Mister Exam
Other calculators
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How to use it?
- Equation:
-
Equation 18x+9=32x+14
- Equation sqrt(4x^2-9x+2)=x-2
- Equation 7x+|-10+2x|x=-4
- Equation 6x^2-42x+72=6(x-4)
- Express {x} in terms of y where:
- -1*x-19*y=-10
- 5*x+5*y=20
- 4*x-16*y=6
- sqrt(x^2+y^2+4x-6y-12)+(x^2+10x-27)=8-|y-9|+|y-3|
- Derivative of:
- -4
- Integral of d{x}:
-
-4
- Limit of the function:
- -4
-
Identical expressions
- 7x+|- ten +2x|x=- four
- 7x plus module of minus 10 plus 2x|x equally minus 4
- 7x plus module of minus ten plus 2x|x equally minus four
-
Similar expressions
- 7x+|-10-2x|x=-4
- 7x+|+10+2x|x=-4
- 7x+|-10+2x|x=+4
- 7x-|-10+2x|x=-4
7x+|-10+2x|x=-4 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
For every modulo expressions in the equation
allow cases, when this expressions ">=0" or "<0",
solve the resulting equation.
1.
$$2 x - 10 \geq 0$$
or
$$5 \leq x \wedge x < \infty$$
we get the equation
$$x \left(2 x - 10\right) + 7 x + 4 = 0$$
after simplifying we get
$$x \left(2 x - 10\right) + 7 x + 4 = 0$$
the solution in this interval:
$$x_{1} = \frac{3}{4} - \frac{\sqrt{23} i}{4}$$
but x1 not in the inequality interval
$$x_{2} = \frac{3}{4} + \frac{\sqrt{23} i}{4}$$
but x2 not in the inequality interval
2.
$$2 x - 10 < 0$$
or
$$-\infty < x \wedge x < 5$$
we get the equation
$$x \left(10 - 2 x\right) + 7 x + 4 = 0$$
after simplifying we get
$$x \left(10 - 2 x\right) + 7 x + 4 = 0$$
the solution in this interval:
$$x_{3} = \frac{17}{4} - \frac{\sqrt{321}}{4}$$
$$x_{4} = \frac{17}{4} + \frac{\sqrt{321}}{4}$$
but x4 not in the inequality interval
The final answer:
$$x_{1} = \frac{17}{4} - \frac{\sqrt{321}}{4}$$
allow cases, when this expressions ">=0" or "<0",
solve the resulting equation.
1.
$$2 x - 10 \geq 0$$
or
$$5 \leq x \wedge x < \infty$$
we get the equation
$$x \left(2 x - 10\right) + 7 x + 4 = 0$$
after simplifying we get
$$x \left(2 x - 10\right) + 7 x + 4 = 0$$
the solution in this interval:
$$x_{1} = \frac{3}{4} - \frac{\sqrt{23} i}{4}$$
but x1 not in the inequality interval
$$x_{2} = \frac{3}{4} + \frac{\sqrt{23} i}{4}$$
but x2 not in the inequality interval
2.
$$2 x - 10 < 0$$
or
$$-\infty < x \wedge x < 5$$
we get the equation
$$x \left(10 - 2 x\right) + 7 x + 4 = 0$$
after simplifying we get
$$x \left(10 - 2 x\right) + 7 x + 4 = 0$$
the solution in this interval:
$$x_{3} = \frac{17}{4} - \frac{\sqrt{321}}{4}$$
$$x_{4} = \frac{17}{4} + \frac{\sqrt{321}}{4}$$
but x4 not in the inequality interval
The final answer:
$$x_{1} = \frac{17}{4} - \frac{\sqrt{321}}{4}$$
Sum and product of roots
[src]
sum
_____ 17 \/ 321 -- - ------- 4 4
$$\frac{17}{4} - \frac{\sqrt{321}}{4}$$
=
_____ 17 \/ 321 -- - ------- 4 4
$$\frac{17}{4} - \frac{\sqrt{321}}{4}$$
product
_____ 17 \/ 321 -- - ------- 4 4
$$\frac{17}{4} - \frac{\sqrt{321}}{4}$$
=
_____ 17 \/ 321 -- - ------- 4 4
$$\frac{17}{4} - \frac{\sqrt{321}}{4}$$
17/4 - sqrt(321)/4
Rapid solution
[src]
_____
17 \/ 321
x1 = -- - -------
4 4
$$x_{1} = \frac{17}{4} - \frac{\sqrt{321}}{4}$$
x1 = 17/4 - sqrt(321)/4