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Identical expressions
- fourteen , three (x- sixteen)(x+ thirty-nine)= zero
- 14,3(x minus 16)(x plus 39) equally 0
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Similar expressions
- 14,3(x-16)(x-39)=0
- 14,3(x+16)(x+39)=0
14,3(x-16)(x+39)=0 equation
The teacher will be very surprised to see your correct solution 😉
The solution
You have entered
[src]
143*(x - 16)
------------*(x + 39) = 0
10
$$\frac{143 \left(x - 16\right)}{10} \left(x + 39\right) = 0$$
Detail solution
Expand the expression in the equation
$$\frac{143 \left(x - 16\right)}{10} \left(x + 39\right) = 0$$
We get the quadratic equation
$$\frac{143 x^{2}}{10} + \frac{3289 x}{10} - \frac{44616}{5} = 0$$
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
$$a = \frac{143}{10}$$
$$b = \frac{3289}{10}$$
$$c = - \frac{44616}{5}$$
, then
Because D > 0, then the equation has two roots.
or
$$x_{1} = 16$$
$$x_{2} = -39$$
$$\frac{143 \left(x - 16\right)}{10} \left(x + 39\right) = 0$$
We get the quadratic equation
$$\frac{143 x^{2}}{10} + \frac{3289 x}{10} - \frac{44616}{5} = 0$$
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
$$a = \frac{143}{10}$$
$$b = \frac{3289}{10}$$
$$c = - \frac{44616}{5}$$
, then
D = b^2 - 4 * a * c =
(3289/10)^2 - 4 * (143/10) * (-44616/5) = 2474329/4
Because D > 0, then the equation has two roots.
x1 = (-b + sqrt(D)) / (2*a)
x2 = (-b - sqrt(D)) / (2*a)
or
$$x_{1} = 16$$
$$x_{2} = -39$$
Sum and product of roots
[src]
sum
-39 + 16
$$-39 + 16$$
=
-23
$$-23$$
product
-39*16
$$- 624$$
=
-624
$$-624$$
-624