The roots of $7x^2 + x - 5 = 0$ are $a$ and $b.$ Compute $(a - 4)(b - 4).$
+580 Using the quadratic formula gives:
\(_{1,\:2}=\frac{-1\pm \sqrt{1^2-4\cdot \:7\left(-5\right)}}{2\cdot \:7}\)
\(x_1=\frac{-1+\sqrt{141}}{2\cdot \:7},\:x_2=\frac{-1-\sqrt{141}}{2\cdot \:7}\)
\(x=\frac{-1+\sqrt{141}}{14},\:x=\frac{-1-\sqrt{141}}{14}\)
So if a and b are roots:
\((\frac{-1+\sqrt{141}}{14}-4)(\frac{-1-\sqrt{141}}{14}-4)\)
(EDIT: Just you are calc if you don't know how to solve!!!)
Just solve that and you got the answer 
-Vinculum
+2668 Expand: \((a-4)(b-4) = ab - 4a - 4b + 16 = ab - 4(a+b) + 16 \)
By Vieta's formula, we know that \(ab = -{5 \over 7 }\) and \(a+b = -{1 \over 7}\)
Now, plugging this in, we get: \(-{5 \over 7} -4(-{1 \over 7})+16 = {-{5 \over 7}} + { 4 \over 7}+16 = \color{brown}\boxed{15{ 6 \over 7 }}\)
