Formation and Solution of Quadratic Equations Mathematics Problem (Math Problem Sample)

Formation and Solution of Quadratic Equations
Step 1: Formation of a quadratic equation
We are given x2+4y2=40 and xy=6. Here,
xy=6 implies that y=6x.
So, equation x2+4y2=40 can be written in terms of x as
x2+46x2=40 or x2+144x2=40 or x4-40x2+144=0.
x4-40x2+144=0 can be expressed as a quadratic equation in x2. Let z=x2 so that
z2-40z+144=0.
Step 2: Solution of a quadratic equation
To solve the above quadratic equation by factorization (Argyros, 1985), proceed as follows
z2-40z+144=z-36z-4=0.
This gives values of z as; z=4, and z=36. Since z=x2 we have the four values of x as
x=-6, x=-2, x=2, x=6.
The corresponding values of y, using y=6x are
y=-1 when x=-6, y= -3 when x=-2,
and y=3 when x=2, y=1 when x=6.
Step 3: Computation of the required result
We are required to find the value of x+2y. We have four possibilities