Answer
Verified
411k+ 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
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
Change the following sentences into negative and interrogative class 10 english CBSE
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
How do you graph the function fx 4x class 9 maths CBSE
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
Application to your principal for the character ce class 8 english CBSE
Write a letter to the principal requesting him to grant class 10 english CBSE