Regents Exam Algebra 1 August 2016 Problem number 19.
What are the solutions to the equation 3x squared plus 10x equals 8?
In this problem we are asked to determine the solutions of the quadratic equation.
We can find the solutions in a couple of ways we can go ahead and graph the function and
find the solutions by using a graphing calculator or we can attempt to factor out the quadratic
function and use the zero product property to find the solutions.
Let's fist solve the problem by graphing the function.
We can find the solutions by graphing in a couple of ways we can graph two distinct functions,
the functions can be obtained by using the expression on the left and right side of the
equation in this case we can graph the function y equals 3x squared plus 10x and graph the
function y equals 8 on the same graph, doing that we obtain the following.
Now we need to determine where the first function intersects the second function, we can determine
this by inspection, looking at the graphs we see that both functions intersect at the
point two thirds 8 and negative four 8.
We can also find the solutions by rewriting the equation and setting it equal to zero,
so we subtract 8 from both sides of the equation and simplify, we then graph the function 3x
squared plus 10x minus 8, the solutions to the equation will be the points where the
function crosses the x-axis, graphing the function we see that this function crosses
the x-axis at the same points as the previous graph.
Either of the two methods will work if you are trying to find the solutions of an equation,
you can graph the expressions on the left and right side of the equation and find the
points where both curves intersect or you can set the equation equal to zero and find
the points where the curve crosses the x-axis.
An alternative way of finding the solutions of a quadratic equation is by factoring.
We start by rewriting the equation by setting it equal to zero so we subtract 8 from both
sides and simplify.
Notice that we have a quadratic trinomial with a leading coefficient that is greater
than 1 so we will use the method of decomposition and grouping to factor out the quadratic trinomial.
We first need to multiply the constant term in this case negative 8 by the leading coefficient
in this case positive 3 obtaining negative 24.
We then need to find two integers that multiply to negative 24 and whose sum is the coefficient
of the linear term in this case 10.
We can systematically go through all the possible combinations until we find a perfect match.
The integers that multiply to 24 are 1 and 24, 2 and 12, 3 and 8, and 4 and 6, we can
obtain negative 24 by making one of the integers negative, since we want the sum to add up
to the positive 10 we should use the integers 2 and 12 and make 2 a negative integer so
that we obtain positive 10 when we add the integers together.
Now that we have found our integers we proceed with the decomposition step and decompose
the linear term, in this case 10x, as a sum of these two integers in this case negative
2x and positive 12x, now we proceed with the grouping step so we are going to group a decomposed
term with a non-decomposed term in this case we will group together 3x squared and negative
2x and group together 12x and negative 8.
Next we factor out the GCF from each grouping, for the first grouping it is x and for the
second grouping it is 4.
Doing that we obtain the following, notice that we now have a common factor between both
terms in this case the quantity 3x minus 2 this is a good sign, the presence of a common
factor indicates that we are on the right track.
The next step is to factor out this quantity from both terms as follows.
We now have the complete factored form of the quadratic trinomial, the final step is
to use the zero product property, set each factor equal to zero and solve for x.
Doing that we obtain x is equal to two thirds and negative 4.
Looking at the options we see that the first option is a perfect match.
So option 1 is our answer.
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