Answer
385.8k+ views
Hint: We first make the coefficient of \[{x^2}\] as 1 by dividing the complete equation by the coefficient of \[{x^2}\]. Then shift the constant value to the right hand side of the equation. Add the square of half value of coefficient of ‘x’ on both sides of the equation. Afterwards we can simplify this using some simple algebraic identity and by taking LCM we will get the desired result.
Complete step-by-step solution:
Given, \[{x^2} - 5x + 8 = 0\].
We can see that the coefficient of \[{x^2}\] is 1. So no need to divide the equation by the coefficient of \[{x^2}\].
The next step is we need to shift the constant terms to the right hand side of the equation,
\[{x^2} - 5x = - 8{\text{ }} - - - - (1)\].
Now we can see that the coefficient of ‘x’ is \[ - 5\]. We divide the coefficient of ‘x’ by 2 and we square it.
\[{\left( {\dfrac{{ - 5}}{2}} \right)^2} = \dfrac{{25}}{4}\].
Now we need to add \[\dfrac{{25}}{4}\] on both sides of the equation (1).
\[ \Rightarrow {x^2} - 5x + \dfrac{{25}}{4} = - 8 + \dfrac{{25}}{4}\]
We know the algebraic identity \[{(a - b)^2} = {a^2} - 2ab + {b^2}\]. Comparing this with the left hand side of an equation we have \[a = x\] and \[b = \dfrac{5}{2}\].
\[ \Rightarrow {\left( {x - \dfrac{5}{2}} \right)^2} = - 8 + \dfrac{{25}}{4}\]
\[ \Rightarrow {\left( {x - \dfrac{5}{2}} \right)^2} = \dfrac{{ - 32 + 25}}{4}\]
\[ \Rightarrow {\left( {x - \dfrac{5}{2}} \right)^2} = \dfrac{{ - 7}}{4}\]
Taking square root on both side we have,
\[ \Rightarrow \left( {x - \dfrac{5}{2}} \right) = \pm \sqrt {\dfrac{{ - 7}}{4}} \]
\[ \Rightarrow \left( {x - \dfrac{5}{2}} \right) = \pm \dfrac{{i\sqrt 7 }}{2}\]
That is we have two roots,
\[ \Rightarrow \left( {x - \dfrac{5}{2}} \right) = \dfrac{{i\sqrt 7 }}{2}\] and \[\left( {x - \dfrac{5}{2}} \right) = - \dfrac{{i\sqrt 7 }}{2}\]
\[ \Rightarrow x = \dfrac{{i\sqrt 7 }}{2} + \dfrac{5}{2}\] and \[x = - \dfrac{{i\sqrt 7 }}{2} + \dfrac{5}{2}\],
\[ \Rightarrow x = \dfrac{{i\sqrt 7 + 5}}{2}\] and \[x = \dfrac{{ - i\sqrt 7 + 5}}{2}\], is the required solution.
Note: Since we have a polynomial of degree two and hence it is called quadratic polynomial. If we have a polynomial of degree ‘n’ then we have ‘n’ roots. In the given problem we have a degree equal to 2. Hence the number of roots are 2. Also keep in mind when shifting values from one side of the equation t0 another side of the equation, always change sign from positive to negative and vice-versa.
Complete step-by-step solution:
Given, \[{x^2} - 5x + 8 = 0\].
We can see that the coefficient of \[{x^2}\] is 1. So no need to divide the equation by the coefficient of \[{x^2}\].
The next step is we need to shift the constant terms to the right hand side of the equation,
\[{x^2} - 5x = - 8{\text{ }} - - - - (1)\].
Now we can see that the coefficient of ‘x’ is \[ - 5\]. We divide the coefficient of ‘x’ by 2 and we square it.
\[{\left( {\dfrac{{ - 5}}{2}} \right)^2} = \dfrac{{25}}{4}\].
Now we need to add \[\dfrac{{25}}{4}\] on both sides of the equation (1).
\[ \Rightarrow {x^2} - 5x + \dfrac{{25}}{4} = - 8 + \dfrac{{25}}{4}\]
We know the algebraic identity \[{(a - b)^2} = {a^2} - 2ab + {b^2}\]. Comparing this with the left hand side of an equation we have \[a = x\] and \[b = \dfrac{5}{2}\].
\[ \Rightarrow {\left( {x - \dfrac{5}{2}} \right)^2} = - 8 + \dfrac{{25}}{4}\]
\[ \Rightarrow {\left( {x - \dfrac{5}{2}} \right)^2} = \dfrac{{ - 32 + 25}}{4}\]
\[ \Rightarrow {\left( {x - \dfrac{5}{2}} \right)^2} = \dfrac{{ - 7}}{4}\]
Taking square root on both side we have,
\[ \Rightarrow \left( {x - \dfrac{5}{2}} \right) = \pm \sqrt {\dfrac{{ - 7}}{4}} \]
\[ \Rightarrow \left( {x - \dfrac{5}{2}} \right) = \pm \dfrac{{i\sqrt 7 }}{2}\]
That is we have two roots,
\[ \Rightarrow \left( {x - \dfrac{5}{2}} \right) = \dfrac{{i\sqrt 7 }}{2}\] and \[\left( {x - \dfrac{5}{2}} \right) = - \dfrac{{i\sqrt 7 }}{2}\]
\[ \Rightarrow x = \dfrac{{i\sqrt 7 }}{2} + \dfrac{5}{2}\] and \[x = - \dfrac{{i\sqrt 7 }}{2} + \dfrac{5}{2}\],
\[ \Rightarrow x = \dfrac{{i\sqrt 7 + 5}}{2}\] and \[x = \dfrac{{ - i\sqrt 7 + 5}}{2}\], is the required solution.
Note: Since we have a polynomial of degree two and hence it is called quadratic polynomial. If we have a polynomial of degree ‘n’ then we have ‘n’ roots. In the given problem we have a degree equal to 2. Hence the number of roots are 2. Also keep in mind when shifting values from one side of the equation t0 another side of the equation, always change sign from positive to negative and vice-versa.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)