How do you solve #2x^2+7x+9=0# algebraically?

1 Answer
Nov 30, 2017

Answer:

In fraction form:

#\rightarrow x=\frac{i\sqrt{23}}{4}-\frac{7}{4}#

Simplified form:

#\rightarrow x\approx -1.75\pm 1.19895788i#

Explanation:

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.