How to solve quadratic equation by completing the square?
Answers
Solve 2x2 – 5x + 1 = 0 by completing the square.
There is one extra step for solving this equation, because the leading coefficient is not 1; I'll first have to divide through to convert the leading coefficient to 1. Here's my process:
2x2 – 5x + 1 = 0
\small{ x^2 - \dfrac{5}{2}x + \dfrac{1}{2} = 0 }x
2
−
2
5
x+
2
1
=0
\small{ x^2 - \dfrac{5}{2}x = -\dfrac{1}{2} }x
2
−
2
5
x=−
2
1
Now that I've got all the terms with variables on one side, with the strictly-numerical term on the other side, I'm ready to complete the square on the left-hand side. First, I take the linear term's coefficient (complete with its sign), –(5/2), and multiply by one-half, and square:
\small{ \left(\dfrac{1}{2}\right)\left(-\dfrac{5}{2}\right) = \color{blue}{-\dfrac{5}{4}} }(
2
1
)(−
2
5
)=−
4
5
\small{ \left(\color{blue}{-\dfrac{5}{4}}\right)^2 = \color{red}{\dfrac{25}{16}} }(−
4
5
)
2
=
16
25
Then I add this new value to both sides, convert to squared-binomial form on the left-hand side, and solve:
\small{ x^2 - \dfrac{5}{2}x \color{red}{+ \dfrac{25}{16}} = -\dfrac{1}{2} \color{red}{+ \dfrac{25}{16}} }x
2
−
2
5
x+
16
25
=−
2
1
+
16
25
\small{ \left(x \color{blue}{- \dfrac{5}{4}}\right)^2 = \dfrac{17}{16} }(x−
4
5
)
2
=
16
17
\small{ \sqrt{\left(x - \dfrac{5}{4}\right)^2\,} = \pm \sqrt{\dfrac{17}{16}\,} }
(x−
4
5
)
2
=±
16
17
\small{ x - \dfrac{5}{4} = \pm \dfrac{\sqrt{17\,}}{4} }x−
4
5
=±
4
17
\small{ x = \dfrac{5}{4} \pm \dfrac{\sqrt{17\,}}{4} }x=
4
5
±
4
17
The two terms on the right-hand side of the last line above can be combined over a common denominator, and this is often ("usually"?) how the answer will be written, especially if the instructions for the exercise included the stipulation to "simplify" the final answer:
4
5±
17
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