using the above formulae find the square of given binomials 2x+6
Answers
Answer:
Learn how to solve quadratic equations like x^2=36 or (x-2)^2=49.
Answer:
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− 25 x+ 21 =0
\small{ x^2 - \dfrac{5}{2}x = -\dfrac{1}{2} }x
2 −25 x=− 21
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}} }(
21 )(− 25 )=− 45
\small{ \left(\color{blue}{-\dfrac{5}{4}}\right)^2 = \color{red}{\dfrac{25}{16}} }(−
45 ) 2 = 1625
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− 25 x+ 1625 =− 21 + 1625
\small{ \left(x \color{blue}{- \dfrac{5}{4}}\right)^2 = \dfrac{17}{16} }(x−
45 ) 2 = 1617
\small{ \sqrt{\left(x - \dfrac{5}{4}\right)^2\,} = \pm \sqrt{\dfrac{17}{16}\,} }
(x− 45 ) 2 =± 1617
\small{ x - \dfrac{5}{4} = \pm \dfrac{\sqrt{17\,}}{4} }x−
45 =±417
\small{ x = \dfrac{5}{4} \pm \dfrac{\sqrt{17\,}}{4} }x=
45 ± 417
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:
\mathbf{\color{purple}{\small{ \mathit{x} = \dfrac{5 \pm \sqrt{17\,}}{4} }}}x=
45±17
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