Find point of intersection of 20/7 x^2-260/7 x=96/5 x-80

Answer 1

Starting with the given equation:

20/7 x^2 - 260/7 x = 96/5 x - 80

We want to solve for x to find the point(s) of intersection. First, we'll move all the x terms to one side and all the constants to the other side:

20/7 x^2 - 260/7 x - 96/5 x + 80 = 0

Now, we'll combine like terms:

20/7 x^2 - 692/35 x + 80 = 0

To solve this quadratic equation, we can use the quadratic formula:

x = (-b ± sqrt(b^2 - 4ac)) / 2a

Where a = 20/7, b = -692/35, and c = 80:

x = (-(-692/35) ± sqrt((-692/35)^2 - 4(20/7)(80))) / (2(20/7))

Simplifying:

x = (692/35 ± sqrt(1984/1225)) / (40/7)

x = (692/35 ± 8/35 √31) / (40/7)

Now, we can simplify further:

x = (7/5)(692/35 ± 8/35 √31)

x = (14/5)(346/35 ± 4/35 √31)

x = (14/5)(10 ± √31)

So the two points of intersection are:

(14/5)(10 + √31) and (14/5)(10 - √31)