Answer:
Step-by-step explanation:
For this qestion u will do step b
See explanation below.
Explanation:
The 'difference between roots and factors of an equation' is not a straightforward question. Let's define both to establish the link between the two..
Assume we have some function of a single variable
x
;
we'll call this
f
(
x
)
Then we can form an equation:
f
(
x
)
=
0
Then the "roots" of this equation are all the values of
x
that satisfy that equation. Remember that these values may be real and/or imaginary.
Now, up to this point we have not assumed anything about
f
x
)
. To consider factors, we now need to assume that
f
(
x
)
=
g
(
x
)
⋅
h
(
x
)
.
That is that
f
(
x
)
factorises into some functions
g
(
x
)
×
h
(
x
)
If we recall our equation:
f
(
x
)
=
0
Then we can now say that either
g
(
x
)
=
0
or
h
(
x
)
=
0
.. and thus show the link between the roots and factors of an equation.
[NB: A simple example of these general principles would be where
f
(
x
)
is a quadratic function that factorises into two linear factors.
3(x)+3=36
-3 -3
3(x)=33
3 3
x=11
Let
![f(x)=\sec^{-1}x](https://tex.z-dn.net/?f=f%28x%29%3D%5Csec%5E%7B-1%7Dx)
. Then
![\sec f(x)=x](https://tex.z-dn.net/?f=%5Csec%20f%28x%29%3Dx)
, and differentiating both sides with respect to
![x](https://tex.z-dn.net/?f=x)
gives
![(\sec f(x))'=\sec f(x)\tan f(x)\,f'(x)=1](https://tex.z-dn.net/?f=%28%5Csec%20f%28x%29%29%27%3D%5Csec%20f%28x%29%5Ctan%20f%28x%29%5C%2Cf%27%28x%29%3D1)
![f'(x)=\dfrac1{\sec f(x)\tan f(x)}](https://tex.z-dn.net/?f=f%27%28x%29%3D%5Cdfrac1%7B%5Csec%20f%28x%29%5Ctan%20f%28x%29%7D)
Now, when
![x=\sqrt2](https://tex.z-dn.net/?f=x%3D%5Csqrt2)
, you get
![(\sec^{-1})'(\sqrt2)=f'(\sqrt2)=\dfrac1{\sec\left(\sec^{-1}\sqrt2\right)\tan\left(\sec^{-1}\sqrt2\right)}](https://tex.z-dn.net/?f=%28%5Csec%5E%7B-1%7D%29%27%28%5Csqrt2%29%3Df%27%28%5Csqrt2%29%3D%5Cdfrac1%7B%5Csec%5Cleft%28%5Csec%5E%7B-1%7D%5Csqrt2%5Cright%29%5Ctan%5Cleft%28%5Csec%5E%7B-1%7D%5Csqrt2%5Cright%29%7D)
You have
![\sec^{-1}\sqrt2=\dfrac\pi4](https://tex.z-dn.net/?f=%5Csec%5E%7B-1%7D%5Csqrt2%3D%5Cdfrac%5Cpi4)
, so
![\sec\left(\sec^{-1}\sqrt2\right)=\sqrt2](https://tex.z-dn.net/?f=%5Csec%5Cleft%28%5Csec%5E%7B-1%7D%5Csqrt2%5Cright%29%3D%5Csqrt2)
and
![\tan\left(\sec^{-1}\sqrt2\right)=1](https://tex.z-dn.net/?f=%5Ctan%5Cleft%28%5Csec%5E%7B-1%7D%5Csqrt2%5Cright%29%3D1)
. So