Casey W. answered 06/16/15
Tutor
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Mathematics (and Science) Instruction by a Mathematician!
In general for a monic quadratic equation of the form:
x^2+bx=c, we can follow Mark's technique of writing
adding (b/2)^2 to both sides will make the left hand side a perfect square, (x+b/2)^2
x^2 + bx + (b/2)^2 = c + (b/2)^2
(x+b/2)^2 = c +(b/2)^2
And so we can then apply a square root (don't forget to introduce the +/-, obtained by expanding an absolute value), and subtracting b/2 from both sides:
(x+b/2) = +/- \sqrt{c+(b/2)^2}
x = -b/2 +/- \sqrt{c+(b/2)^2}
x^2+bx=c, we can follow Mark's technique of writing
adding (b/2)^2 to both sides will make the left hand side a perfect square, (x+b/2)^2
x^2 + bx + (b/2)^2 = c + (b/2)^2
(x+b/2)^2 = c +(b/2)^2
And so we can then apply a square root (don't forget to introduce the +/-, obtained by expanding an absolute value), and subtracting b/2 from both sides:
(x+b/2) = +/- \sqrt{c+(b/2)^2}
x = -b/2 +/- \sqrt{c+(b/2)^2}
Both of your problems above can be viewed as equations with c=0, and b= (5/6) and -7, respectively.
The term we wish to add to form a perfect square is (b/2)^2, which we must also subtract off in these problems, since we really have expressions, not equations.
r^2+(5/6)r + (5/6)^2 -(5/6)^2 = (r+5/6)^2 - (5/6)^2
p^2-7p+(7/2)^2 -(7/2)^2 = (p+7/2)^2 - (7/2)^2
Hope that helps!