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- Equation:
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Equation (x+3)(x-4)-18=0
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Equation tgx=-1
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Identical expressions
- (x+ three)(x- four)- eighteen = zero
- (x plus 3)(x minus 4) minus 18 equally 0
- (x plus three)(x minus four) minus eighteen equally zero
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- (x+3)(x-4)-18=O
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Similar expressions
- (x+3)(x-4)+18=0
- (x-3)(x-4)-18=0
- (x+3)*(x-4)-18=0
- (x+3)(x+4)-18=0
(x+3)(x-4)-18=0 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Expand the expression in the equation
$$\left(x - 4\right) \left(x + 3\right) - 18 = 0$$
We get the quadratic equation
$$x^{2} - x - 30 = 0$$
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
$$a = 1$$
$$b = -1$$
$$c = -30$$
, then
Because D > 0, then the equation has two roots.
or
$$x_{1} = 6$$
$$x_{2} = -5$$
$$\left(x - 4\right) \left(x + 3\right) - 18 = 0$$
We get the quadratic equation
$$x^{2} - x - 30 = 0$$
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
$$a = 1$$
$$b = -1$$
$$c = -30$$
, then
D = b^2 - 4 * a * c =
(-1)^2 - 4 * (1) * (-30) = 121
Because D > 0, then the equation has two roots.
x1 = (-b + sqrt(D)) / (2*a)
x2 = (-b - sqrt(D)) / (2*a)
or
$$x_{1} = 6$$
$$x_{2} = -5$$
Sum and product of roots
[src]
sum
-5 + 6
$$-5 + 6$$
=
1
$$1$$
product
-5*6
$$- 30$$
=
-30
$$-30$$
-30
The graph