Question

How do you solve ${y^2} + 3y + 2 = 0$ ?

Answer 1

To order to determine the solve the above quadratic question use the Splitting up the middle term method by splitting the middle term$3y$into $y + 2y$and then take common $y$from the first two terms and 2 from the last two terms and then find the common binomial parenthesis. Equating all these with 0 you’ll get your desired solution for $y$.

Complete step by step solution:
Given a quadratic equation ${y^2} + 3y + 2$,let it be $f(y)$
$f(y) = {y^2} + 3y + 2 = 0$

Comparing the equation with the standard Quadratic equation $a{y^2} + by + c$
a becomes 1

b becomes 3

And c becomes 2

To find the quadratic factorization we’ll use splitting up the middle term method

So first calculate the product of coefficient of ${y^2}$and the constant term which comes to be
$= 2 \times 1 = 2$

Now the second Step is to find the 2 factors of the number 2 such that the whether addition or subtraction of those numbers is equal to the middle term or coefficient of y and the product of those factors results in the value of constant .

So if we factorize 2 ,the answer comes to be 2 and 1 as $2 + 1 = 3$ that is the middle term . and $1 \times 2 = 2$ which is perfectly equal to the constant value.

Now writing the middle term sum of the factors obtained ,so equation $f(x)$ becomes
$f(y) = {y^2} + y + 2y + 2 = 0$

Now taking common from the first 2 terms and last 2 terms
$\Rightarrow y(y + 1) + 2(y + 1) = 0$

Finding the common binomial parenthesis, the equation becomes
$
\Rightarrow (y + 1)(y + 2) = 0 \\
y + 1 = 0 \\
y = - 1 \\
and \\
y + 2 = 0 \\
y = - 2 \\

$Value of$y$can be$ - 2, - 1$

Therefore, the solution to the quadratic equation ${y^2} + 3y + 2 = 0$is $y = - 2, - 1$

Note: You can also alternatively use a direct method which uses Quadratic Formula to find both roots of a quadratic equation as

${y_1} = \dfrac{{ - b + \sqrt {{b^2} - 4ac} }}{{2a}}$ and ${y_2} = \dfrac{{ - b - \sqrt {{b^2} - 4ac} }}{{2a}}$
${y_1},{y_2}$are root/solution to quadratic equation $a{y^2} + by + c$