Answer

Verified

383.1k+ views

**Hint:**Suppose we have a quadratic equation ${{x}^{2}}+bx+c=0$. Then we add a term ${{\left( \dfrac{b}{2} \right)}^{2}}$ on both sides of the equation to obtain a perfect square that consists of x. This is how we use the completing square method.

**Complete step-by-step solution:**

The given equation is a quadratic equation in one variable. Quadratic equation is an equation in which the degree of the polynomial involved in the equation is equal to 2 (i.e. the higher power of the variable is 2).

To solve a quadratic equation means to find the value of the variable for which it satisfies the given equation. There are many ways to solve a quadratic equation in one variable. One of those ways is completing the square method.

In the completing square method we try to prepare a perfect square of the given equation.

Suppose we have a quadratic equation ${{x}^{2}}+bx+c=0$, then we add a term ${{\left( \dfrac{b}{2} \right)}^{2}}$ on both sides of the equation.

Therefore, we get that ${{x}^{2}}+bx+{{\left( \dfrac{b}{2} \right)}^{2}}+c={{\left( \dfrac{b}{2} \right)}^{2}}$

This equation can be further written as ${{x}^{2}}+bx+{{\left( \dfrac{b}{2} \right)}^{2}}={{\left( \dfrac{b}{2} \right)}^{2}}-c$ ….. (i)

Here, we can see that the expression ${{x}^{2}}+bx+{{\left( \dfrac{b}{2} \right)}^{2}}$ forms a perfect square such that ${{x}^{2}}+bx+{{\left( \dfrac{b}{2} \right)}^{2}}={{\left( x+\dfrac{b}{2} \right)}^{2}}$

Then substitute this value in equation (i).

$\Rightarrow {{\left( x+\dfrac{b}{2} \right)}^{2}}={{\left( \dfrac{b}{2} \right)}^{2}}-c$

Therefore, we can obtain a perfect square in this way.

Let us now perform the same method for the given equation.

Add ${{\left( -\dfrac{5}{2} \right)}^{2}}=\dfrac{25}{4}$ on both sides of the equation ${{x}^{2}}-5x+\dfrac{25}{4}+1=\dfrac{25}{4}$

$\Rightarrow {{\left( x-\dfrac{5}{2} \right)}^{2}}=\dfrac{25}{4}-1$

$\Rightarrow {{\left( x-\dfrac{5}{2} \right)}^{2}}=\dfrac{21}{4}$

Now, we can take square roots on both sides.

$\Rightarrow \left( x-\dfrac{5}{2} \right)=\pm \sqrt{\dfrac{21}{4}}$

$\Rightarrow x-\dfrac{5}{2}=\pm \dfrac{\sqrt{21}}{2}$

$\Rightarrow x=\dfrac{5\pm \sqrt{21}}{2}$

Then, this means that $x=\dfrac{5+\sqrt{21}}{2}$ or $x=\dfrac{5-\sqrt{21}}{2}$.

**Note:**When we have a quadratic equation in the form of $a{{x}^{2}}+bx+c=0$, we have to add the term $\dfrac{1}{a}{{\left( \dfrac{b}{2} \right)}^{2}}$ on both the sides of the equation.

Otherwise, we can divide the whole equation by a factor ‘a’ and then add ${{\left( \dfrac{b}{2} \right)}^{2}}$ on both sides.

Recently Updated Pages

How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE

Why Are Noble Gases NonReactive class 11 chemistry CBSE

Let X and Y be the sets of all positive divisors of class 11 maths CBSE

Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE

Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE

Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE

Trending doubts

Which are the Top 10 Largest Countries of the World?

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

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Guru Purnima speech in English in 100 words class 7 english CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Change the following sentences into negative and interrogative class 10 english CBSE

Difference Between Plant Cell and Animal Cell

Three liquids are given to you One is hydrochloric class 11 chemistry CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE