Mascarfi wrote:

Can you tell how you got here "(3x+19)(x+9) = 0" without spending too much time?

I got to the part when we know that there are more than one solutions for two, but in order to answer the question I had to find the two solution, wich took me a lot of time.

Thanks.

2 ways:

--- roots of \(ax^2+bx+c = 0\) are , \(x_1\), \(x_2\)= \(\frac{-b\pm\sqrt{b^2-4ac}}{2a}\) or

--- you have been given, \(ax^2+bx+c = 0\) ---> \(x_1\) + \(x_2\) (sum of the roots) = -b/a and \(x_1\)*\(x_2\)(product of the roots) = c/a

For this question,

Method 1: \(3x^2+ 46x+171=0\) ---->\(x_1\), \(x_2\)= \(\frac{-b\pm\sqrt{b^2-4ac}}{2a}\) = \(\frac{-46\pm\sqrt{46^2-4*3*171}}{2*3}\),

\(x_1\), \(x_2\)= -19/3 or -3.

Method 2: \(x_1\), \(x_2\) = -46/3 and \(x_1\), \(x_2\)=171/3=57

\(x_1\), \(x_2\)= \(\frac{-b\pm\sqrt{b^2-4ac}}{2a}\)

From this, use the formula \((a-b)^2 = (a+b)^2-4ab\) to calculate a-b and then solve this equation with a+b to get the 2 required values.

Both these methods are lengthy. You should practice more for solving quadratic equations. One way I looked at it was that all 3 coefficients (a,b and c) of the quadratic equation are positive, the 'x' coefficient (i.e. b) could only have been formed by adding 2 positive numbers (as the product of these 2 numbers was positive = 171*3).

Then, I broke down \(3*171\)= \(3^3*19\)and saw that I will 27 and 19 to give me 46 and \(27*19\)= \(171*3\).