Respuesta :
when you use all prime, non-one numbers for b and c, it cannot be factored most of the time
example
1x^2+7x+5
a=1
b=7
c=5
x=[tex] \frac{-b+/- \sqrt{b^2-4ac} }{2a} [/tex]
x=[tex] \frac{-7+/- \sqrt{7^2-4(1)(5)} }{2(1)} [/tex]
x=[tex] \frac{-7+/- \sqrt{49-20} }{2} [/tex]
x=[tex] \frac{-7+/- \sqrt{29} }{2} [/tex]
x=[tex] \frac{-7+ \sqrt{29} }{2} [/tex] or [tex] \frac{-7- \sqrt{29} }{2} [/tex]
example
1x^2+7x+5
a=1
b=7
c=5
x=[tex] \frac{-b+/- \sqrt{b^2-4ac} }{2a} [/tex]
x=[tex] \frac{-7+/- \sqrt{7^2-4(1)(5)} }{2(1)} [/tex]
x=[tex] \frac{-7+/- \sqrt{49-20} }{2} [/tex]
x=[tex] \frac{-7+/- \sqrt{29} }{2} [/tex]
x=[tex] \frac{-7+ \sqrt{29} }{2} [/tex] or [tex] \frac{-7- \sqrt{29} }{2} [/tex]
Part 1: x^2 - 5x -3 =2 where a is equal to 1, b is equal to -5 and c is equal to -5. It cannot be factored because we need 2 integers that will yield a product equal to -5 and the sum of the integers is equal to -5. Using the quadratic formula, we will obtain answers equal to:
x1 = (5+3√5)/2
x2 = (5-3√5)/2
x1 = (5+3√5)/2
x2 = (5-3√5)/2