Can someone pwease help mwe wit dis?
2 answers:
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The extraneous solution of the given radical equation is x = 3
<h3>Finding the extraneous solution:</h3>
Here we have the radical equation:
We want to get the extraneous solution, to get it, we first need to square both sides of the equation, so we get:
Now we got a quadratic equation, we can rewrite it as:
To solve this we use Bhaskara's formula, we will get:
Then the two solutions are:
x = (15 + 9)/2 = 12
x = (15 - 9)/2 = 3
<h3>Which one is the extraneous solution?</h3>Let's evaluate both solutions in our original equation and let's see which is the one that does not work:
for x = 12 we have:
This is true.
For x = 3 we have:
This is false, so the extraneous solution is x = 3.
If you want to learn more about extraneous solutions, you can read:
Answer: H =
Step-by-step explanation:
Step 1: Add h to both sides.
hx+x²+h=−h+x+h
hx+x²+h=x
Step 2: Add -x^2 to both sides.
hx+x²+h+−x²=x+−x²
hx+h=−x²+x
Step 3: Factor out variable h.
h(x+1)=−x²+x
Step 4: Divide both sides by x+1
Answer:
Is in the attached image.
Hope this helped
-Adan
