What is the question you are asking?
<h2>
The required 'option a) - 9 , 
, 4' is correct.</h2>
Step-by-step explanation:
We have,
f(x) = (x + 9)(x − 4)(6x + 1)
To find, all zeroes of the given equation = ?
∴ f(x) = (x + 9)(x − 4)(6x + 1)
⇒ (x + 9)(x − 4)(6x + 1) = 0
⇒ x + 9 = 0 or, x − 4 = 0 or, 6x + 1 = 0
⇒ x + 9 = 0 ⇒ x = - 9
⇒ x − 4 = 0 ⇒ x = 4
⇒ 6x + 1 = 0
⇒ 6x = - 1
⇒ x =
∴ x = - 9 , , 4
Thus, the required 'option a) - 9 , , 4' is correct.
- I’m assuming the expression implies the following:
(1-n)/(n^2 - 1)
The steps goes as follows:
= -(n-1)/(n^2-1) (factored a negative out)
= -(n-1)/[(n+1)(n-1)] (Factored denominator into a difference of squares, but don’t just blindly accept the fact, expand it and you should get the original denominator)
At this point, the (n-1) terms cancel, as you should know: a real number, besides 0, over itself is 1.
= -1/(n+1) , which is the answer
Note: You could also see it as 1/-(n+1) or 1/(-n-1)
Answer:
The values of p in the equation are 0 and 6
Step-by-step explanation:
First, you have to make the denominators the same. to do that, first factor 2p^2-7p-4 = \left(2p+1\right)\left(p-4\right)2p
2
−7p−4=(2p+1)(p−4)
So then the equation looks like:
\frac{p}{2p+1}-\frac{2p^2+5}{(2p+1)(p-4)}=-\frac{5}{p-4}
2p+1
p
−
(2p+1)(p−4)
2p
2
+5
=−
p−4
5
To make the denominators equal, multiply 2p+1 with p-4 and p-4 with 2p+1:
\frac{p^2-4p}{(2p+1)(p-4)}-\frac{2p^2+5}{(2p+1)(p-4)}=-\frac{10p+5}{(p-4)(2p+1)}
(2p+1)(p−4)
p
2
−4p
−
(2p+1)(p−4)
2p
2
+5
=−
(p−4)(2p+1)
10p+5
Since, this has an equal sign we 'get rid of' or 'forget' the denominator and only solve the numerator.
(p^2-4p)-(2p^2+5)=-(10p+5)(p
2
−4p)−(2p
2
+5)=−(10p+5)
Now, solve like a normal equation. Solve (p^2-4p)-(2p^2+5)(p
2
−4p)−(2p
2
+5) first:
(p^2-4p)-(2p^2+5)=-p^2-4p-5(p
2
−4p)−(2p
2
+5)=−p
2
−4p−5
-p^2-4p-5=-10p+5−p
2
−4p−5=−10p+5
Combine like terms:
-p^2-4p+0=-10p−p
2
−4p+0=−10p
-p^2+6p=0−p
2
+6p=0
Factor:
p=0, p=6p
