Mister Exam
Other calculators
-
How to use it?
- Equation:
-
Equation (2*x-6)^2-4*x^2=0
-
Equation 3*x^2+2=-5*x
-
Equation x^3+9*x^2+27*x+27=0
-
Equation x^2-7*x+6=0
- Express {x} in terms of y where:
- -20*x+6*y=-12
- -11*x+16*y=-5
- -12*x-9*y=-3
- -6*x-2*y=7
- Factor polynomial:
- x^3+5*x^2
- Limit of the function:
-
x^3+5*x^2
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Identical expressions
- x^ three + five *x^ two = nine *x+ forty-five
- x cubed plus 5 multiply by x squared equally 9 multiply by x plus 45
- x to the power of three plus five multiply by x to the power of two equally nine multiply by x plus forty minus five
- x3+5*x2=9*x+45
- x³+5*x²=9*x+45
- x to the power of 3+5*x to the power of 2=9*x+45
- x^3+5x^2=9x+45
- x3+5x2=9x+45
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Similar expressions
- 1.9x+45.6=x*25
- x^3+5*x^2-9x+45=0
- x^3+5*x^2=9*x-45
- x^3+5x^2=9x+45
- x^3-5*x^2=9*x+45
- (x^3)+(11*x^2)+(39*x)+45=0
x^3+5*x^2=9*x+45 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given the equation:
$$x^{3} + 5 x^{2} = 9 x + 45$$
transform
$$\left(- 9 x + \left(\left(5 x^{2} + \left(x^{3} - 27\right)\right) - 45\right)\right) + 27 = 0$$
or
$$\left(- 9 x + \left(\left(5 x^{2} + \left(x^{3} - 3^{3}\right)\right) - 5 \cdot 3^{2}\right)\right) + 3 \cdot 9 = 0$$
$$- 9 \left(x - 3\right) + \left(5 \left(x^{2} - 3^{2}\right) + \left(x^{3} - 3^{3}\right)\right) = 0$$
$$- 9 \left(x - 3\right) + \left(\left(x - 3\right) \left(\left(x^{2} + 3 x\right) + 3^{2}\right) + 5 \left(x - 3\right) \left(x + 3\right)\right) = 0$$
Take common factor -3 + x from the equation
we get:
$$\left(x - 3\right) \left(\left(5 \left(x + 3\right) + \left(\left(x^{2} + 3 x\right) + 3^{2}\right)\right) - 9\right) = 0$$
or
$$\left(x - 3\right) \left(x^{2} + 8 x + 15\right) = 0$$
then:
$$x_{1} = 3$$
and also
we get the equation
$$x^{2} + 8 x + 15 = 0$$
This equation is of the form
A quadratic equation can be solved
using the discriminant.
The roots of the quadratic equation:
$$x_{2} = \frac{\sqrt{D} - b}{2 a}$$
$$x_{3} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = 1$$
$$b = 8$$
$$c = 15$$
, then
Because D > 0, then the equation has two roots.
or
$$x_{2} = -3$$
$$x_{3} = -5$$
The final answer for x^3 + 5*x^2 - 9*x - 45 = 0:
$$x_{1} = 3$$
$$x_{2} = -3$$
$$x_{3} = -5$$
$$x^{3} + 5 x^{2} = 9 x + 45$$
transform
$$\left(- 9 x + \left(\left(5 x^{2} + \left(x^{3} - 27\right)\right) - 45\right)\right) + 27 = 0$$
or
$$\left(- 9 x + \left(\left(5 x^{2} + \left(x^{3} - 3^{3}\right)\right) - 5 \cdot 3^{2}\right)\right) + 3 \cdot 9 = 0$$
$$- 9 \left(x - 3\right) + \left(5 \left(x^{2} - 3^{2}\right) + \left(x^{3} - 3^{3}\right)\right) = 0$$
$$- 9 \left(x - 3\right) + \left(\left(x - 3\right) \left(\left(x^{2} + 3 x\right) + 3^{2}\right) + 5 \left(x - 3\right) \left(x + 3\right)\right) = 0$$
Take common factor -3 + x from the equation
we get:
$$\left(x - 3\right) \left(\left(5 \left(x + 3\right) + \left(\left(x^{2} + 3 x\right) + 3^{2}\right)\right) - 9\right) = 0$$
or
$$\left(x - 3\right) \left(x^{2} + 8 x + 15\right) = 0$$
then:
$$x_{1} = 3$$
and also
we get the equation
$$x^{2} + 8 x + 15 = 0$$
This equation is of the form
a*x^2 + b*x + c = 0
A quadratic equation can be solved
using the discriminant.
The roots of the quadratic equation:
$$x_{2} = \frac{\sqrt{D} - b}{2 a}$$
$$x_{3} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = 1$$
$$b = 8$$
$$c = 15$$
, then
D = b^2 - 4 * a * c =
(8)^2 - 4 * (1) * (15) = 4
Because D > 0, then the equation has two roots.
x2 = (-b + sqrt(D)) / (2*a)
x3 = (-b - sqrt(D)) / (2*a)
or
$$x_{2} = -3$$
$$x_{3} = -5$$
The final answer for x^3 + 5*x^2 - 9*x - 45 = 0:
$$x_{1} = 3$$
$$x_{2} = -3$$
$$x_{3} = -5$$
Vieta's Theorem
it is reduced cubic equation
$$p x^{2} + q x + v + x^{3} = 0$$
where
$$p = \frac{b}{a}$$
$$p = 5$$
$$q = \frac{c}{a}$$
$$q = -9$$
$$v = \frac{d}{a}$$
$$v = -45$$
Vieta Formulas
$$x_{1} + x_{2} + x_{3} = - p$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = q$$
$$x_{1} x_{2} x_{3} = v$$
$$x_{1} + x_{2} + x_{3} = -5$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = -9$$
$$x_{1} x_{2} x_{3} = -45$$
$$p x^{2} + q x + v + x^{3} = 0$$
where
$$p = \frac{b}{a}$$
$$p = 5$$
$$q = \frac{c}{a}$$
$$q = -9$$
$$v = \frac{d}{a}$$
$$v = -45$$
Vieta Formulas
$$x_{1} + x_{2} + x_{3} = - p$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = q$$
$$x_{1} x_{2} x_{3} = v$$
$$x_{1} + x_{2} + x_{3} = -5$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = -9$$
$$x_{1} x_{2} x_{3} = -45$$
Sum and product of roots
[src]
sum
-5 - 3 + 3
$$\left(-5 - 3\right) + 3$$
=
-5
$$-5$$
product
-5*(-3)*3
$$3 \left(- -15\right)$$
=
45
$$45$$
45
The graph