Question

How do you solve using the completing square method \[4{x^2} + 4x + 1 = 49\] ?

Answer 1

Hint: We first make the coefficient of \[{x^2}\]as 1 by dividing the complete equation by the coefficient. Then shift the constant value to the right hand side of the equation. Add the square of half value of coefficient of x on both sides of the equation. Use the identity \[{(a + b)^2} = {a^2} + {b^2} + 2ab\] to write the left side of the equation. Take the square root on both sides of the equation and write the value by solving the equation. Take LCM on the left side and cross multiply values.

Complete step by step solution:
We are given the quadratic equation \[4{x^2} + 4x + 1 = 49\] … (1)
We will shift all constant values to left hand side first and calculate the constant
\[ \Rightarrow 4{x^2} + 4x + 1 - 49 = 0\]
\[ \Rightarrow 4{x^2} + 4x - 48 = 0\]
Divide whole equation by 4
\[ \Rightarrow {x^2} + x - 12 = 0\]
We can write quadratic equation as \[{x^2} + x - 12 = 0\]
We compare the equation with general quadratic equation i.e. \[a{x^2} + bx + c = 0\]
Coefficient of \[{x^2}\]is 1, coefficient of x is +1 and constant value is -12.
Since we see in equation (1) coefficient of \[{x^2}\]is 1
We will shift the constant term on the right hand side and will proceed with next steps.
\[ \Rightarrow {x^2} + x = 12\]
The coefficient of x is 1, we divide the number by 2 and square it i.e. \[{\left( {\dfrac{1}{2}} \right)^2} = \dfrac{1}{4}\]
Add this term to both sides of the equation
\[ \Rightarrow {x^2} + x + \dfrac{1}{4} = 12 + \dfrac{1}{4}\]
Take LCM on right hand side of the equation
\[ \Rightarrow {x^2} + x + \dfrac{1}{4} = \dfrac{{48 + 1}}{4}\]
We can write left hand side of the equation using the identity \[{(a + b)^2} = {a^2} + {b^2} + 2ab\] and simplify right hand side by simply adding the terms
\[ \Rightarrow {\left( {x + \dfrac{1}{2}} \right)^2} = \dfrac{{49}}{4}\]
Take square root on both sides of the equation
\[ \Rightarrow \sqrt {{{\left( {x + \dfrac{1}{2}} \right)}^2}} = \sqrt {\dfrac{{49}}{4}} \]
We know \[4 = {2^2};49 = {7^2}\]
\[ \Rightarrow \sqrt {{{\left( {x + \dfrac{1}{2}} \right)}^2}} = \sqrt {\dfrac{{{7^2}}}{{{2^2}}}} \]
Cancel square root by square power on both sides of the equation
\[ \Rightarrow \left( {x + \dfrac{1}{2}} \right) = \pm \dfrac{7}{2}\]
Now equate left side with right side of the equation
\[ \Rightarrow x + \dfrac{1}{2} = \dfrac{7}{2}\]and \[x + \dfrac{1}{2} = - \dfrac{7}{2}\]
Shift constant values to right side
\[ \Rightarrow x = \dfrac{7}{2} - \dfrac{1}{2}\]and \[x = - \dfrac{7}{2} - \dfrac{1}{2}\]
\[ \Rightarrow x = \dfrac{6}{2}\]and \[x = - \dfrac{8}{2}\]
\[ \Rightarrow x = 3\] and \[x = - 4\]

\[\therefore \] The solution of \[4{x^2} + 4x + 1 = 49\] is \[x = 3\] and \[x = - 4\].

Note: Many students make mistakes by squaring the value of b and then divide by 2 instead they have to first divide by 2 and then square it. Also, keep in mind when shifting values from one side of the equation to another side of the equation, always change sign from positive to negative and vice-versa. Also, always pay attention to the coefficient of ‘x’ as it signifies which squaring identity we apply here.