using quadratic formula, find the roots of the following equations 3x^2_17x+25=0
Answers
Step-by-step explanation:
Given:-
3x^2_17x+25=0
To find:-
Find the roots of the following equations 3x^2_17x+25=0 ,using quadratic formula
Solution:-
Given quadratic equation is 3x^2-17x+25=0
On Comparing this with the standard quadratic equation ax^2+bx+c = 0
a = 3
b = -17
c = 25
We know that
The quadratic formula for solving x of
ax^2+bx+c = 0
x = [-b±√(b^2-4ac)]/2a
=>x = [-(-17)±√{(-17)^2-4(3)(25)}]/(2×3)
=>x = [17±√(289-300)]/6
=>x =[17±√-11]/6
=> x = (17+√-11)/6 or (17-√-11)/6.
Answer:-.
The roots of the given quadratic equation are
(17+√-11)/6 and (17-√-11)/6
or
(17+√11 i)/6 and (17-√11 i)/6
Used formulae:-
Quadratic Formula:-
x = [-b±√(b^2-4ac)]/2a
- the standard quadratic equation ax^2+bx+c = 0
- i^2 = -1
- Quadratic formula is also called Sridharacharya formula.
Answer:
The roots are: (17+13.74i)/6 & (17-13.74i)/6
Step-by-step explanation:
3x²-17x+25=0
Here, a = 3 ; b = - 17 & c = 25
r1 = (-b+D) / 2a
r2 = (-b-D) / 2a
where, D = √(b²-4ac)
D = √289-300
= √-189
= 13.74i [where i=√-1]
r1 = (17+13.74i)/6
r2 = (17-13.74i)/6