Verified Answer :

apologiabiology

use quadratic formula
if you had ax^2+bx+c=0, then
x=\(\frac{-b+/- \sqrt{b^{2}-4ac} }{2a}\)
a=1
b=?
c=34
subsitute
\(\frac{-b+/- \sqrt{b^{2}-4(1)(34)} }{2(1)}\)=5+/-3i
\(\frac{-b+/- \sqrt{b^{2}-136} }{2}\)=5+/-3i
make 5+/-3 into fraction over 2,(10+/-6i)/2
\(\frac{-b+/- \sqrt{b^{2}-136} }{2}\)=(10+/-6i)/2
multiply both sides by 2
\(-b+/- \sqrt{b^{2}-136}\)=10+/-6i
we conclude that -b=10
b=-10

ok so equaton is
x^2-10x+34


sebastiann73

The roots of  x² -bx+34=0  are 5 ± 3i.  Then b = ?

-------------------

Vieta's formulas :

5 - 3i + 5 + 3i = b ⇒ b =10 

Now, we have the equation:

x² -10x + 34 = 0 






Related Questions