Mister Exam
Other calculators
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How to use it?
- Equation:
-
Equation 4-6*(x+2)=3-5*x
-
Equation 1/(x-3)^2-3/(x-3)-4=0
-
Equation 2-3(7+2x)=11
-
Equation x^2-2x+1=0
- Express {x} in terms of y where:
- 17*x-20*y=6
- 5*x-13*y=17
- 6*x+17*y=3
- 1*x+4*y=20
- Graphing y =:
- (x-3)(x+4)
- Integral of d{x}:
-
x(1-x)
-
Identical expressions
- (x- three)(x+ four)=x(one -x)
- (x minus 3)(x plus 4) equally x(1 minus x)
- (x minus three)(x plus four) equally x(one minus x)
- x-3x+4=x1-x
-
Similar expressions
- (-2*x^x-3*x+4)-(-2*x^x+7*x-6)=0
- (0.7-x)=1.96^2*((x*(1-x))/80)
- 144*x^2+225*(1-x)^2-360*x*(1-x)=0
- (x-3)*(x+4)-18=0
- (x-3)(x+4)=x(1+x)
- 6x-5/7=2x-1/3+2x1-x/21
- (x-3)(x-4)=x(1-x)
- (x+3)(x+4)=x(1-x)
- (x+2)(x+2)=(1-x)(1-x)
- x*(x-3)*x+4*x-7=0
(x-3)(x+4)=x(1-x) equation
The teacher will be very surprised to see your correct solution 😉
The solution
You have entered
[src]
(x - 3)*(x + 4) = x*(1 - x)
$$\left(x - 3\right) \left(x + 4\right) = x \left(1 - x\right)$$
Detail solution
Move right part of the equation to
left part with negative sign.
The equation is transformed from
$$\left(x - 3\right) \left(x + 4\right) = x \left(1 - x\right)$$
to
$$- x \left(1 - x\right) + \left(x - 3\right) \left(x + 4\right) = 0$$
Expand the expression in the equation
$$- x \left(1 - x\right) + \left(x - 3\right) \left(x + 4\right) = 0$$
We get the quadratic equation
$$2 x^{2} - 12 = 0$$
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
$$a = 2$$
$$b = 0$$
$$c = -12$$
, then
Because D > 0, then the equation has two roots.
or
$$x_{1} = \sqrt{6}$$
$$x_{2} = - \sqrt{6}$$
left part with negative sign.
The equation is transformed from
$$\left(x - 3\right) \left(x + 4\right) = x \left(1 - x\right)$$
to
$$- x \left(1 - x\right) + \left(x - 3\right) \left(x + 4\right) = 0$$
Expand the expression in the equation
$$- x \left(1 - x\right) + \left(x - 3\right) \left(x + 4\right) = 0$$
We get the quadratic equation
$$2 x^{2} - 12 = 0$$
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
$$a = 2$$
$$b = 0$$
$$c = -12$$
, then
D = b^2 - 4 * a * c =
(0)^2 - 4 * (2) * (-12) = 96
Because D > 0, then the equation has two roots.
x1 = (-b + sqrt(D)) / (2*a)
x2 = (-b - sqrt(D)) / (2*a)
or
$$x_{1} = \sqrt{6}$$
$$x_{2} = - \sqrt{6}$$
Sum and product of roots
[src]
sum
___ ___ - \/ 6 + \/ 6
$$- \sqrt{6} + \sqrt{6}$$
=
0
$$0$$
product
___ ___ -\/ 6 *\/ 6
$$- \sqrt{6} \sqrt{6}$$
=
-6
$$-6$$
-6
Rapid solution
[src]
___ x1 = -\/ 6
$$x_{1} = - \sqrt{6}$$
___ x2 = \/ 6
$$x_{2} = \sqrt{6}$$
x2 = sqrt(6)
The graph