Mister Exam
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How to use it?
- Equation:
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Equation (2log3*2log3)*(8sinx−√3)−7log3*(8sinx−√3)+6=0
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Equation tg(3x+(pi)/6)=1
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Equation 20*x^2-21*x+22=17x^2-75x-218
- Equation 2x+1/3-4x-x^2/12=x^2-4/9
- Express {x} in terms of y where:
- 8*x-7*y=-13
- 15*x-15*y=16
- 9*x+18*y=6
- (x^2+y^2-1)^3+x^2*y^3=0
- Factor polynomial:
- x^2-4/9
-
Identical expressions
- two x+ one / three - four x-x^ two / twelve =x^2-4/ nine
- 2x plus 1 divide by 3 minus 4x minus x squared divide by 12 equally x squared minus 4 divide by 9
- two x plus one divide by three minus four x minus x to the power of two divide by twelve equally x squared minus 4 divide by nine
- 2x+1/3-4x-x2/12=x2-4/9
- 2x+1/3-4x-x²/12=x²-4/9
- 2x+1/3-4x-x to the power of 2/12=x to the power of 2-4/9
- 2x+1 divide by 3-4x-x^2 divide by 12=x^2-4 divide by 9
-
Similar expressions
- 2x-1/3-4x-x^2/12=x^2-4/9
- 2x+1/3+4x-x^2/12=x^2-4/9
- 2x+1/3-4x+x^2/12=x^2-4/9
- 2x+1/3-4x-x^2/12=x^2+4/9
2x+1/3-4x-x^2/12=x^2-4/9 equation
The teacher will be very surprised to see your correct solution 😉
The solution
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[src]
2
x 2 4
2*x + 1/3 - 4*x - -- = x - -
12 9
$$- \frac{x^{2}}{12} + \left(- 4 x + \left(2 x + \frac{1}{3}\right)\right) = x^{2} - \frac{4}{9}$$
Detail solution
Move right part of the equation to
left part with negative sign.
The equation is transformed from
$$- \frac{x^{2}}{12} + \left(- 4 x + \left(2 x + \frac{1}{3}\right)\right) = x^{2} - \frac{4}{9}$$
to
$$\left(\frac{4}{9} - x^{2}\right) + \left(- \frac{x^{2}}{12} + \left(- 4 x + \left(2 x + \frac{1}{3}\right)\right)\right) = 0$$
Expand the expression in the equation
$$\left(\frac{4}{9} - x^{2}\right) + \left(- \frac{x^{2}}{12} + \left(- 4 x + \left(2 x + \frac{1}{3}\right)\right)\right) = 0$$
We get the quadratic equation
$$- \frac{13 x^{2}}{12} - 2 x + \frac{7}{9} = 0$$
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
$$a = - \frac{13}{12}$$
$$b = -2$$
$$c = \frac{7}{9}$$
, then
Because D > 0, then the equation has two roots.
or
$$x_{1} = - \frac{2 \sqrt{597}}{39} - \frac{12}{13}$$
$$x_{2} = - \frac{12}{13} + \frac{2 \sqrt{597}}{39}$$
left part with negative sign.
The equation is transformed from
$$- \frac{x^{2}}{12} + \left(- 4 x + \left(2 x + \frac{1}{3}\right)\right) = x^{2} - \frac{4}{9}$$
to
$$\left(\frac{4}{9} - x^{2}\right) + \left(- \frac{x^{2}}{12} + \left(- 4 x + \left(2 x + \frac{1}{3}\right)\right)\right) = 0$$
Expand the expression in the equation
$$\left(\frac{4}{9} - x^{2}\right) + \left(- \frac{x^{2}}{12} + \left(- 4 x + \left(2 x + \frac{1}{3}\right)\right)\right) = 0$$
We get the quadratic equation
$$- \frac{13 x^{2}}{12} - 2 x + \frac{7}{9} = 0$$
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
$$a = - \frac{13}{12}$$
$$b = -2$$
$$c = \frac{7}{9}$$
, then
D = b^2 - 4 * a * c =
(-2)^2 - 4 * (-13/12) * (7/9) = 199/27
Because D > 0, then the equation has two roots.
x1 = (-b + sqrt(D)) / (2*a)
x2 = (-b - sqrt(D)) / (2*a)
or
$$x_{1} = - \frac{2 \sqrt{597}}{39} - \frac{12}{13}$$
$$x_{2} = - \frac{12}{13} + \frac{2 \sqrt{597}}{39}$$
Vieta's Theorem
rewrite the equation
$$- \frac{x^{2}}{12} + \left(- 4 x + \left(2 x + \frac{1}{3}\right)\right) = x^{2} - \frac{4}{9}$$
of
$$a x^{2} + b x + c = 0$$
as reduced quadratic equation
$$x^{2} + \frac{b x}{a} + \frac{c}{a} = 0$$
$$x^{2} + \frac{24 x}{13} - \frac{28}{39} = 0$$
$$p x + q + x^{2} = 0$$
where
$$p = \frac{b}{a}$$
$$p = \frac{24}{13}$$
$$q = \frac{c}{a}$$
$$q = - \frac{28}{39}$$
Vieta Formulas
$$x_{1} + x_{2} = - p$$
$$x_{1} x_{2} = q$$
$$x_{1} + x_{2} = - \frac{24}{13}$$
$$x_{1} x_{2} = - \frac{28}{39}$$
$$- \frac{x^{2}}{12} + \left(- 4 x + \left(2 x + \frac{1}{3}\right)\right) = x^{2} - \frac{4}{9}$$
of
$$a x^{2} + b x + c = 0$$
as reduced quadratic equation
$$x^{2} + \frac{b x}{a} + \frac{c}{a} = 0$$
$$x^{2} + \frac{24 x}{13} - \frac{28}{39} = 0$$
$$p x + q + x^{2} = 0$$
where
$$p = \frac{b}{a}$$
$$p = \frac{24}{13}$$
$$q = \frac{c}{a}$$
$$q = - \frac{28}{39}$$
Vieta Formulas
$$x_{1} + x_{2} = - p$$
$$x_{1} x_{2} = q$$
$$x_{1} + x_{2} = - \frac{24}{13}$$
$$x_{1} x_{2} = - \frac{28}{39}$$
Sum and product of roots
[src]
sum
_____ _____ 12 2*\/ 597 12 2*\/ 597 - -- + --------- + - -- - --------- 13 39 13 39
$$\left(- \frac{2 \sqrt{597}}{39} - \frac{12}{13}\right) + \left(- \frac{12}{13} + \frac{2 \sqrt{597}}{39}\right)$$
=
-24 ---- 13
$$- \frac{24}{13}$$
product
/ _____\ / _____\ | 12 2*\/ 597 | | 12 2*\/ 597 | |- -- + ---------|*|- -- - ---------| \ 13 39 / \ 13 39 /
$$\left(- \frac{12}{13} + \frac{2 \sqrt{597}}{39}\right) \left(- \frac{2 \sqrt{597}}{39} - \frac{12}{13}\right)$$
=
-28 ---- 39
$$- \frac{28}{39}$$
-28/39
Rapid solution
[src]
_____
12 2*\/ 597
x1 = - -- + ---------
13 39
$$x_{1} = - \frac{12}{13} + \frac{2 \sqrt{597}}{39}$$
_____
12 2*\/ 597
x2 = - -- - ---------
13 39
$$x_{2} = - \frac{2 \sqrt{597}}{39} - \frac{12}{13}$$
x2 = -2*sqrt(597)/39 - 12/13