Posted by KELLY on Tuesday, December 21, 2010 at 5:33pm.
In the quadratic equation
ax²+bx+c=0
b/a = -(sum of the two roots)
c/a = product of the two roots.
Assume a=1, then
b=-(7-3i+7+3i)=-14
c=(7-3i)*(7+3i)=49+9=58
So the quadratic equation is
x²-14x+58=0
Check by solving the equation and you should get back 7±3i as the roots.
x = 7 + 3i, and X = 7 - 3i.
We can use either of the 2 values of X:
X = 7 + 3i,
(X - 7) = 3i,
Square both sides:
X^2 --14X + 49 = 9(-1),
X^2 - 14X + 49 +9 = 0,
X^2 - 14X + 58 = 0.
Related Questions
Pre-Calculus/Trig Check please and help - Could someone please check the first ...
Algebra II (check) - Assume that no denominator equals 0. 2+i/1-3i = 2+i/1-3i...
algebra - Can someone assist help me with the following? Simplify the expression...
Pre-calculus-check answers - Write the polynomial equation of least degree that ...
algebra 2 - what is the quadratic function that has the roots (4+3i) and (4-3i)
Algebra - Suppose that a polynomial function of degree 5 with rational ...
Math- Algebra II - Simplify (7-3i)^2. So would I do this: (7-3i) (7-3i) then ...
math help people lik really desperate!! - ok i have to divide complex numbers: ...
check my work please! - ok i have to divide complex numbers: question is: 1-2i+i...
maths - If (e^x)(sin3x)=Im ((e^x)(e^i3x)) integrate (e^x)(sin3x) I get the ...
For Further Reading