Question 1115736
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First I'm going to alter {{{p^2-10p+c=0}}} so that it's in a more familiar form
p^2-10p+c=0
x^2-10x+c=0 ... replace every 'p' with 'x'
1x^2-10x+c=0 ... rewrite x^2 as 1x^2
1x^2+(-10)x+c=0 ... rewrite the "1x^2-10x" as "1x^2+(-10)x"


Note how {{{1x^2+(-10)x+c=0}}} is in the form {{{ax^2+bx+c=0}}}


It might be better to line up the terms like so
1x^2+(-10)x+c=0
ax^2+bx+c=0


Which helps us see that 
a = 1
b = -10
c = unknown for now


Now turn to the discriminant formula
{{{d = b^2 - 4ac}}}
Exactly one rational solution occurs when {{{d = 0}}}


So we'll plug in a = 1, b = -10, and d = 0. Then we'll solve for c
{{{d = b^2 - 4ac}}}


{{{0 = (-10)^2 - 4(1)c}}}


{{{0 = 100-4c}}}


{{{4c = 100}}}


{{{c = 100/4}}}


{{{c = 25}}} (this is the final answer)


So the equation {{{p^2-10p+c=0}}} becomes {{{p^2-10p+25=0}}}


I'll leave it as an exercise for you to confirm that {{{p^2-10p+25=0}}} has exactly one solution, and to find the solution. Though this is optional, its good to practice. Hint: Quadratic Formula
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