Question 153536
{{{x^2 + 3x - 4}}}
Solving this means finding the roots, or finding
the values of {{{x}}} that make {{{x^2 + 3x - 4 = 0}}}
true.
If I just say, off the top of my head, {{{x = 2}}} for
a solution, then I'm saying {{{2^2 + 3*2 - 4 = 0}}} and
{{{4 + 6 - 4 = 0}}}
{{{10 - 4 = 0}}}
{{{6 = 0}}}
That's not true, so {{{x = 2}}} is not a solution
One way to solve this is by completing the square.
First add {{{4}}} to both sides
{{{x^2 + 3x = 4}}}
Now take one-half of the coefficient of the {{{x}}} term,
square it, and add it to both sides, like this:
{{{x^2 + 3x + (3/2)^2 = 4 + (3/2)^2}}}
{{{x^2 + 3x + (9/4) = (16/4) + (9/4)}}}
{{{x^2 + 3x + (9/4) = 25/4}}}
This is the same as
{{{(x + (3/2))^2 = 25/4}}}
Now take the square root of both sides
{{{x + (3/2) = 5/2}}}
and also
{{{x + (3/2) = -(5/2)}}} (since {{{25/4}}} has a + and a - square root)
Subtract {{{3/2}}} from both sides
{{{x = (5/2) - (3/2)}}}
{{{x = 1}}}
And also
{{{x = -(5/2) - (3/2)}}}
{{{x = -4}}}
So, the solutions are {{{x = 1}}} and {{{x = -4}}}
Plug these values into the equation to see if they
are roots, for instance:
{{{(-4)^2 + 3*(-4) - 4 = 0}}}
{{{16 - 12 - 4 = 0}}}
{{{16 - 16 = 0}}}
{{{0 = 0}}}
So, {{{x = -4}}} is a solution