SOLUTION: Hello: Thank you so much for offering the free tutoring. My exercise is: x^2+4x+4=7. I have gotten to: x^2+4x+4=7=0 x^2+4x+(-3)=0 now I am stuck. Thank you for

Algebra ->  Radicals -> SOLUTION: Hello: Thank you so much for offering the free tutoring. My exercise is: x^2+4x+4=7. I have gotten to: x^2+4x+4=7=0 x^2+4x+(-3)=0 now I am stuck. Thank you for       Log On


   

Question 100472: Hello:
Thank you so much for offering the free tutoring.
My exercise is:
x^2+4x+4=7.
I have gotten to:
x^2+4x+4=7=0
x^2+4x+(-3)=0
now I am stuck.
Thank you for your help.
Answer by bucky(2189) About Me  (Show Source):
You can put this solution on YOUR website!
The most common ways to solve a quadratic equation are by graphing, factoring, and completing
the square. Graphing has the disadvantage of being slow and requiring the scale to be
big enough so you can read it accurately. Factoring into a "nice" set of factors is not
always possible. (That is the case with this problem.) But completing the square can be
counted on to give you an answer. Using the quadratic formula is just a convenient
form of completing the square.
.
The quadratic formula says that if you can arrange the quadratic equation into the standard form:
.
ax%5E2+%2B+bx+%2B+c+=+0
.
The solutions will be given by the equation:
.
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
.
You have worked your problem down to the form:
.
x%5E2%2B4x%2B%28-3%29=0
.
Let's just remove the parentheses on the left side to make the equation:
.
x%5E2%2B4x-3=0
.
Notice that this compares to the standard form in which a = +1, b = +4, and c = -3.
Therefore, you can go to the solutions equation and substitute these values of a, b, and c
into it to get the answers for x.
.
Start with the solutions equation:
.
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
.
and substitute +1 for a, +4 for b, and -3 for c to get:
.
x+=+%28-%284%29+%2B-+sqrt%28+%284%29%5E2-4%2A1%2A%28-3%29+%29%29%2F%282%2A1%29+
.
within the radical sign note that 4%5E2=+16 and -4%2A1%2A%28-3%29+=+12. Substituting
these into the radical simplifies the equation to:
.
x+=+%28-%284%29+%2B-+sqrt%28++16%2B12%29%29%2F%282%2A1%29+
.
and the terms in the radical combine to 28 to give:
.
x+=+%28-%284%29+%2B-+sqrt%2828%29%29%2F%282%2A1%29+
.
Note that the radical can be simplified. By the rules of radicals:
sqrt%2828%29+=+sqrt%284%2A7%29+=+sqrt%284%29%2Asqrt%287%29+=+2%2Asqrt%287%29
.
Therefore, sqrt%2828%29 can be replaced by 2%2Asqrt%287%29 and the answer equation
simplifies to:
.
x+=+%28-%284%29+%2B-+2%2Asqrt%287%29%29%2F%282%2A1%29+
.
Now, remove the parentheses from -(4) to make it just -1 and multiply the 2*1 in the denominator
to get just 2. With these changes the answer equation simplifies to:
.
x+=+%28-4+%2B-+2%2Asqrt%287%29%29%2F2+
.
So the two answers to your problem are:
.
x+=+%28-4+%2B+2%2Asqrt%287%29%29%2F2+ and x+=+%28-4+-+2%2Asqrt%287%29%29%2F2+
.
You can approximate these answers by using a calculator to find that sqrt%287%29+=+2.645751311
and substituting that into the two answers results in:
.

.
for one answer and
.

.
for the other answer.
.
In summary, the two answers are x = 0.645751311 and x = -4.645751311
.
Hope this helps you to understand the problem a little better and gets you familiar with
using the quadratic formula as a method you can count on to solve a quadratic equation
that is giving you trouble.