Answer
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Hint: Here first of all we will find the equivalent equation of the first two terms and will find the middle term and then will place the value in the given equation by completing the square concepts on both the sides.
Complete step-by-step solution:
Here, to find the perfect square, find the half of the middle term, add it to x and square it and keep the constant on the end.
$ {x^2} + 5x \to {\left( {x + \dfrac{5}{2}} \right)^2} + 2 $
Take away the term previously halved and squared.
$ {\left( {x + \dfrac{5}{2}} \right)^2} - \dfrac{{25}}{4} + \dfrac{8}{4} = 0 $
Simplify the above expression, turn the extra constant to have the same denominator.
$ {\left( {x + \dfrac{5}{2}} \right)^2} - \dfrac{{17}}{4} = 0 $
Take the negative term on the opposite side of the equation, when the term is moved from one side to another.
$ {\left( {x + \dfrac{5}{2}} \right)^2} = \dfrac{{17}}{4} $
Take the square root on both sides of the equation.
$ \sqrt {{{\left( {x + \dfrac{5}{2}} \right)}^2}} = \sqrt {\dfrac{{17}}{4}} $
Square and square root cancel each other on the left hand side of the equation.
$ \Rightarrow \left( {x + \dfrac{5}{2}} \right) = \pm \sqrt {\dfrac{{17}}{4}} $
We have put plus or minus signs with the square root, since the square of negative term or the positive term always gives the positive term.
Now, move constant on the right hand side of the equation. When any term is moved from one side to another, then the sign of the term is also changed. Positive terms become negative and vice-versa.
$ \Rightarrow x = \pm \sqrt {\dfrac{{17}}{4}} - \dfrac{5}{2} $
Hence, the solution of the given expression is –
$ \Rightarrow x = \sqrt {\dfrac{{17}}{4}} - \dfrac{5}{2} $ or $ \Rightarrow x = - \sqrt {\dfrac{{17}}{4}} - \dfrac{5}{2} $
This is the required solution.
Note: Always remember that when the square of a negative number is taken, it always gives the positive term. Square is the number multiplied by itself twice. Square and square root always cancel each other. Always remember when you move any term from one side to the another, then the sign of the term also changes. Positive term is changed to negative and the negative term becomes positive.
Complete step-by-step solution:
Here, to find the perfect square, find the half of the middle term, add it to x and square it and keep the constant on the end.
$ {x^2} + 5x \to {\left( {x + \dfrac{5}{2}} \right)^2} + 2 $
Take away the term previously halved and squared.
$ {\left( {x + \dfrac{5}{2}} \right)^2} - \dfrac{{25}}{4} + \dfrac{8}{4} = 0 $
Simplify the above expression, turn the extra constant to have the same denominator.
$ {\left( {x + \dfrac{5}{2}} \right)^2} - \dfrac{{17}}{4} = 0 $
Take the negative term on the opposite side of the equation, when the term is moved from one side to another.
$ {\left( {x + \dfrac{5}{2}} \right)^2} = \dfrac{{17}}{4} $
Take the square root on both sides of the equation.
$ \sqrt {{{\left( {x + \dfrac{5}{2}} \right)}^2}} = \sqrt {\dfrac{{17}}{4}} $
Square and square root cancel each other on the left hand side of the equation.
$ \Rightarrow \left( {x + \dfrac{5}{2}} \right) = \pm \sqrt {\dfrac{{17}}{4}} $
We have put plus or minus signs with the square root, since the square of negative term or the positive term always gives the positive term.
Now, move constant on the right hand side of the equation. When any term is moved from one side to another, then the sign of the term is also changed. Positive terms become negative and vice-versa.
$ \Rightarrow x = \pm \sqrt {\dfrac{{17}}{4}} - \dfrac{5}{2} $
Hence, the solution of the given expression is –
$ \Rightarrow x = \sqrt {\dfrac{{17}}{4}} - \dfrac{5}{2} $ or $ \Rightarrow x = - \sqrt {\dfrac{{17}}{4}} - \dfrac{5}{2} $
This is the required solution.
Note: Always remember that when the square of a negative number is taken, it always gives the positive term. Square is the number multiplied by itself twice. Square and square root always cancel each other. Always remember when you move any term from one side to the another, then the sign of the term also changes. Positive term is changed to negative and the negative term becomes positive.
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