=0
- 2x^2+9*x-8=0
2x^2-9*x-8=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
$$a = 2$$
$$b = -9$$
$$c = -8$$
, then
Because D > 0, then the equation has two roots.
or
$$x_{1} = \frac{9}{4} + \frac{\sqrt{145}}{4}$$
$$x_{2} = \frac{9}{4} - \frac{\sqrt{145}}{4}$$
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
$$a = 2$$
$$b = -9$$
$$c = -8$$
, then
D = b^2 - 4 * a * c =
(-9)^2 - 4 * (2) * (-8) = 145
Because D > 0, then the equation has two roots.
x1 = (-b + sqrt(D)) / (2*a)
x2 = (-b - sqrt(D)) / (2*a)
or
$$x_{1} = \frac{9}{4} + \frac{\sqrt{145}}{4}$$
$$x_{2} = \frac{9}{4} - \frac{\sqrt{145}}{4}$$
Vieta's Theorem
rewrite the equation
$$\left(2 x^{2} - 9 x\right) - 8 = 0$$
of
$$a x^{2} + b x + c = 0$$
as reduced quadratic equation
$$x^{2} + \frac{b x}{a} + \frac{c}{a} = 0$$
$$x^{2} - \frac{9 x}{2} - 4 = 0$$
$$p x + q + x^{2} = 0$$
where
$$p = \frac{b}{a}$$
$$p = - \frac{9}{2}$$
$$q = \frac{c}{a}$$
$$q = -4$$
Vieta Formulas
$$x_{1} + x_{2} = - p$$
$$x_{1} x_{2} = q$$
$$x_{1} + x_{2} = \frac{9}{2}$$
$$x_{1} x_{2} = -4$$
$$\left(2 x^{2} - 9 x\right) - 8 = 0$$
of
$$a x^{2} + b x + c = 0$$
as reduced quadratic equation
$$x^{2} + \frac{b x}{a} + \frac{c}{a} = 0$$
$$x^{2} - \frac{9 x}{2} - 4 = 0$$
$$p x + q + x^{2} = 0$$
where
$$p = \frac{b}{a}$$
$$p = - \frac{9}{2}$$
$$q = \frac{c}{a}$$
$$q = -4$$
Vieta Formulas
$$x_{1} + x_{2} = - p$$
$$x_{1} x_{2} = q$$
$$x_{1} + x_{2} = \frac{9}{2}$$
$$x_{1} x_{2} = -4$$
Sum and product of roots
[src]
sum
_____ _____ 9 \/ 145 9 \/ 145 - - ------- + - + ------- 4 4 4 4
$$\left(\frac{9}{4} - \frac{\sqrt{145}}{4}\right) + \left(\frac{9}{4} + \frac{\sqrt{145}}{4}\right)$$
=
9/2
$$\frac{9}{2}$$
product
/ _____\ / _____\ |9 \/ 145 | |9 \/ 145 | |- - -------|*|- + -------| \4 4 / \4 4 /
$$\left(\frac{9}{4} - \frac{\sqrt{145}}{4}\right) \left(\frac{9}{4} + \frac{\sqrt{145}}{4}\right)$$
=
-4
$$-4$$
-4
Rapid solution
[src]
_____
9 \/ 145
x1 = - - -------
4 4
$$x_{1} = \frac{9}{4} - \frac{\sqrt{145}}{4}$$
_____
9 \/ 145
x2 = - + -------
4 4
$$x_{2} = \frac{9}{4} + \frac{\sqrt{145}}{4}$$
x2 = 9/4 + sqrt(145)/4