Courses
Courses for Kids
Free study material
Offline Centres
More

How do you solve ${\left( {2x + 1} \right)^2} = {\left( {x + 2} \right)^2}$?

seo-qna
Last updated date: 23rd Feb 2024
Total views: 338.4k
Views today: 10.38k
IVSAT 2024
Answer
VerifiedVerified
338.4k+ views
Hint: The above question is based on the concept of equations having variables on both the sides in the form of quadratic equations. The main approach towards solving the problem is to the sum of squares of two terms on the right-hand side and the left-hand side of the equation and then further simplifying it and getting the value of x.

Complete step by step solution:
In the above given equation, it contains variables on both sides. We first need to solve the sum of squares of two terms present on the left-hand side and right-hand side.
The formula for sum of squares of two terms a and b are as follows:
${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$
Now applying this to the left-hand side we get,
\[{\left( {2x + 1} \right)^2} = 4{x^2} + 4x + 1\]
And then applying it to the term present in the right-hand side we get,
\[{\left( {x + 2} \right)^2} = {x^2} + 4x + 4\]
Now equating both the terms of left-hand side and right-hand side we get,
\[4{x^2} + 4x + 1 = {x^2} + 4x + 4\]
Now subtracting with the term \[{x^2}\] on both the side s we get,
\[
\Rightarrow 4{x^2} + 4x + 1 - {x^2} = {x^2} + 4x + 4 - {x^2} \\
\Rightarrow 3{x^2} + 4x + 1 = 4x + 4 \\
\]
Now subtracting 4x on both the sides we get ,
\[3{x^2} + 1 = 4\]
Then again subtract it with 4
$
\Rightarrow 3{x^2} + 1 - 4 = 4 - 4 \\
\Rightarrow 3{x^2} - 3 = 0 \\
$
Now by getting the common factor out we get,
\[
\Rightarrow 3\left( {{x^2} - 1} \right) = 0 \\
\Rightarrow 3\left( {x + 1} \right)\left( {x - 1} \right) = 0 \\
\]
Now by equating each term with zero we get,
\[
x - 1 = 0 \Rightarrow x = 1 \\
x + 1 = 0 \Rightarrow x = - 1 \\
\]


Note: An important thing to note is that we get two values of x i.e. 1 and -1 .We can cross check the values whether it is correct or not by substituting the values 1 and -1 in the above equation and we get the same terms on the left hand and right hand side then the values are correct.