How to Find The Roots of this Quadratic Given Sum & Product

Question

My question is:

The sum of the roots of a quadratic is $55/72$, and the products of the roots is $-25/12$. Find the roots.

How I'm trying to do it so far: (Also, please correct my thought process if it's wrong. I really want to become better at this, and also want to be guided in the right direction-- not just given the right answer).

We can express the factored form of this quadratic as,

$(x+r)(x-s)$

$x^2 - sx + rx - sr$,

$x^2 -x(r+s) - sr$,

$x^2 - x(55/72) - (25/12)$

or I can write it as,

$72x^2 - 55x - 150$

From here on, I'd have to factor this, and the way the AoPS book has structured the Quadratics section is to avoid pure trial and error, a brute force approach to find the roots of this quadratic. Instead it focuses on clever algebraic manipulation, so I'm trying to figure out how to do that in order to get the roots.

Help would be much appreciated. Thanks!

Answer 1

In the same spirit as voldemort, let the two roots be $a$ and $b$. So, what you are given is $$a+b=\frac{55}{72}$$ $$a\times b=-\frac{25}{12}$$ But, you can use identities $$(a-b)^2=a^2+b^2-2a b=(a+b)^2-4a b$$ So, $$(a-b)^2=\Big(\frac{55}{72}\Big)^2+4\times\frac{25}{12}=\frac{46225}{5184}=\Big(\frac{215}{72}\Big)^2$$ For making life easier, suppose $a>b$; so now $$a+b=\frac{55}{72}$$ $$a-b=\frac{215}{72}$$

Answer 2

Your way is correct, but there is another way to do this.

Let the two roots be $a,b$. Also, assume that $a>b$. (since the product of the roots is negative, and the sum is positive, there is no issue about complex roots).

Now, you are given $a+b$ and $ab$.

Notice that $(a-b)^2=(a+b)^2-4ab$. So, $a-b=?$ After that can you find $a$ and $b$?