Answer

Verified

341.4k+ 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.

Recently Updated Pages

The branch of science which deals with nature and natural class 10 physics CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

Define absolute refractive index of a medium

Find out what do the algal bloom and redtides sign class 10 biology CBSE

Prove that the function fleft x right xn is continuous class 12 maths CBSE

Find the values of other five trigonometric functions class 10 maths CBSE

Trending doubts

How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE

Distinguish between the reserved forests and protected class 10 biology CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Give simple chemical tests to distinguish between the class 12 chemistry CBSE

Difference Between Plant Cell and Animal Cell

Which of the following books is not written by Harshavardhana class 6 social science CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

In which states of India are mango showers common What class 9 social science CBSE

What Made Mr Keesing Allow Anne to Talk in Class class 10 english CBSE