Mister Exam
Other calculators
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How to use it?
- Equation:
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Equation 4*x^2-25=0
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Equation 8x-15=-12+5x
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Equation x^2-x-7=0
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Equation y+30*5=450
- Express {x} in terms of y where:
- 14*x+3*y=-1
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- 18*x-13*y=1
- -4*x-8*y=1
-
Identical expressions
- (3x- one)(-x- four)=v
- (3x minus 1)( minus x minus 4) equally v
- (3x minus one)( minus x minus four) equally v
- 3x-1-x-4=v
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Similar expressions
- (3x-1)(x-4)=v
- (3x+1)(-x-4)=v
- (3x-1)(-x+4)=v
(3x-1)(-x-4)=v equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given the linear equation:
Expand brackets in the left part
Looking for similar summands in the left part:
Move free summands (without v)
from left part to right part, we given:
$$\left(- x - 4\right) \left(3 x - 1\right) + 1 = v + 1$$
Move the summands with the unknown v
from the right part to the left part:
$$- v + \left(- x - 4\right) \left(3 x - 1\right) + 1 = 1$$
Divide both parts of the equation by (1 - v + (-1 + 3*x)*(-4 - x))/v
We get the answer: v = 4 - 11*x - 3*x^2
(3*x-1)*(-x-4) = v
Expand brackets in the left part
3*x-1-x-4 = v
Looking for similar summands in the left part:
(-1 + 3*x)*(-4 - x) = v
Move free summands (without v)
from left part to right part, we given:
$$\left(- x - 4\right) \left(3 x - 1\right) + 1 = v + 1$$
Move the summands with the unknown v
from the right part to the left part:
$$- v + \left(- x - 4\right) \left(3 x - 1\right) + 1 = 1$$
Divide both parts of the equation by (1 - v + (-1 + 3*x)*(-4 - x))/v
v = 1 / ((1 - v + (-1 + 3*x)*(-4 - x))/v)
We get the answer: v = 4 - 11*x - 3*x^2
The graph
Rapid solution
[src]
2 2 v1 = 4 - 11*re(x) - 3*re (x) + 3*im (x) + I*(-11*im(x) - 6*im(x)*re(x))
$$v_{1} = i \left(- 6 \operatorname{re}{\left(x\right)} \operatorname{im}{\left(x\right)} - 11 \operatorname{im}{\left(x\right)}\right) - 3 \left(\operatorname{re}{\left(x\right)}\right)^{2} - 11 \operatorname{re}{\left(x\right)} + 3 \left(\operatorname{im}{\left(x\right)}\right)^{2} + 4$$
v1 = i*(-6*re(x)*im(x) - 11*im(x)) - 3*re(x)^2 - 11*re(x) + 3*im(x)^2 + 4
Sum and product of roots
[src]
sum
2 2 4 - 11*re(x) - 3*re (x) + 3*im (x) + I*(-11*im(x) - 6*im(x)*re(x))
$$i \left(- 6 \operatorname{re}{\left(x\right)} \operatorname{im}{\left(x\right)} - 11 \operatorname{im}{\left(x\right)}\right) - 3 \left(\operatorname{re}{\left(x\right)}\right)^{2} - 11 \operatorname{re}{\left(x\right)} + 3 \left(\operatorname{im}{\left(x\right)}\right)^{2} + 4$$
=
2 2 4 - 11*re(x) - 3*re (x) + 3*im (x) + I*(-11*im(x) - 6*im(x)*re(x))
$$i \left(- 6 \operatorname{re}{\left(x\right)} \operatorname{im}{\left(x\right)} - 11 \operatorname{im}{\left(x\right)}\right) - 3 \left(\operatorname{re}{\left(x\right)}\right)^{2} - 11 \operatorname{re}{\left(x\right)} + 3 \left(\operatorname{im}{\left(x\right)}\right)^{2} + 4$$
product
2 2 4 - 11*re(x) - 3*re (x) + 3*im (x) + I*(-11*im(x) - 6*im(x)*re(x))
$$i \left(- 6 \operatorname{re}{\left(x\right)} \operatorname{im}{\left(x\right)} - 11 \operatorname{im}{\left(x\right)}\right) - 3 \left(\operatorname{re}{\left(x\right)}\right)^{2} - 11 \operatorname{re}{\left(x\right)} + 3 \left(\operatorname{im}{\left(x\right)}\right)^{2} + 4$$
=
2 2 4 - 11*re(x) - 3*re (x) + 3*im (x) - I*(11 + 6*re(x))*im(x)
$$- i \left(6 \operatorname{re}{\left(x\right)} + 11\right) \operatorname{im}{\left(x\right)} - 3 \left(\operatorname{re}{\left(x\right)}\right)^{2} - 11 \operatorname{re}{\left(x\right)} + 3 \left(\operatorname{im}{\left(x\right)}\right)^{2} + 4$$
4 - 11*re(x) - 3*re(x)^2 + 3*im(x)^2 - i*(11 + 6*re(x))*im(x)