Divide 8 into 2 parts such that 7 times the square of the smaller part exceeds twoce the square of the other part by 13
a + b = 8
7 a^2 = 2 b^2 + 13
7 (8 - b)^2 = 2 b^2 + 13
448 - 112 b + 7 b^2 = 2 b^2 + 13
5 b^2 - 112 b + 435 = 0
smaller part --- x
larger part ---- 8 - x
7x^2 > 2(8-x)^2 by 13
7x^2 = 2(8-x)^2 + 13
7x^2 = 128 - 32x + 2x^2 + 13
5x^2 + 32x - 141 = 0
(x - 3)(5x + 47) = 0
x = 3 or x = -47/5
the two parts are either 3 and 5
or
they are -47/5 and 87/5
To solve this problem, let's assume that one part of the number is "x" and the other part is "8-x".
According to the problem statement, we need to find two numbers such that 7 times the square of the smaller part exceeds twice the square of the other part by 13.
Mathematically, this can be written as:
7x^2 = 2(8-x)^2 + 13
To solve this equation, we can follow these steps:
Step 1: Expand and simplify:
7x^2 = 2(64 - 16x + x^2) + 13
7x^2 = 128 - 32x + 2x^2 + 13
Step 2: Combine like terms:
7x^2 = 2x^2 - 32x + 141
Step 3: Move all terms to one side of the equation to set it equal to zero:
7x^2 - 2x^2 + 32x - 141 = 0
Step 4: Combine like terms:
5x^2 + 32x - 141 = 0
Now, we have a quadratic equation. To solve it, we can either factor, complete the square, or use the quadratic formula.
In this case, the quadratic equation doesn't easily factorize, so let's use the quadratic formula:
x = (-b ± sqrt(b^2 - 4ac)) / (2a)
For our equation, a = 5, b = 32, and c = -141.
Substituting these values into the quadratic formula, we get:
x = (-32 ± sqrt(32^2 - 4 * 5 * -141)) / (2 * 5)
Simplifying this expression will give us the two possible values for x.