Bunuel wrote:
Baten80 wrote:
In the equation x^2+bx+12=0, x is variable and b is a constant. What is the value of b?
(1) x-3 is a factor of x^2+bx+12.
(2) 4 is a root of x^2+bx+12=0
You don't really need Viete's formula for the roots of a quadratic equation.
(1) x-3 is a factor of x^2+bx+12 --> simply means that x=3 is a root of x^2+bx+12=0 (if x-3 is a factor of x^2+bx+12 then x^2+bx+12=(x-3)(x-k)=0, for some k, so x=3 is on of the roots of the equation) --> substitute x=3: 3^2+3b+12=0 --> b=-7. Sufficient.
(2) 4 is a
root of x^2+bx+12=0 --> the same here, substitute x=4: 4^2+4b+12=0 --> b=-7. Sufficient.
Answer: D.
Hello Bunuel,
why do you conclude that 4 is a root of the equation if the statement just says it is
a factor of (the statement from the page) ... My confusion is that x could be = 12 and 4 would still be a factor of the entire equation.
2) 4 is a factor of \(x^2+bx+12=0\) ...
if \(x = 12\), the equation would still be a multiple of 4 and:
\((12)^2+12b+12=0\)
\(156=-12b\)
\(-13=b\) which is not \(= -7\) and therefore it could be "not sufficient". What am I missing?
UPDATE: I looked for the question online and it seems that it has a mistake here, it shouldn't be "factor" ir should say "root". Source:
https://books.google.com.mx/books?id=ss ... 3F&f=falsePlease someone correct that, otherwise it is very confusing.