Answer
Verified
384.6k+ 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.
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
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
At which age domestication of animals started A Neolithic class 11 social science CBSE
Which are the Top 10 Largest Countries of the World?
Give 10 examples for herbs , shrubs , climbers , creepers
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Difference Between Plant Cell and Animal Cell
Write a letter to the principal requesting him to grant class 10 english CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE