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Equation (5x+2)(-x-4)=0
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Equation 2-3*(2*x+2)=5-4*x
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Identical expressions
- (3x- two)*(3x- two)+(4x- five)^ two =10x+ twenty-one
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Similar expressions
- x^2-10*x+21=0
- (3*x-2)*(3*x+2)+(4*x-5)^2=10*x+21
- (3x-2)*(3x-2)+(4x+5)^2=10x+21
- (3x-2)*(3x+2)+(4x-5)^2=10x+21
- (3x+2)*(3x-2)+(4x-5)^2=10x+21
- (3x-2)*(3x-2)+(4x-5)^2=10x-21
- (3x-2)*(3x-2)-(4x-5)^2=10x+21
(3x-2)*(3x-2)+(4x-5)^2=10x+21 equation
The teacher will be very surprised to see your correct solution 😉
The solution
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[src]
2 (3*x - 2)*(3*x - 2) + (4*x - 5) = 10*x + 21
$$\left(3 x - 2\right) \left(3 x - 2\right) + \left(4 x - 5\right)^{2} = 10 x + 21$$
Detail solution
Move right part of the equation to
left part with negative sign.
The equation is transformed from
$$\left(3 x - 2\right) \left(3 x - 2\right) + \left(4 x - 5\right)^{2} = 10 x + 21$$
to
$$\left(- 10 x - 21\right) + \left(\left(3 x - 2\right) \left(3 x - 2\right) + \left(4 x - 5\right)^{2}\right) = 0$$
Expand the expression in the equation
$$\left(- 10 x - 21\right) + \left(\left(3 x - 2\right) \left(3 x - 2\right) + \left(4 x - 5\right)^{2}\right) = 0$$
We get the quadratic equation
$$25 x^{2} - 62 x + 8 = 0$$
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
$$a = 25$$
$$b = -62$$
$$c = 8$$
, then
Because D > 0, then the equation has two roots.
or
$$x_{1} = \frac{\sqrt{761}}{25} + \frac{31}{25}$$
Simplify
$$x_{2} = \frac{31}{25} - \frac{\sqrt{761}}{25}$$
Simplify
left part with negative sign.
The equation is transformed from
$$\left(3 x - 2\right) \left(3 x - 2\right) + \left(4 x - 5\right)^{2} = 10 x + 21$$
to
$$\left(- 10 x - 21\right) + \left(\left(3 x - 2\right) \left(3 x - 2\right) + \left(4 x - 5\right)^{2}\right) = 0$$
Expand the expression in the equation
$$\left(- 10 x - 21\right) + \left(\left(3 x - 2\right) \left(3 x - 2\right) + \left(4 x - 5\right)^{2}\right) = 0$$
We get the quadratic equation
$$25 x^{2} - 62 x + 8 = 0$$
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
$$a = 25$$
$$b = -62$$
$$c = 8$$
, then
D = b^2 - 4 * a * c =
(-62)^2 - 4 * (25) * (8) = 3044
Because D > 0, then the equation has two roots.
x1 = (-b + sqrt(D)) / (2*a)
x2 = (-b - sqrt(D)) / (2*a)
or
$$x_{1} = \frac{\sqrt{761}}{25} + \frac{31}{25}$$
Simplify
$$x_{2} = \frac{31}{25} - \frac{\sqrt{761}}{25}$$
Simplify
Sum and product of roots
[src]
sum
_____ _____
31 \/ 761 31 \/ 761
0 + -- - ------- + -- + -------
25 25 25 25
$$\left(0 + \left(\frac{31}{25} - \frac{\sqrt{761}}{25}\right)\right) + \left(\frac{\sqrt{761}}{25} + \frac{31}{25}\right)$$
=
62 -- 25
$$\frac{62}{25}$$
product
/ _____\ / _____\ |31 \/ 761 | |31 \/ 761 | 1*|-- - -------|*|-- + -------| \25 25 / \25 25 /
$$1 \cdot \left(\frac{31}{25} - \frac{\sqrt{761}}{25}\right) \left(\frac{\sqrt{761}}{25} + \frac{31}{25}\right)$$
=
8/25
$$\frac{8}{25}$$
8/25
Rapid solution
[src]
_____
31 \/ 761
x1 = -- - -------
25 25
$$x_{1} = \frac{31}{25} - \frac{\sqrt{761}}{25}$$
_____
31 \/ 761
x2 = -- + -------
25 25
$$x_{2} = \frac{\sqrt{761}}{25} + \frac{31}{25}$$
The graph