Mister Exam
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How to use it?
- Equation:
-
Equation x^2-5x+9=0
- Equation −2∣x+4∣=3−4x-2
- Equation sin(6πsinx)+√3cos(6πsinx)=2
-
Equation ctg(x)=-sqrt3/3
- Express {x} in terms of y where:
- 9*x+18*y=6
- (x^2+y^2-1)^3+x^2*y^3=0
- -1*x-19*y=-10
- 5*x+5*y=20
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Identical expressions
- − two ∣x+ four ∣= three −4x- two
- −2∣x plus 4∣ equally 3−4x minus 2
- − two ∣x plus four ∣ equally three −4x minus two
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Similar expressions
- −2∣x+4∣=3−4x+2
- −2∣x-4∣=3−4x-2
−2∣x+4∣=3−4x-2 equation
The teacher will be very surprised to see your correct solution 😉
The solution
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[src]
-2*|x + 4| = 3 - 4*x - 2
$$- 2 \left|{x + 4}\right| = \left(3 - 4 x\right) - 2$$
Detail solution
For every modulo expressions in the equation
allow cases, when this expressions ">=0" or "<0",
solve the resulting equation.
1.
$$x + 4 \geq 0$$
or
$$-4 \leq x \wedge x < \infty$$
we get the equation
$$4 x - 2 \left(x + 4\right) - 1 = 0$$
after simplifying we get
$$2 x - 9 = 0$$
the solution in this interval:
$$x_{1} = \frac{9}{2}$$
2.
$$x + 4 < 0$$
or
$$-\infty < x \wedge x < -4$$
we get the equation
$$4 x - 2 \left(- x - 4\right) - 1 = 0$$
after simplifying we get
$$6 x + 7 = 0$$
the solution in this interval:
$$x_{2} = - \frac{7}{6}$$
but x2 not in the inequality interval
The final answer:
$$x_{1} = \frac{9}{2}$$
allow cases, when this expressions ">=0" or "<0",
solve the resulting equation.
1.
$$x + 4 \geq 0$$
or
$$-4 \leq x \wedge x < \infty$$
we get the equation
$$4 x - 2 \left(x + 4\right) - 1 = 0$$
after simplifying we get
$$2 x - 9 = 0$$
the solution in this interval:
$$x_{1} = \frac{9}{2}$$
2.
$$x + 4 < 0$$
or
$$-\infty < x \wedge x < -4$$
we get the equation
$$4 x - 2 \left(- x - 4\right) - 1 = 0$$
after simplifying we get
$$6 x + 7 = 0$$
the solution in this interval:
$$x_{2} = - \frac{7}{6}$$
but x2 not in the inequality interval
The final answer:
$$x_{1} = \frac{9}{2}$$
Sum and product of roots
[src]
sum
9/2
$$\frac{9}{2}$$
=
9/2
$$\frac{9}{2}$$
product
9/2
$$\frac{9}{2}$$
=
9/2
$$\frac{9}{2}$$
9/2