How can you solve by complete square method \[{{x}^{2}}-5x+5=0\]
Answer
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460.2k+ views
Hint: We solve this problem by using the complete square method. Complete square method is that we add some number on both sides so that it gets converted to a square of some linear polynomial and some constant. That is we get in the form
\[{{\left( x+a \right)}^{2}}=b\]
Then we can find the value of \['x'\] easily by applying the square root on both sides.
Complete step by step answer:
We are given that the quadratic equation as
\[\Rightarrow {{x}^{2}}-5x+5=0\]
We know that the square of sum of two numbers as
\[{{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}\]
Let us convert the given equation in this form as
\[\Rightarrow {{\left( x \right)}^{2}}+2\left( \dfrac{-5}{2} \right)\left( x \right)+5=0\]
Here we have two terms from the formula
So, by adding \[{{\left( \dfrac{-5}{2} \right)}^{2}}\] on both sides we get
\[\Rightarrow {{\left( x \right)}^{2}}+2\left( \dfrac{-5}{2} \right)\left( x \right)+{{\left( \dfrac{-5}{2} \right)}^{2}}+5={{\left( \dfrac{-5}{2} \right)}^{2}}\]
Now, by using the formula of square of sum of two numbers we get
\[\begin{align}
& \Rightarrow {{\left( x+\left( \dfrac{-5}{2} \right) \right)}^{2}}=\dfrac{25}{4}-5 \\
& \Rightarrow {{\left( x-\dfrac{5}{2} \right)}^{2}}=\dfrac{5}{4} \\
\end{align}\]
Now by applying square root on both sides we get
\[\begin{align}
& \Rightarrow \left( x-\dfrac{5}{2} \right)=\pm \dfrac{\sqrt{5}}{2} \\
& \Rightarrow x=\dfrac{5\pm \sqrt{5}}{2} \\
\end{align}\]
Here, we get two values of \['x'\] that is one for positive sign and other for negative sign as
\[\Rightarrow x=\dfrac{5+\sqrt{5}}{2},\dfrac{5-\sqrt{5}}{2}\]
Therefore, the roots of given equation \[{{x}^{2}}-5x+5=0\] are \[\dfrac{5+\sqrt{5}}{2},\dfrac{5-\sqrt{5}}{2}\].
Note: We can solve this problem in another way using the formula.
The formula of roots of equation of the form \[a{{x}^{2}}+x+c=0\] is given as
\[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]
We are given that the equation as
\[\Rightarrow {{x}^{2}}-5x+5=0\]
By using the formula of roots we get
\[\begin{align}
& \Rightarrow x=\dfrac{-\left( -5 \right)\pm \sqrt{{{\left( -5 \right)}^{2}}-4\left( 1 \right)\left( 5 \right)}}{2\left( 1 \right)} \\
& \Rightarrow x=\dfrac{5\pm \sqrt{5}}{2} \\
\end{align}\]
Therefore, the roots of given equation \[{{x}^{2}}-5x+5=0\] are \[\dfrac{5+\sqrt{5}}{2},\dfrac{5-\sqrt{5}}{2}\]
This process may not be correct for the given question because we are asked to use the complete square method. So, we need to solve that process then we can check whether the roots we get in that procedure are correct or not using the formula method.
\[{{\left( x+a \right)}^{2}}=b\]
Then we can find the value of \['x'\] easily by applying the square root on both sides.
Complete step by step answer:
We are given that the quadratic equation as
\[\Rightarrow {{x}^{2}}-5x+5=0\]
We know that the square of sum of two numbers as
\[{{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}\]
Let us convert the given equation in this form as
\[\Rightarrow {{\left( x \right)}^{2}}+2\left( \dfrac{-5}{2} \right)\left( x \right)+5=0\]
Here we have two terms from the formula
So, by adding \[{{\left( \dfrac{-5}{2} \right)}^{2}}\] on both sides we get
\[\Rightarrow {{\left( x \right)}^{2}}+2\left( \dfrac{-5}{2} \right)\left( x \right)+{{\left( \dfrac{-5}{2} \right)}^{2}}+5={{\left( \dfrac{-5}{2} \right)}^{2}}\]
Now, by using the formula of square of sum of two numbers we get
\[\begin{align}
& \Rightarrow {{\left( x+\left( \dfrac{-5}{2} \right) \right)}^{2}}=\dfrac{25}{4}-5 \\
& \Rightarrow {{\left( x-\dfrac{5}{2} \right)}^{2}}=\dfrac{5}{4} \\
\end{align}\]
Now by applying square root on both sides we get
\[\begin{align}
& \Rightarrow \left( x-\dfrac{5}{2} \right)=\pm \dfrac{\sqrt{5}}{2} \\
& \Rightarrow x=\dfrac{5\pm \sqrt{5}}{2} \\
\end{align}\]
Here, we get two values of \['x'\] that is one for positive sign and other for negative sign as
\[\Rightarrow x=\dfrac{5+\sqrt{5}}{2},\dfrac{5-\sqrt{5}}{2}\]
Therefore, the roots of given equation \[{{x}^{2}}-5x+5=0\] are \[\dfrac{5+\sqrt{5}}{2},\dfrac{5-\sqrt{5}}{2}\].
Note: We can solve this problem in another way using the formula.
The formula of roots of equation of the form \[a{{x}^{2}}+x+c=0\] is given as
\[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]
We are given that the equation as
\[\Rightarrow {{x}^{2}}-5x+5=0\]
By using the formula of roots we get
\[\begin{align}
& \Rightarrow x=\dfrac{-\left( -5 \right)\pm \sqrt{{{\left( -5 \right)}^{2}}-4\left( 1 \right)\left( 5 \right)}}{2\left( 1 \right)} \\
& \Rightarrow x=\dfrac{5\pm \sqrt{5}}{2} \\
\end{align}\]
Therefore, the roots of given equation \[{{x}^{2}}-5x+5=0\] are \[\dfrac{5+\sqrt{5}}{2},\dfrac{5-\sqrt{5}}{2}\]
This process may not be correct for the given question because we are asked to use the complete square method. So, we need to solve that process then we can check whether the roots we get in that procedure are correct or not using the formula method.
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