show that p= 4/5q if the equation px^2+3px+p+q=0 where p is not equal to 0 , has two equal real roots
for 2 equal roots, b^2 - 4ac = 0
9p^2 - 4(p)(p+q) = 0
9p^2 - 4p^2 - 4pq = 0
5p^2 - 4pq = 0
p(5p - 4q) = 0
p = 0 or p = 4q/5
but p≠0 or else we don't have a quadratic,
so
p = (4/5)q
To show that the equation \(px^2 + 3px + p + q = 0\) has two equal real roots, we need to find the conditions under which this occurs.
If a quadratic equation has two equal real roots, it means that the discriminant (\(b^2 - 4ac\)) is equal to zero.
In our equation, the discriminant is given by \((3p)^2 - 4p(p+q)\). Let's set this expression equal to zero and solve for \(p\):
\((3p)^2 - 4p(p+q) = 0\)
Simplifying the equation, we get:
\(9p^2 - 4p^2 - 4pq = 0\)
Combining like terms:
\(5p^2 - 4pq = 0\)
Factoring out a \(p\) from both terms:
\(p(5p - 4q) = 0\)
To determine the values of \(p\) that satisfy this equation, we have two possibilities:
1. \(p = 0\), but we are given that \(p\) is not equal to zero.
2. \(5p - 4q = 0\)
Solving the second equation for \(p\) will give us the desired expression for \(p\) in terms of \(q\):
\(5p = 4q\)
Dividing both sides by 5:
\(p = \frac{4}{5}q\)
Therefore, if \(p = \frac{4}{5}q\), the equation \(px^2 + 3px + p + q = 0\) will have two equal real roots.