factorise x square +4x + a square
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rAmmoHan11:
yes . it helped me
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Use completing the square to solve x2 – 4x – 8 = 0.
As noted above, this quadratic does not factor, so I can't solve the equation by factoring. And they haven't given me the equation in a form that is ready to square-root. But there is a way for me to manipulate the quadratic to put it into that ready-for-square-rooting form, so I can solve.
First, I put the loose number on the other side of the equation:
x2 – 4x – 8 = 0
x2 – 4x = 8
Then I look at the coefficient of the x-term, which is –4 in this case. I take half of this number (including the sign), which gives me –2. (I need to keep track of this value. It will simplify my work later on.)
Then I square this value to get +4, and add this squared value to both sides of the equation:
x2 – 4x + 4 = 8 + 4
x2 – 4x + 4 = 12
This process creates a quadratic expression that is a perfect square on the left-hand side of the equation. I can factor, or I can simply replace the quadratic with the squared-binomial form, which is the variable, x, together with the one-half number that I got before (and noted that I'd need later), which was –2. Either way, I get the square-rootable equation:
(x – 2)2 = 12
(I know it's a "–2" inside the parentheses because half of –4 was –2. By noting the sign when I'm finding one-half of the coefficient, I help keep myself from messing up the sign later, when I'm converting to squared-binomial form.)
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(By the way, this process is called "completing the square" because we add a term to convert the quadratic expression into something that factors as the square of a binomial; that is, we've "completed" the expression to create a perfect-square binomial.)
Now I can square-root both sides of the equation, simplify, and solve:
(x – 2)2 = 12
\small{ \sqrt{(x - 2)^2\,} = \pm \sqrt{12\,} }√(x−2)2=±√12
\small{ x - 2 = \pm \sqrt{4 \cdot 3\,} }x−2=±√4⋅3
\small{ x - 2 = \pm 2 \cdot \sqrt{3\,} }x−2=±2⋅√3
\small{ x = 2 \pm 2 \sqrt{3\,} }x=2±2√3
Using this method, I get the same answer as I had before; namely:
\mathbf{\color{purple}{\small{ \mathit{x} = 2 \pm 2 \sqrt{3\,} }}}x=2±2√3
As noted above, this quadratic does not factor, so I can't solve the equation by factoring. And they haven't given me the equation in a form that is ready to square-root. But there is a way for me to manipulate the quadratic to put it into that ready-for-square-rooting form, so I can solve.
First, I put the loose number on the other side of the equation:
x2 – 4x – 8 = 0
x2 – 4x = 8
Then I look at the coefficient of the x-term, which is –4 in this case. I take half of this number (including the sign), which gives me –2. (I need to keep track of this value. It will simplify my work later on.)
Then I square this value to get +4, and add this squared value to both sides of the equation:
x2 – 4x + 4 = 8 + 4
x2 – 4x + 4 = 12
This process creates a quadratic expression that is a perfect square on the left-hand side of the equation. I can factor, or I can simply replace the quadratic with the squared-binomial form, which is the variable, x, together with the one-half number that I got before (and noted that I'd need later), which was –2. Either way, I get the square-rootable equation:
(x – 2)2 = 12
(I know it's a "–2" inside the parentheses because half of –4 was –2. By noting the sign when I'm finding one-half of the coefficient, I help keep myself from messing up the sign later, when I'm converting to squared-binomial form.)
Affiliate
ASK
(By the way, this process is called "completing the square" because we add a term to convert the quadratic expression into something that factors as the square of a binomial; that is, we've "completed" the expression to create a perfect-square binomial.)
Now I can square-root both sides of the equation, simplify, and solve:
(x – 2)2 = 12
\small{ \sqrt{(x - 2)^2\,} = \pm \sqrt{12\,} }√(x−2)2=±√12
\small{ x - 2 = \pm \sqrt{4 \cdot 3\,} }x−2=±√4⋅3
\small{ x - 2 = \pm 2 \cdot \sqrt{3\,} }x−2=±2⋅√3
\small{ x = 2 \pm 2 \sqrt{3\,} }x=2±2√3
Using this method, I get the same answer as I had before; namely:
\mathbf{\color{purple}{\small{ \mathit{x} = 2 \pm 2 \sqrt{3\,} }}}x=2±2√3
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