Find the complex roots of r²-14r+49=0?

Question

I tried factoring it since it was in quadratic form and got:

(r-7)(r-7)

and then found the roots to both be r=7, but in the back of the book it shows the answers to be r=2 and r=7. What did I do wrong?

Answer 1

Hi,

Your book probably made an error or you might be looking at a different problem's answer. The Quadratic Equation that you were given is indeed factored as (r-7)(r-7) since when we use the FOIL Method we get back to the original Quadratic Equation.

Try using FOIL on your answer and I know you will get back to the original problem!

I hope that helps you out! Please let me know if you have any other questions!

Sincerely,

Andrew

Answer 2

You've found a REAL root, r = 7. To find the imaginary root consider the quadratic formula.

r = -b +/- sqrt ( b^2 - 4ac) / 2a

the values for a, b , c are:

a = 1 ; b = -14 ; c = 49

plug these into the quad. equation:

r = -(-14) +/- sqrt ( (-14)^2 - 4(1)(49) ) / 2(1)

simplify ...

r = 14 +/- sqrt ( 196 - 196 ) / 2

r = 7. book must be wrong.

Answer 3

r²-14r+49=0
(r-7)(r-7)=0
r= 7

Make be there must be an error in printing, but your ans is correct.

Answer 4

You did nothing wrong. Your answer is correct.

For the book answer to be correct, the problem would have to have been r^2 -9r+14=0.

Answer 5

r²-14r+49=0
r²-7r-7r+49=0
r(r-7)-7(r-7)=0
(r-7)(r-7)=0
so r=7
is correct