Mister Exam
Other calculators
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How to use it?
- Equation:
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Equation 1/x^2-1/x-6=0
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Equation x^2-8=0
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Equation x^2=-3
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Equation x^4-2*x^3+x^2-1=0
- Express {x} in terms of y where:
- -9*x-12*y=-19
- 6*x+7*y=-10
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- Graphing y =:
- x^4-13x^2+36
-
Identical expressions
- x^ four -13x^ two + thirty-six = zero
- x to the power of 4 minus 13x squared plus 36 equally 0
- x to the power of four minus 13x to the power of two plus thirty minus six equally zero
- x4-13x2+36=0
- x⁴-13x²+36=0
- x to the power of 4-13x to the power of 2+36=0
- x^4-13x^2+36=O
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Similar expressions
- x^4+13x^2+36=0
- x^4-13x^2-36=0
x^4-13x^2+36=0 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given the equation:
$$\left(x^{4} - 13 x^{2}\right) + 36 = 0$$
Do replacement
$$v = x^{2}$$
then the equation will be the:
$$v^{2} - 13 v + 36 = 0$$
This equation is of the form
A quadratic equation can be solved
using the discriminant.
The roots of the quadratic equation:
$$v_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$v_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = 1$$
$$b = -13$$
$$c = 36$$
, then
Because D > 0, then the equation has two roots.
or
$$v_{1} = 9$$
$$v_{2} = 4$$
The final answer:
Because
$$v = x^{2}$$
then
$$x_{1} = \sqrt{v_{1}}$$
$$x_{2} = - \sqrt{v_{1}}$$
$$x_{3} = \sqrt{v_{2}}$$
$$x_{4} = - \sqrt{v_{2}}$$
then:
$$x_{1} = $$
$$\frac{0}{1} + \frac{9^{\frac{1}{2}}}{1} = 3$$
$$x_{2} = $$
$$\frac{\left(-1\right) 9^{\frac{1}{2}}}{1} + \frac{0}{1} = -3$$
$$x_{3} = $$
$$\frac{0}{1} + \frac{4^{\frac{1}{2}}}{1} = 2$$
$$x_{4} = $$
$$\frac{\left(-1\right) 4^{\frac{1}{2}}}{1} + \frac{0}{1} = -2$$
$$\left(x^{4} - 13 x^{2}\right) + 36 = 0$$
Do replacement
$$v = x^{2}$$
then the equation will be the:
$$v^{2} - 13 v + 36 = 0$$
This equation is of the form
a*v^2 + b*v + c = 0
A quadratic equation can be solved
using the discriminant.
The roots of the quadratic equation:
$$v_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$v_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = 1$$
$$b = -13$$
$$c = 36$$
, then
D = b^2 - 4 * a * c =
(-13)^2 - 4 * (1) * (36) = 25
Because D > 0, then the equation has two roots.
v1 = (-b + sqrt(D)) / (2*a)
v2 = (-b - sqrt(D)) / (2*a)
or
$$v_{1} = 9$$
$$v_{2} = 4$$
The final answer:
Because
$$v = x^{2}$$
then
$$x_{1} = \sqrt{v_{1}}$$
$$x_{2} = - \sqrt{v_{1}}$$
$$x_{3} = \sqrt{v_{2}}$$
$$x_{4} = - \sqrt{v_{2}}$$
then:
$$x_{1} = $$
$$\frac{0}{1} + \frac{9^{\frac{1}{2}}}{1} = 3$$
$$x_{2} = $$
$$\frac{\left(-1\right) 9^{\frac{1}{2}}}{1} + \frac{0}{1} = -3$$
$$x_{3} = $$
$$\frac{0}{1} + \frac{4^{\frac{1}{2}}}{1} = 2$$
$$x_{4} = $$
$$\frac{\left(-1\right) 4^{\frac{1}{2}}}{1} + \frac{0}{1} = -2$$
Rapid solution
[src]
x1 = -3
$$x_{1} = -3$$
x2 = -2
$$x_{2} = -2$$
x3 = 2
$$x_{3} = 2$$
x4 = 3
$$x_{4} = 3$$
x4 = 3
Sum and product of roots
[src]
sum
-3 - 2 + 2 + 3
$$\left(\left(-3 - 2\right) + 2\right) + 3$$
=
0
$$0$$
product
-3*(-2)*2*3
$$3 \cdot 2 \left(- -6\right)$$
=
36
$$36$$
36
The graph