Mister Exam
Other calculators
-
How to use it?
- Equation:
-
Equation sin(x)=cos(x)
-
Equation cos(x)=(-1/2)
-
Equation sin(x-1)=0
-
Equation 4*x^2-25=0
- Express {x} in terms of y where:
- 12*x+5*y=7
- -10*x+14*y=20
- -8*x-17*y=-12
- -6*x+17*y=-15
-
Identical expressions
- (x+ three)*(two x+ one)/x^2- four = zero
- (x plus 3) multiply by (2x plus 1) divide by x squared minus 4 equally 0
- (x plus three) multiply by (two x plus one) divide by x squared minus four equally zero
- (x+3)*(2x+1)/x2-4=0
- x+3*2x+1/x2-4=0
- (x+3)*(2x+1)/x²-4=0
- (x+3)*(2x+1)/x to the power of 2-4=0
- (x+3)(2x+1)/x^2-4=0
- (x+3)(2x+1)/x2-4=0
- x+32x+1/x2-4=0
- x+32x+1/x^2-4=0
- (x+3)*(2x+1)/x^2-4=O
- (x+3)*(2x+1) divide by x^2-4=0
-
Similar expressions
- (x+3)*(2x+1)/x^2+4=0
- (x-3)*(2x+1)/x^2-4=0
- (x+3)*(2x-1)/x^2-4=0
(x+3)*(2x+1)/x^2-4=0 equation
The teacher will be very surprised to see your correct solution 😉
The solution
You have entered
[src]
(x + 3)*(2*x + 1)
----------------- - 4 = 0
2
x
$$-4 + \frac{\left(x + 3\right) \left(2 x + 1\right)}{x^{2}} = 0$$
Detail solution
Given the equation:
$$-4 + \frac{\left(x + 3\right) \left(2 x + 1\right)}{x^{2}} = 0$$
transform:
Take common factor from the equation
$$- \frac{2 x^{2} - 7 x - 3}{x^{2}} = 0$$
the denominator
$$x$$
then
Because the right side of the equation is zero, then the solution of the equation is exists if at least one of the multipliers in the left side of the equation equal to zero.
We get the equations
$$- 2 x^{2} + 7 x + 3 = 0$$
solve the resulting equation:
1.
$$- 2 x^{2} + 7 x + 3 = 0$$
This equation is of the form
A quadratic equation can be solved
using the discriminant.
The roots of the quadratic equation:
$$x_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$x_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = -2$$
$$b = 7$$
$$c = 3$$
, then
Because D > 0, then the equation has two roots.
or
$$x_{1} = \frac{7}{4} - \frac{\sqrt{73}}{4}$$
$$x_{2} = \frac{7}{4} + \frac{\sqrt{73}}{4}$$
but
The final answer:
$$x_{1} = \frac{7}{4} - \frac{\sqrt{73}}{4}$$
$$x_{2} = \frac{7}{4} + \frac{\sqrt{73}}{4}$$
$$-4 + \frac{\left(x + 3\right) \left(2 x + 1\right)}{x^{2}} = 0$$
transform:
Take common factor from the equation
$$- \frac{2 x^{2} - 7 x - 3}{x^{2}} = 0$$
the denominator
$$x$$
then
x is not equal to 0
Because the right side of the equation is zero, then the solution of the equation is exists if at least one of the multipliers in the left side of the equation equal to zero.
We get the equations
$$- 2 x^{2} + 7 x + 3 = 0$$
solve the resulting equation:
1.
$$- 2 x^{2} + 7 x + 3 = 0$$
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:
$$x_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$x_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = -2$$
$$b = 7$$
$$c = 3$$
, then
D = b^2 - 4 * a * c =
(7)^2 - 4 * (-2) * (3) = 73
Because D > 0, then the equation has two roots.
x1 = (-b + sqrt(D)) / (2*a)
x2 = (-b - sqrt(D)) / (2*a)
or
$$x_{1} = \frac{7}{4} - \frac{\sqrt{73}}{4}$$
$$x_{2} = \frac{7}{4} + \frac{\sqrt{73}}{4}$$
but
x is not equal to 0
The final answer:
$$x_{1} = \frac{7}{4} - \frac{\sqrt{73}}{4}$$
$$x_{2} = \frac{7}{4} + \frac{\sqrt{73}}{4}$$
Sum and product of roots
[src]
sum
____ ____ 7 \/ 73 7 \/ 73 - - ------ + - + ------ 4 4 4 4
$$\left(\frac{7}{4} - \frac{\sqrt{73}}{4}\right) + \left(\frac{7}{4} + \frac{\sqrt{73}}{4}\right)$$
=
7/2
$$\frac{7}{2}$$
product
/ ____\ / ____\ |7 \/ 73 | |7 \/ 73 | |- - ------|*|- + ------| \4 4 / \4 4 /
$$\left(\frac{7}{4} - \frac{\sqrt{73}}{4}\right) \left(\frac{7}{4} + \frac{\sqrt{73}}{4}\right)$$
=
-3/2
$$- \frac{3}{2}$$
-3/2
Rapid solution
[src]
____
7 \/ 73
x1 = - - ------
4 4
$$x_{1} = \frac{7}{4} - \frac{\sqrt{73}}{4}$$
____
7 \/ 73
x2 = - + ------
4 4
$$x_{2} = \frac{7}{4} + \frac{\sqrt{73}}{4}$$
x2 = 7/4 + sqrt(73)/4