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# How do you solve ${{x}^{2}}-5x+1=0$ by completing the square?

Last updated date: 12th Aug 2024
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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.