Question

How do you solve the \[5{x^2} - 30x + 5 = 0\] using completely the square?

Answer 1

Hint: In this question, we should solve the quadratic equation using complete square, first take out the common term from the equation here it will be 5 then the equation becomes \[{x^2} - 6x + 1 = 0\], next transform the equation such that the constant term is on the right side, and divide both sides with the coefficient of \[{x^2}\]term i.e., 1, now add the square of the half of the coefficient to both sides, here it is \[{\left( {\dfrac{6}{2}} \right)^2}\] which will be equal to 9 ,now factor the square of the binomial on the left side, and take the square root on both sides, and solve for required \[x\].

Complete step by step solution:
Completing the Square is a method used to solve a quadratic equation by changing the form of the equation so that the left side is a perfect square trinomial.
To solve \[a{x^2} + bx + c = 0\]by completing the square:
      1. Transform the equation so that the constant term,\[c\], is alone on the right side.
      2. If \[a\] the leading coefficient (the coefficient of the \[{x^2}\] term), is not equal to 1, divide both sides by \[a\].
      3. Add the square of half the coefficient of the \[x\]-term, \[{\left( {\dfrac{b}{{2a}}} \right)^2}\] to both sides of the equation.
      4. Factor the left side as the square of a binomial.
      5. Take the square root of both sides. (Remember: \[{\left( {x + q} \right)^2} = r\] is equivalent to \[x + q = \sqrt r \].)
      6. Solve for \[x\].
Now given quadratic equation, \[5{x^2} - 30x + 5 = 0\],
Taking common term i.e., 5 from the equation we get,
\[ \Rightarrow 5\left( {{x^2} - 6x + 1} \right) = 0\],
Now the equation becomes,
\[ \Rightarrow {x^2} - 6x + 1 = 0\],
Now transforming the equation we get,
\[ \Rightarrow {x^2} - 6x = - 1\],
Now comparing this to \[{a^2} + 2ab + {b^2}\] format, we do the following re-grouping, we further add and subtract \[{\left( {\dfrac{6}{2}} \right)^2}\], which is equal to 9, from both the sides we get,
\[ \Rightarrow \]\[{x^2} - 6x + 9 = - 1 + 9\]
Now we can see that the half of the expression represents a perfect square, we get,
\[ \Rightarrow \]\[{\left( {x - 3} \right)^2} = 8\],
Now taking out the square we get,
\[ \Rightarrow \]\[\left( {x - 3} \right) = \pm \sqrt 8 \],
Now simplifying we get,
\[ \Rightarrow x - 3 = \pm 2\sqrt 2 \],
Now taking the constant term to right hand side we get,
\[ \Rightarrow \]\[x = 3 \pm 2\sqrt 2 \],
The value for\[x\] is \[3 \pm 2\sqrt 2 \].

The value of \[x\]when the quadratic equation \[5{x^2} - 30x + 5 = 0\] is solved by completely the square will be equal to \[3 \pm 2\sqrt 2 \].

Note:
In these type of questions, we can solve by using quadratic formula i.e., \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\], but we should keep in mind that we can also solve the equation using completely the square, and we can cross check the values of \[x\] by using the above formula. Also we should always convert the coefficient of \[{x^2} = 1\], to easily solve the equation by this method.