equation
- Thread starter jshaziza
- Start date
Hello, jshaziza!
I bet that the "30" is the problem number . . .
We have: \(\displaystyle \:y^2\,+\,13y\,+\,30\:=\:-10\)
Then: \(\displaystyle \:y^2\,+\,13y\,+\,40\:=\:0\)
. . which factors: \(\displaystyle \
y\,+\,5)(y\,+\,8)\:=\:0\)
. . and has roots: \(\displaystyle \:y\:=\:-5,\,-8\)
I bet that the "30" is the problem number . . .
We have: \(\displaystyle \:y^2\,+\,13y\,+\,30\:=\:-10\)
Then: \(\displaystyle \:y^2\,+\,13y\,+\,40\:=\:0\)
. . which factors: \(\displaystyle \
y\,+\,5)(y\,+\,8)\:=\:0\). . and has roots: \(\displaystyle \:y\:=\:-5,\,-8\)
And if 30 is in fact part of the problem: you can complete the square:
\(\displaystyle 3y^{2}\,+\,39y\,+\,91 = 0\)
Divide off by 3:
\(\displaystyle \frac{3y^{2}\,+\,39y\,+\,91}{3} = 0\)
\(\displaystyle y^{2}\,+\,13y = -\frac{91}{3}\)
Add \(\displaystyle (\frac{1}{2}b)^{2}\) to both sides.
\(\displaystyle y^{2}\,+\,13y\,+\,\frac{169}{4} = -\frac{91}{3} + \frac{169}{4}\)
Simplify the right hand side, and convert the left hand side to squared form.
\(\displaystyle (y\,+\,\frac{13}{2})^{2} = \frac{143}{12}\)
\(\displaystyle y = -\frac{13}{2} (+/-)\sqrt{\frac{143}{12}}\)
Cheers!
John.
\(\displaystyle 3y^{2}\,+\,39y\,+\,91 = 0\)
Divide off by 3:
\(\displaystyle \frac{3y^{2}\,+\,39y\,+\,91}{3} = 0\)
\(\displaystyle y^{2}\,+\,13y = -\frac{91}{3}\)
Add \(\displaystyle (\frac{1}{2}b)^{2}\) to both sides.
\(\displaystyle y^{2}\,+\,13y\,+\,\frac{169}{4} = -\frac{91}{3} + \frac{169}{4}\)
Simplify the right hand side, and convert the left hand side to squared form.
\(\displaystyle (y\,+\,\frac{13}{2})^{2} = \frac{143}{12}\)
\(\displaystyle y = -\frac{13}{2} (+/-)\sqrt{\frac{143}{12}}\)
Cheers!
John.