We’ll solve this two ways: by completing the square and using the quadratic formula. Here’s the first way:
#2x^2+7x+9=0#
First divide everything by #2#, since #a# cannot have a coefficient:
#\rightarrow x^2+\frac{7}{2}x+\frac{9}{2}=0#
Now, move #c# to the RHS:
#\rightarrow x^2+\frac{7}{2}x=-\frac{9}{2}#
Now add #(\frac{b}{2})^2# to both sides. Here, #b=7#.
#\rightarrow x^2+\frac{7}{2}x+\frac{49}{16}=\frac{49}{16}-\frac{9}{2}#
Simplify the RHS:
#\rightarrow x^2+\frac{7}{2}x+\frac{49}{16}=-\frac{23}{16}#
Factor the LHS into #(x+\frac{b}{2})^2#:
#\rightarrow (x+\frac{\frac{7}{2}}{2})^2=-\frac{23}{16}#
#\rightarrow (x+\frac{7}{4})^2=-\frac{23}{16}#
Take the square root of both sides:
#\rightarrow x+\frac{7}{4}=\sqrt{-\frac{23}{16}}#
Isolate #x#:
#\rightarrow x=\sqrt{-\frac{23}{16}}-\frac{7}[4}#
#\rightarrow x=i\sqrt{\frac{23}{16}}-\frac{7}{4}#
#\rightarrow x=i\frac{\sqrt{23}}{4}-\frac{7}{4}#
#\rightarrow x\approx-1.75\pm 1.19895788i#
Here’s the second way, using the quadratic formula:
#2x^2+7x+9=0#
First, we need to identify #a#, #b#, and #c#:
#a=2#
#b=7#
#c=9#
Now, plug them into the formula:
#x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}#
#\rightarrow x=\frac{-7\pm\sqrt{7^2-4(2)(9)}}{2(2)}#
#\rightarrow x=\frac{-7\pm\sqrt{49-72}}{4}#
#\rightarrow x=\frac{-7\pm\sqrt{-23}}{4}#
#\rightarrow x=\frac{-7\pm i\sqrt{23}}{4}#
Rearranging yields:
#\rightarrow x=\frac{i\sqrt{23}}{4}-\frac{7}{4}#
#\rightarrow x\approx-1.75\pm 1.19895788i#
It’s the same answer we got by completing the square, so we know the answer is correct.
