The roots of 7x^2 + x - 5 = 0 are a and b Compute (a - 4)(b - 4)
Thank you! (it's not 80/7)
By the quadratic formula, a and b are \(\frac{-1 \pm \sqrt{141}}{4}\)
Plug these into (a - 4)(b - 4), and you get 103/7
There are two methods to solve this.
First method:
Expand:
\((a-4)(b-4)=ab-4a-4b+16=ab-4(a+b)+16\)
Then, if a and b are the roots of \(7x^2+x-5=0\), then by Vietas' formulae:
\(a+b=-\frac{1}{7}\) and \(ab=-\frac{5}{7}\)
Therefore:
\((a-4)(b-4)=-\frac{5}{7}-4(-\frac{1}{7})+16=\frac{111}{7}\) is the desired answer.
Second method:
If a and b are the roots of \(7x^2+x-5=0\)
Then, we do the following substitution: let \(y=x-4\), hence \(x=y+4\)
So:
\(7(y+4)^2+(y+4)-5=0 \iff 7(y^2+8y+16)+y+4-5=7y^2+56y+112+y-1=7y^2+57y+111\)
Therefore, the product of the roots (Which is now (a-4)(b-4), as we transformed the quadratic with the roots a and b to another quadratic with roots a-4 and b-4.) is again, by Vietas' formulae: \(\frac{111}{7}\).
