SOLUTION: So i wanted to solve this problem wicht is f(x)=x^2 + 7 Please i really need help

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Question 695470: So i wanted to solve this problem wicht is f(x)=x^2 + 7
Please i really need help

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
+f%28x%29+=+x%5E2+%2B+7+
Can you read this OK?
It says: " There is a function called +f%28x%29+ which
equals +x%5E2+%2B+7+.
-------------------
You can plug in a value for +x+ on both sides like this:
+f%282%29+=+2%5E2+%2B+7+
+f%282%29+=+4+%2B+7+
+f%282%29+=+11+
-------------
When you say " Solve the problem " , I assume you want
to find the "roots" of the equation. The roots are any and
all values of +x+ for which +f%28x%29+=+0+
--------------
+f%28x%29+=+0+
+x%5E2+%2B+7+=+0+
+x%5E2+=+-7+
Do you understand imaginary numbers? You need to use the fact
that +i+=+sqrt%28-1%29+.
+x%5E2+=+-7+
+x%5E2+=+-1%2A7+
Take the square root of both sides
+x+=+sqrt%28-1%29+%2A+sqrt%287%29+
There are two square roots of 7. They are
+sqrt%287%29+ and +-sqrt%287%29+, another answer is
+x+=+sqrt%28-1%29+%2A+%28+-sqrt%287%29+%29+
-------------------------
And, since +i+=+sqrt%28-1%29+,
+x+=+sqrt%287%29%2Ai+
+x+=+-sqrt%287%29%2Ai+
-----------------
So, you have two imaginary solutions to
+x%5E2+%2B+7+=+0+
This is always the case when the plot of the
parabola ( which the function is ) does not
cross the x-axis.
-----------------
If the function +f%28x%29+ did cross the x-axis, then
the solutions would be:
+x+ = value of x at one of the crossings
+x+ = value of x at the other crossing
---------------------------------------
To demonstrate, here's two plots, +f%28x%29+=+x%5E2+%2B+7+
and a parabola which actually crosses the x-axis.
Your function floats above the x-axis, so it can't
have "real" roots. They must be "imaginary"
+graph%28+400%2C+400%2C+-8%2C+8%2C+-20%2C+20%2C+x%5E2+%2B+7%2C+x%5E2+-+7+%29+
Just keep learning how to talk about equations
before solving them. Them key is knowing exactly
what the words mean.
Good luck