Mister Exam
Other calculators
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How to use it?
- Equation:
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Equation x/4+x=4
-
Equation x^2+11=0
- Equation x+y=5
-
Equation 1*(x-1)*(x-3)=0
- Express {x} in terms of y where:
- -11*x-3*y=14
- -1*x-17*y=-12
- 20*x+18*y=-9
- -2*x+20*y=-2
-
Identical expressions
- (x- two)^ two - eight *(|x- two |)+ fifteen = zero
- (x minus 2) squared minus 8 multiply by ( module of x minus 2|) plus 15 equally 0
- (x minus two) to the power of two minus eight multiply by ( module of x minus two |) plus fifteen equally zero
- (x-2)2-8*(|x-2|)+15=0
- x-22-8*|x-2|+15=0
- (x-2)²-8*(|x-2|)+15=0
- (x-2) to the power of 2-8*(|x-2|)+15=0
- (x-2)^2-8(|x-2|)+15=0
- (x-2)2-8(|x-2|)+15=0
- x-22-8|x-2|+15=0
- x-2^2-8|x-2|+15=0
- (x-2)^2-8*(|x-2|)+15=O
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Similar expressions
- (x+2)^2-8*(|x-2|)+15=0
- (x-2)^2+8*(|x-2|)+15=0
- (x-2)^2-8*(|x-2|)-15=0
- (x-2)^2-8*(|x+2|)+15=0
(x-2)^2-8*(|x-2|)+15=0 equation
The teacher will be very surprised to see your correct solution 😉
The solution
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[src]
2 (x - 2) - 8*|x - 2| + 15 = 0
$$\left(x - 2\right)^{2} - 8 \left|{x - 2}\right| + 15 = 0$$
Detail solution
For every modulo expressions in the equation
allow cases, when this expressions ">=0" or "<0",
solve the resulting equation.
1.
$$x - 2 \geq 0$$
or
$$2 \leq x \wedge x < \infty$$
we get the equation
$$\left(x - 2\right)^{2} - 8 \left(x - 2\right) + 15 = 0$$
after simplifying we get
$$- 8 x + \left(x - 2\right)^{2} + 31 = 0$$
the solution in this interval:
$$x_{1} = 5$$
$$x_{2} = 7$$
2.
$$x - 2 < 0$$
or
$$-\infty < x \wedge x < 2$$
we get the equation
$$- 8 \cdot \left(2 - x\right) + \left(x - 2\right)^{2} + 15 = 0$$
after simplifying we get
$$8 x + \left(x - 2\right)^{2} - 1 = 0$$
the solution in this interval:
$$x_{3} = -3$$
$$x_{4} = -1$$
The final answer:
$$x_{1} = 5$$
$$x_{2} = 7$$
$$x_{3} = -3$$
$$x_{4} = -1$$
allow cases, when this expressions ">=0" or "<0",
solve the resulting equation.
1.
$$x - 2 \geq 0$$
or
$$2 \leq x \wedge x < \infty$$
we get the equation
$$\left(x - 2\right)^{2} - 8 \left(x - 2\right) + 15 = 0$$
after simplifying we get
$$- 8 x + \left(x - 2\right)^{2} + 31 = 0$$
the solution in this interval:
$$x_{1} = 5$$
$$x_{2} = 7$$
2.
$$x - 2 < 0$$
or
$$-\infty < x \wedge x < 2$$
we get the equation
$$- 8 \cdot \left(2 - x\right) + \left(x - 2\right)^{2} + 15 = 0$$
after simplifying we get
$$8 x + \left(x - 2\right)^{2} - 1 = 0$$
the solution in this interval:
$$x_{3} = -3$$
$$x_{4} = -1$$
The final answer:
$$x_{1} = 5$$
$$x_{2} = 7$$
$$x_{3} = -3$$
$$x_{4} = -1$$
Rapid solution
[src]
x1 = -3
$$x_{1} = -3$$
x2 = -1
$$x_{2} = -1$$
x3 = 5
$$x_{3} = 5$$
x4 = 7
$$x_{4} = 7$$
Sum and product of roots
[src]
sum
0 - 3 - 1 + 5 + 7
$$\left(\left(\left(-3 + 0\right) - 1\right) + 5\right) + 7$$
=
8
$$8$$
product
1*-3*-1*5*7
$$1 \left(-3\right) \left(-1\right) 5 \cdot 7$$
=
105
$$105$$
105
The graph