Mister Exam
Other calculators
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How to use it?
- Equation:
-
Equation sin^2((x/4)+(pi/4))*sin^2((x/4)-(pi/4))=0,375sin^2(-(pi/4))
-
Equation 2tgx–9ctgx+3=0
- Equation s-0,522s=3,585
- Equation (c-4)(c+4)
- Express {x} in terms of y where:
- 9*x+18*y=6
- 15*x-15*y=16
- (x^2+y^2-1)^3+x^2*y^3=0
- -1*x-19*y=-10
-
Identical expressions
- (c- four)(c+ four)
- (c minus 4)(c plus 4)
- (c minus four)(c plus four)
- c-4c+4
-
Similar expressions
- (c+4)(c+4)
- (c-4)(c-4)
(c-4)(c+4) equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Expand the expression in the equation
$$\left(c - 4\right) \left(c + 4\right) = 0$$
We get the quadratic equation
$$c^{2} - 16 = 0$$
This equation is of the form
A quadratic equation can be solved
using the discriminant.
The roots of the quadratic equation:
$$c_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$c_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = 1$$
$$b = 0$$
$$c = -16$$
, then
Because D > 0, then the equation has two roots.
or
$$c_{1} = 4$$
$$c_{2} = -4$$
$$\left(c - 4\right) \left(c + 4\right) = 0$$
We get the quadratic equation
$$c^{2} - 16 = 0$$
This equation is of the form
a*c^2 + b*c + c = 0
A quadratic equation can be solved
using the discriminant.
The roots of the quadratic equation:
$$c_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$c_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = 1$$
$$b = 0$$
$$c = -16$$
, then
D = b^2 - 4 * a * c =
(0)^2 - 4 * (1) * (-16) = 64
Because D > 0, then the equation has two roots.
c1 = (-b + sqrt(D)) / (2*a)
c2 = (-b - sqrt(D)) / (2*a)
or
$$c_{1} = 4$$
$$c_{2} = -4$$
Sum and product of roots
[src]
sum
-4 + 4
$$-4 + 4$$
=
0
$$0$$
product
-4*4
$$- 16$$
=
-16
$$-16$$
-16