Mister Exam
Other calculators
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How to use it?
- Equation:
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Equation (x-3)*(x+4)=0
-
Equation x^3-x^2-5=0
-
Equation x-2(3x+2)=16
- Equation x^2-1=1+(1/2)*y
- Express {x} in terms of y where:
- -10*x+16*y=-6
- -13*x+9*y=1
- -10*x+9*y=-6
- -17*x+13*y=-3
- Graphing y =:
- (x-3)*(x+4)
-
Identical expressions
- (x- three)*(x+ four)= zero
- (x minus 3) multiply by (x plus 4) equally 0
- (x minus three) multiply by (x plus four) equally zero
- (x-3)(x+4)=0
- x-3x+4=0
- (x-3)*(x+4)=O
-
Similar expressions
- (x-3)*(x-4)=0
- (x-3)(x+4)=x(1-x)
- (x-3)(x+4)=0
- x*(x-3)*x+4*x-7=0
- (x+3)*(x+4)=0
- 6(2x+0.5)=8x-(3x+4)
(x-3)*(x+4)=0 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Expand the expression in the equation
$$\left(x + 4\right) \left(x - 3\right) + 0 = 0$$
We get the quadratic equation
$$x^{2} + x - 12 = 0$$
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
$$a = 1$$
$$b = 1$$
$$c = -12$$
, then
Because D > 0, then the equation has two roots.
or
$$x_{1} = 3$$
Simplify
$$x_{2} = -4$$
Simplify
$$\left(x + 4\right) \left(x - 3\right) + 0 = 0$$
We get the quadratic equation
$$x^{2} + x - 12 = 0$$
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
$$a = 1$$
$$b = 1$$
$$c = -12$$
, then
D = b^2 - 4 * a * c =
(1)^2 - 4 * (1) * (-12) = 49
Because D > 0, then the equation has two roots.
x1 = (-b + sqrt(D)) / (2*a)
x2 = (-b - sqrt(D)) / (2*a)
or
$$x_{1} = 3$$
Simplify
$$x_{2} = -4$$
Simplify
Sum and product of roots
[src]
sum
0 - 4 + 3
$$\left(-4 + 0\right) + 3$$
=
-1
$$-1$$
product
1*-4*3
$$1 \left(-4\right) 3$$
=
-12
$$-12$$
-12
The graph