Question 366757
There is a short way and a long way to solve this equation.<br>
The short way requires that we can recognize that this equation is in quadratic form for (2x+5). If this in not clear to you then perhaps using a temporary variable will help:
Let q = (2x+5)
Subsituting this into your equation we get:
{{{q^2 - q - 6 = 0}}}
This is clearly a quadratic equation. We can solve it by factoring:
{{{(q -3)(q+2) = 0}}}
Then use the Zero Product property which tells us that this (or any) product can be zero <i>only</i> if one of the factors is zero. So:
q - 3 = 0 or q + 2 = 0
Solving these we get:
q = 3 or q = -2
Of course we want to know what x is not q. So we substitute back in for q:
2x + 5 = 3 or 2x + 5 = -2
Solving these we get:
x = -1 or x = -7/2
After a few of these "quadratic form" equations you will not longer need the temporary variable. You will be able to go straight from
{{{(2x+5)^2 - (2x+5)-6=0}}}
to
{{{((2x+5) - 3)((2x + 5) + 2) = 0}}}
which simplifies to:
{{{(2x+2)(2x + 7) = 0}}}
then
2x+2 = 0 or 2x+7 = 0
etc.<br>
The long way is to simplify your equation first. Squaring 2x+5 correctly givess us:
{{{4x^2 + 20x + 25 - (2x + 5) - 6 = 0}}}
Combining like terms we get:
{{{4x^2 +18x + 14 = 0}}}
Now we can factor. First the Greatest Common Factor:
{{{2(2x^2 + 9x + 7) = 0}}}
Then the trinomial:
{{{2(2x + 7)(x + 1) = 0}}}
Then the Zero Product Property:
2 = 0 or 2x+7 = 0 or x+1 = 0
The first equation has no solution. From the other two equations we get:
x = -7/2 or x = -1 (which are the solutions we got the short way)