Solve the equation.
b^2+12b+36=48
using quadratic formula before that subtract 36 from 48
b^2 + 12b + 36-48 = 0.
b^2 + 12b - 12 = 0
Use Quadratic formula:
b = (-B +- sqrt(B^2-4AC))/2A.
b = (-12 +- sqrt(12^2+48))/2 = ?.
You should get 0.928, and -12.928.
Note: 48 is positive, because we are subtracting a -48.
find the orthogonal trajectories of the family of circles x2+y2+2fy+1=0 f being a parameter
To solve the equation b^2 + 12b + 36 = 48, we can follow these steps:
Step 1: Move all terms to one side of the equation to set it equal to zero.
b^2 + 12b + 36 - 48 = 0
Simplifying the equation, we get:
b^2 + 12b - 12 = 0
Step 2: Now, we can either factor the quadratic or use the quadratic formula to find the solutions. Let's factor the quadratic equation.
The factors of the quadratic expression (b^2 + 12b - 12) are:
(b + 6)(b + 2)
Setting each factor equal to zero individually, we have:
b + 6 = 0 or b + 2 = 0
Step 3: Solve for b in each equation:
For b + 6 = 0, subtract 6 from both sides:
b = -6
For b + 2 = 0, subtract 2 from both sides:
b = -2
Step 4: The solutions to the equation are b = -6 and b = -2.