Explain the "Completing The Square" method to solve quadratic equations with examples.
Give detailed step by step explanation. Give at least 2-3 examples. Class 10 CBSE
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x² – 4x – 8 = 0
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.)
Now I can square-root both sides of the equation, simplify, and solve:

Using this method, I get
x=2±23
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.)
Now I can square-root both sides of the equation, simplify, and solve:
Using this method, I get
x=2±23
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GovindKrishnan:
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Explain the "Completing The Square" method to solve quadratic equations with examples.
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