Mister Exam
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How to use it?
- Equation:
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Equation (x+2)^2=(1-x)^2
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Equation 2*cos(x)+1=0
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Equation (-5*x-3)*(2*x-1)=0
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Equation x^2-8x+15=0
- Express {x} in terms of y where:
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Identical expressions
- (six *x- one)/(five -x)=(six *x+ three)/ nine + four / three
- (6 multiply by x minus 1) divide by (5 minus x) equally (6 multiply by x plus 3) divide by 9 plus 4 divide by 3
- (six multiply by x minus one) divide by (five minus x) equally (six multiply by x plus three) divide by nine plus four divide by three
- (6x-1)/(5-x)=(6x+3)/9+4/3
- 6x-1/5-x=6x+3/9+4/3
- (6*x-1) divide by (5-x)=(6*x+3) divide by 9+4 divide by 3
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Similar expressions
- (6*x-1)/(5-x)=(6*x-3)/9+4/3
- (6*x-1)/(5+x)=(6*x+3)/9+4/3
- (6*x-1)/(5-x)=(6*x+3)/9-4/3
- (6*x+1)/(5-x)=(6*x+3)/9+4/3
(6*x-1)/(5-x)=(6*x+3)/9+4/3 equation
The teacher will be very surprised to see your correct solution 😉
The solution
You have entered
[src]
6*x - 1 6*x + 3 4 ------- = ------- + - 5 - x 9 3
$$\frac{6 x - 1}{5 - x} = \frac{6 x + 3}{9} + \frac{4}{3}$$
Detail solution
Given the equation:
$$\frac{6 x - 1}{5 - x} = \frac{6 x + 3}{9} + \frac{4}{3}$$
Multiply the equation sides by the denominators:
5 - x
we get:
$$6 x - 1 = \left(5 - x\right) \left(\frac{6 x + 3}{9} + \frac{4}{3}\right)$$
$$6 x - 1 = - \frac{\left(x - 5\right) \left(2 x + 5\right)}{3}$$
Move right part of the equation to
left part with negative sign.
The equation is transformed from
$$6 x - 1 = - \frac{\left(x - 5\right) \left(2 x + 5\right)}{3}$$
to
$$\frac{2 x^{2}}{3} + \frac{13 x}{3} - \frac{28}{3} = 0$$
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
$$a = \frac{2}{3}$$
$$b = \frac{13}{3}$$
$$c = - \frac{28}{3}$$
, then
Because D > 0, then the equation has two roots.
or
$$x_{1} = - \frac{13}{4} + \frac{\sqrt{393}}{4}$$
$$x_{2} = - \frac{\sqrt{393}}{4} - \frac{13}{4}$$
$$\frac{6 x - 1}{5 - x} = \frac{6 x + 3}{9} + \frac{4}{3}$$
Multiply the equation sides by the denominators:
5 - x
we get:
$$6 x - 1 = \left(5 - x\right) \left(\frac{6 x + 3}{9} + \frac{4}{3}\right)$$
$$6 x - 1 = - \frac{\left(x - 5\right) \left(2 x + 5\right)}{3}$$
Move right part of the equation to
left part with negative sign.
The equation is transformed from
$$6 x - 1 = - \frac{\left(x - 5\right) \left(2 x + 5\right)}{3}$$
to
$$\frac{2 x^{2}}{3} + \frac{13 x}{3} - \frac{28}{3} = 0$$
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
$$a = \frac{2}{3}$$
$$b = \frac{13}{3}$$
$$c = - \frac{28}{3}$$
, then
D = b^2 - 4 * a * c =
(13/3)^2 - 4 * (2/3) * (-28/3) = 131/3
Because D > 0, then the equation has two roots.
x1 = (-b + sqrt(D)) / (2*a)
x2 = (-b - sqrt(D)) / (2*a)
or
$$x_{1} = - \frac{13}{4} + \frac{\sqrt{393}}{4}$$
$$x_{2} = - \frac{\sqrt{393}}{4} - \frac{13}{4}$$
Rapid solution
[src]
_____
13 \/ 393
x1 = - -- + -------
4 4
$$x_{1} = - \frac{13}{4} + \frac{\sqrt{393}}{4}$$
_____
13 \/ 393
x2 = - -- - -------
4 4
$$x_{2} = - \frac{\sqrt{393}}{4} - \frac{13}{4}$$
x2 = -sqrt(393)/4 - 13/4
Sum and product of roots
[src]
sum
_____ _____ 13 \/ 393 13 \/ 393 - -- + ------- + - -- - ------- 4 4 4 4
$$\left(- \frac{\sqrt{393}}{4} - \frac{13}{4}\right) + \left(- \frac{13}{4} + \frac{\sqrt{393}}{4}\right)$$
=
-13/2
$$- \frac{13}{2}$$
product
/ _____\ / _____\ | 13 \/ 393 | | 13 \/ 393 | |- -- + -------|*|- -- - -------| \ 4 4 / \ 4 4 /
$$\left(- \frac{13}{4} + \frac{\sqrt{393}}{4}\right) \left(- \frac{\sqrt{393}}{4} - \frac{13}{4}\right)$$
=
-14
$$-14$$
-14