Mister Exam
Other calculators
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How to use it?
- Equation:
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Equation -x=4-3(3x-8)
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Equation x/9+x=-10/3
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Equation √(x^2-1)=2*x
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Equation 4(x-3)=x+6
- Express {x} in terms of y where:
- -6*x-20*y=8
- -9*x-15*y=-4
- 1*x+5*y=-11
- 12*x-14*y=20
- Graphing y =:
- 4^x-3*2^x+2
- Factor polynomial:
- 4^x-3*2^x+2
-
Identical expressions
- four ^x- three * two ^x+ two = zero
- 4 to the power of x minus 3 multiply by 2 to the power of x plus 2 equally 0
- four to the power of x minus three multiply by two to the power of x plus two equally zero
- 4x-3*2x+2=0
- 4^x-32^x+2=0
- 4x-32x+2=0
- 4^x-3*2^x+2=O
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Similar expressions
- 4^x+3*2^x+2=0
- 4^x-3*2^x-2=0
4^x-3*2^x+2=0 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given the equation:
$$\left(- 3 \cdot 2^{x} + 4^{x}\right) + 2 = 0$$
or
$$\left(- 3 \cdot 2^{x} + 4^{x}\right) + 2 = 0$$
Do replacement
$$v = 2^{x}$$
we get
$$v^{2} - 3 v + 2 = 0$$
or
$$v^{2} - 3 v + 2 = 0$$
This equation is of the form
A quadratic equation can be solved
using the discriminant.
The roots of the quadratic equation:
$$v_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$v_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = 1$$
$$b = -3$$
$$c = 2$$
, then
Because D > 0, then the equation has two roots.
or
$$v_{1} = 2$$
$$v_{2} = 1$$
do backward replacement
$$2^{x} = v$$
or
$$x = \frac{\log{\left(v \right)}}{\log{\left(2 \right)}}$$
The final answer
$$x_{1} = \frac{\log{\left(1 \right)}}{\log{\left(2 \right)}} = 0$$
$$x_{2} = \frac{\log{\left(2 \right)}}{\log{\left(2 \right)}} = 1$$
$$\left(- 3 \cdot 2^{x} + 4^{x}\right) + 2 = 0$$
or
$$\left(- 3 \cdot 2^{x} + 4^{x}\right) + 2 = 0$$
Do replacement
$$v = 2^{x}$$
we get
$$v^{2} - 3 v + 2 = 0$$
or
$$v^{2} - 3 v + 2 = 0$$
This equation is of the form
a*v^2 + b*v + c = 0
A quadratic equation can be solved
using the discriminant.
The roots of the quadratic equation:
$$v_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$v_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = 1$$
$$b = -3$$
$$c = 2$$
, then
D = b^2 - 4 * a * c =
(-3)^2 - 4 * (1) * (2) = 1
Because D > 0, then the equation has two roots.
v1 = (-b + sqrt(D)) / (2*a)
v2 = (-b - sqrt(D)) / (2*a)
or
$$v_{1} = 2$$
$$v_{2} = 1$$
do backward replacement
$$2^{x} = v$$
or
$$x = \frac{\log{\left(v \right)}}{\log{\left(2 \right)}}$$
The final answer
$$x_{1} = \frac{\log{\left(1 \right)}}{\log{\left(2 \right)}} = 0$$
$$x_{2} = \frac{\log{\left(2 \right)}}{\log{\left(2 \right)}} = 1$$
The graph