
Solve the given equation \[{x^2} + 2x + 4 = 0\], by completing the square?
Answer
558.6k+ views
Hint:To complete a square of any quadratic equation you should be aware about the formulae or expansion of \[{(a + b)^2}\] and after you compare your given equation with the expansion of this equation you can have your solution. The expansion of above equation is:
\[{(a + b)^2} = {a^{}} + 2ab + {b^2}\]
Formulae Used:
\[{(a + b)^2} = {a^{}} + 2ab + {b^2}\], \[i = \sqrt {( - 1)} \]
Complete step by step solution:
For the given question we have the quadratic equation as \[{x^2} + 2x +4 = 0\]. Now comparing this equation with the standard equation that is \[{(a + b)^2}\].We get,
\[a = x,\,{b^2} = 4 = {2^2} \\
\Rightarrow b = 2 \]
Comparing with the expansion of the standard equation that is \[{(a + b)^2}\] we get that the
\[2ab\] term is not complete in our equation that is we have to do some subtraction in the given
equation to meet the necessary term; On subtraction “3” on both side of the equation the new equation formed is;
\[
{x^2} + 2x + 4 - 3 = 0 - 3 \\
\Rightarrow {x^2} + 2x + 1 = - 3 \\
\]
On comparing this new equation with the standard equation that is \[{(a + b)^2}\]
We get ;
\[a = x,\,{b^2} = 1 = {1^2} \\
\Rightarrow b = 1\]
Now we can write the equation in square form as;
\[{(x + 1)^2} = {(\sqrt 3 i)^2}\]
Since for any squared equation L.H.S and R.H.S should be positive,here ,$i = \sqrt {( - 1)}$.Now solving this equation for value of “x” we get;taking square root both side of equation we get
\[
\sqrt {{{(x + 1)}^2}} = \sqrt {{{(\sqrt 3 i)}^2}} \\
\Rightarrow(x + 1) = (\sqrt 3 i) \\
\therefore x = \sqrt 3 i - 1 \\
\]
Here for this equation the value of “x” is complex, not real and the obtained value is \[x = \sqrt 3 i - 1\].
Note: For such an equation you have to compare with the standard equation, for that you need to compare first the “a”,”b” then for “2ab” and accordingly you can go with the adjustment needed for exact comparison. Something few comparisons are needed with “a” also but mostly the constant term is needed to be edited after adding or subtracting some quantity on both sides of the equation.
\[{(a + b)^2} = {a^{}} + 2ab + {b^2}\]
Formulae Used:
\[{(a + b)^2} = {a^{}} + 2ab + {b^2}\], \[i = \sqrt {( - 1)} \]
Complete step by step solution:
For the given question we have the quadratic equation as \[{x^2} + 2x +4 = 0\]. Now comparing this equation with the standard equation that is \[{(a + b)^2}\].We get,
\[a = x,\,{b^2} = 4 = {2^2} \\
\Rightarrow b = 2 \]
Comparing with the expansion of the standard equation that is \[{(a + b)^2}\] we get that the
\[2ab\] term is not complete in our equation that is we have to do some subtraction in the given
equation to meet the necessary term; On subtraction “3” on both side of the equation the new equation formed is;
\[
{x^2} + 2x + 4 - 3 = 0 - 3 \\
\Rightarrow {x^2} + 2x + 1 = - 3 \\
\]
On comparing this new equation with the standard equation that is \[{(a + b)^2}\]
We get ;
\[a = x,\,{b^2} = 1 = {1^2} \\
\Rightarrow b = 1\]
Now we can write the equation in square form as;
\[{(x + 1)^2} = {(\sqrt 3 i)^2}\]
Since for any squared equation L.H.S and R.H.S should be positive,here ,$i = \sqrt {( - 1)}$.Now solving this equation for value of “x” we get;taking square root both side of equation we get
\[
\sqrt {{{(x + 1)}^2}} = \sqrt {{{(\sqrt 3 i)}^2}} \\
\Rightarrow(x + 1) = (\sqrt 3 i) \\
\therefore x = \sqrt 3 i - 1 \\
\]
Here for this equation the value of “x” is complex, not real and the obtained value is \[x = \sqrt 3 i - 1\].
Note: For such an equation you have to compare with the standard equation, for that you need to compare first the “a”,”b” then for “2ab” and accordingly you can go with the adjustment needed for exact comparison. Something few comparisons are needed with “a” also but mostly the constant term is needed to be edited after adding or subtracting some quantity on both sides of the equation.
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