Question

For the equation x2 + 3x + j = 0, find all the values of j such that the equation has two real number solutions. Show your work.

Answer 1

let j=ab
a+b=3 check how many solutions exists

Answer 2

j=3a-a^2
put a=0,a=1,a=2,a=3

Answer 3

actually \[b^2-4ac>0 \implies 9-4j>0 \]
\[j<9/4\]

Answer 4

Solve the following for x:\[x^2+3x+j=0\]\[x=\frac{1}{2} \left(-3\pm \sqrt{9-4 j}\right) \]The expression under the radical sign has to be greater than or equal to zero:\[9-4 j\geq 0 \]\[j\leq \frac{9}{4} \]

Answer 5

Thank you guys so much!

Answer 6

so should j also be more than 9/4

Answer 7

i mean\[j \ge -9/4\]

Answer 8

\[-9/4 \le j \le 9/4\]

Answer 9

oh i see nevermind it works for all j<9/4 no other condition

Answer 10

but if j=9/4
only one root twice so shud we conside < or <=
@robtobey