Question

How do you solve ${y^2} - 6y + 8 = 0$?

Answer 1

Hint: Here we will use the sum-product pattern where the given equation will be written in the factored form. Find two numbers $p$ and $q$ that when multiplied together equals $c$ , and when added together equals $b$ . Then, take a term common from the first two terms and another term common from the last two terms. Make factors and keep each factor equal to 0.

Complete step-by-step solution:
sing the Sum-Product pattern,
Firstly, we identify the variables $a,b$ and $c$ from the given equation and then find two numbers $p$ and $q$ that when multiplied together equals $c$ , and when added together equals $b$ .
Given equation: ${y^2} - 6y + 8 = 0$ is of the form $a{y^2} + by + c = 0$ .
Now according to this, we have values of variables as $a = 1,b = - 6,c = 8$ .
Here, the two numbers that when multiplied together equals $c$ , and when added together equals $b$ , are $p = - 4$ and $q = - 2$ .
Now, we will write the equation in factored form. We will rewrite the polynomial splitting the middle term using the two factors found above $p$ and $q$ .
${y^2} - 6y + 8 = 0$
$\Rightarrow$${y^2} - 4y - 2y + 8 = 0$
Add up the first two terms, taking out the common factor,
Add up the last two terms, taking out the common factor,
$\Rightarrow$$y\left( {y - 4} \right) - 2\left( {y - 4} \right) = 0$
Now solving the above equation by taking out the common factor$\left( {y - 4} \right)$.
$\Rightarrow$$\left( {y - 4} \right)\left( {y - 2} \right) = 0$
Now, create two separate equations and then set each factor equal to zero.
$\Rightarrow$$y - 4 = 0$ and $y - 2 = 0$.
Solve the two equations by rearranging them and isolate the variable to find each solution.
$\Rightarrow$$y = 4$ and $y = 2$

Therefore, on solving the given equation ${y^2} - 6y + 8 = 0$ we get the solution $y = 4$ and $y = 2$.

Note: There is also an alternative method to solve the given equation by using the quadratic formula $y = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$, where the values of $a,b$ and $c$ remain same as $a = 1,b = - 6,c = 8$
and after substituting the values in the formula we will get
$y = \dfrac{{ - \left( { - 6} \right) \pm \sqrt {{{\left( { - 6} \right)}^2} - 4\left( 1 \right)\left( 8 \right)} }}{{2\left( 1 \right)}}$
Solving and simplifying the equation, we will get,
$\Rightarrow$\[y = \dfrac{{6 \pm \sqrt {36 - 32} }}{2}\]
$\Rightarrow$\[y = \dfrac{{6 \pm \sqrt 4 }}{2}\]
Simplifying further,
$\Rightarrow$\[y = \dfrac{{6 \pm 2}}{2}\]
To solve for the unknown variable, separate them into two equations, ones with plus and other with minus.
$\Rightarrow$$y = \dfrac{{6 + 2}}{2},\dfrac{{6 - 2}}{2}$
$\Rightarrow$$y = \dfrac{8}{2},\dfrac{4}{2}$
$\Rightarrow$$y = 4,y = 2$
Note that both the methods will give the same solution, it depends on you which method you choose to solve the equation.