Mister Exam
Other calculators
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How to use it?
- Equation:
- Equation x+y
- Equation |x^2+7x|=4x+10
-
Equation 2(x+4)(x+2)=x^2+2x
- Equation √3(cos4x+1)=2(2-cos4x)cos2x
- Express {x} in terms of y where:
- (x^2+y^2-1)^3+x^2*y^3=0
- -1*x-19*y=-10
- 4*x-16*y=6
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Identical expressions
- |x^ two +7x|=4x+ ten
- module of x squared plus 7x| equally 4x plus 10
- module of x to the power of two plus 7x| equally 4x plus ten
- |x2+7x|=4x+10
- |x²+7x|=4x+10
- |x to the power of 2+7x|=4x+10
-
Similar expressions
- |x^2+7x|=4x-10
- 22x=14*(x+10)
- -1/6*x^(3)+3,8*x^(2)+44,64*x+10,9066=0
- |x^2-7x|=4x+10
- p^2+320.44*x+109126.98=0
|x^2+7x|=4x+10 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.
$$x^{2} + 7 x \geq 0$$
or
$$\left(0 \leq x \wedge x < \infty\right) \vee \left(x \leq -7 \wedge -\infty < x\right)$$
we get the equation
$$- 4 x + \left(x^{2} + 7 x\right) - 10 = 0$$
after simplifying we get
$$x^{2} + 3 x - 10 = 0$$
the solution in this interval:
$$x_{1} = -5$$
but x1 not in the inequality interval
$$x_{2} = 2$$
2.
$$x^{2} + 7 x < 0$$
or
$$-7 < x \wedge x < 0$$
we get the equation
$$- 4 x + \left(- x^{2} - 7 x\right) - 10 = 0$$
after simplifying we get
$$- x^{2} - 11 x - 10 = 0$$
the solution in this interval:
$$x_{3} = -10$$
but x3 not in the inequality interval
$$x_{4} = -1$$
The final answer:
$$x_{1} = 2$$
$$x_{2} = -1$$
allow cases, when this expressions ">=0" or "<0",
solve the resulting equation.
1.
$$x^{2} + 7 x \geq 0$$
or
$$\left(0 \leq x \wedge x < \infty\right) \vee \left(x \leq -7 \wedge -\infty < x\right)$$
we get the equation
$$- 4 x + \left(x^{2} + 7 x\right) - 10 = 0$$
after simplifying we get
$$x^{2} + 3 x - 10 = 0$$
the solution in this interval:
$$x_{1} = -5$$
but x1 not in the inequality interval
$$x_{2} = 2$$
2.
$$x^{2} + 7 x < 0$$
or
$$-7 < x \wedge x < 0$$
we get the equation
$$- 4 x + \left(- x^{2} - 7 x\right) - 10 = 0$$
after simplifying we get
$$- x^{2} - 11 x - 10 = 0$$
the solution in this interval:
$$x_{3} = -10$$
but x3 not in the inequality interval
$$x_{4} = -1$$
The final answer:
$$x_{1} = 2$$
$$x_{2} = -1$$