Mister Exam
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How to use it?
- Equation:
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Equation sin(pi*x)=0
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Equation 7x^3-28x=0
- Equation sqrt(x)-4root(4)(x)=5
- Equation 16-3|x|=4
- Express {x} in terms of y where:
- 8*x+7*y=-13
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Identical expressions
- sixteen - three |x|= four
- 16 minus 3 module of x| equally 4
- sixteen minus three module of x| equally four
-
Similar expressions
- 16+3|x|=4
16-3|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.
$$x \geq 0$$
or
$$0 \leq x \wedge x < \infty$$
we get the equation
$$12 - 3 x = 0$$
after simplifying we get
$$12 - 3 x = 0$$
the solution in this interval:
$$x_{1} = 4$$
2.
$$x < 0$$
or
$$-\infty < x \wedge x < 0$$
we get the equation
$$12 - 3 \left(- x\right) = 0$$
after simplifying we get
$$3 x + 12 = 0$$
the solution in this interval:
$$x_{2} = -4$$
The final answer:
$$x_{1} = 4$$
$$x_{2} = -4$$
allow cases, when this expressions ">=0" or "<0",
solve the resulting equation.
1.
$$x \geq 0$$
or
$$0 \leq x \wedge x < \infty$$
we get the equation
$$12 - 3 x = 0$$
after simplifying we get
$$12 - 3 x = 0$$
the solution in this interval:
$$x_{1} = 4$$
2.
$$x < 0$$
or
$$-\infty < x \wedge x < 0$$
we get the equation
$$12 - 3 \left(- x\right) = 0$$
after simplifying we get
$$3 x + 12 = 0$$
the solution in this interval:
$$x_{2} = -4$$
The final answer:
$$x_{1} = 4$$
$$x_{2} = -4$$
Sum and product of roots
[src]
sum
-4 + 4
$$-4 + 4$$
=
0
$$0$$
product
-4*4
$$- 16$$
=
-16
$$-16$$
-16