Question

How do you solve by completing the square ${{x}^{2}}-8x-1=0$?

Answer 1

Hint: In this question, we need to find the value of x from the equation ${{x}^{2}}-8x-1=0$ by completing the square. For this, we will add and subtract term y in the left side of the equation. y for an equation of the form ${{x}^{2}}+bx+c$ is given by ${{\left( \dfrac{b}{2} \right)}^{2}}$. After that, we will use positive y to form a whole square of the form ${{\left( a-b \right)}^{2}}$ and take the left out constant term on the other side. At last, taking square root on both sides and solving for x will give us the value of x.

Complete step by step answer:
Here we are given the equation as ${{x}^{2}}-8x-1=0$. We need to solve it by completing the square. For this, let us add and subtract a certain element y in the left side of the equation. We know that, for any equation of the form ${{x}^{2}}+bx+c=0$ we add and subtract ${{\left( \dfrac{b}{2} \right)}^{2}}$ element to complete the square. As we can see, ${{x}^{2}}-8x-1=0$ is of the form ${{x}^{2}}+bx+c$ where b = -8 and c = -1.
So here the element y will be ${{\left( \dfrac{-8}{2} \right)}^{2}}={{\left( -4 \right)}^{2}}=16$.
So let us add and subtract it on the left side we get ${{x}^{2}}-8x-1+16-16=0\Rightarrow {{x}^{2}}-8x+16-17=0$.
As we can say, ${{x}^{2}}-8x+16$ can be written as ${{\left( x \right)}^{2}}-2\times x\times 4+{{\left( 4 \right)}^{2}}$ which is of the form ${{a}^{2}}-2ab+{{b}^{2}}$ Therefore, it will be equal to ${{\left( a-b \right)}^{2}}$. Hence we have ${{\left( x-4 \right)}^{2}}-17=0$.
Taking 17 to the other side we get ${{\left( x-4 \right)}^{2}}=17$.
Now let us take square root on both sides to solve for x we get \[{{\left( {{\left( x-4 \right)}^{2}} \right)}^{\dfrac{1}{2}}}=\pm {{\left( 17 \right)}^{\dfrac{1}{2}}}\Rightarrow x-4=\pm {{\left( 17 \right)}^{\dfrac{1}{2}}}\].
In radical form we have \[x-4=\pm \sqrt{17}\].
Taking 4 to the other side we have \[x=\pm \sqrt{17}+4\].
So the two values of x become $x=4+\sqrt{17}\text{ and }x=4-\sqrt{17}$.
These are our required values.

Note:
Students should note that, before selecting y make sure that coefficient of ${{x}^{2}}$ is 1. If the coefficient of ${{x}^{2}}$ is then divide the whole equation by a to make the coefficient 1. The equation should be of the form ${{x}^{2}}+bx+c$ to find the element y. Take care if the whole square will be ${{\left( a-b \right)}^{2}},{{\left( a+b \right)}^{2}}$. It depends upon the sign of the coefficient of x.