Question

How do you solve the quadratic ${y^2} + 2y + 1 = 0$ using any method?

Answer 1

Hint: We are given the quadratic equation. We have to find the solution of the quadratic equation. First, we will factorize the equation by applying the algebraic identity. Then, we will set each factor equal to 0 and solve for y.

Complete step by step solution:
We are given the quadratic equation, ${y^2} + 2y + 1 = 0$
First, we will apply the algebraic identity ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$ to the equation.
Here, $a = y$ and $b = 1$
$ \Rightarrow {y^2} + 2y + 1 = {\left( {y + 1} \right)^2}$
Now, set the factor equal to 0.
$ \Rightarrow {\left( {y + 1} \right)^2} = 0$
Solve the equation for y.
$ \Rightarrow y = - 1$

Final Answer: Thus, the solution of the equation is $y = - 1$.

Additional Information:
The factorization of the quadratic equation can be done in many ways such as factorization by splitting the middle term. In middle term splitting, the sum must be equal to 2y and the product is equal to 1. Then, take out the common terms and find the factors of the equation. Then, set each factor equal to zero and solve the equation for the variable. Another method to factorize the equation is by completing the square if it is not possible to apply the algebraic identity directly to the equation.

Note:
In such types of questions students mainly do mistakes while applying the algebraic identity. Students may get confused on which algebraic identity must be applied. In such types of questions, students can also make calculation mistakes while solving for the variable.
Alternate method:
Another method to find the solution of the equation $a{x^2} + bx + c = 0$ is by applying the quadratic formula, $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Here, $a = 1$, $b = 2$ and $c = 1$
Substitute the values into the quadratic formula.
$ \Rightarrow y = \dfrac{{ - 2 \pm \sqrt {{2^2} - 4 \times 1 \times 1} }}{{2 \times 1}}$
On simplifying the expression, we get:
$ \Rightarrow y = \dfrac{{ - 2 \pm \sqrt {4 - 4} }}{2}$
$ \Rightarrow y = \dfrac{{ - 2 \pm 0}}{2}$
$ \Rightarrow y = - 1$
Thus, solution is same using this method also $y = - 1$