1) a
2) c
#1 explanation:
x^4-20x^2=-64
x^4-20x^2+64=0
(find a number that multiplies to the last number aka 64 and adds up to b aka 20)
((x-4)(x-16))^2
x= -2,2,-4,4
same for #2 w different variable and numbers
Step 1:
Start by putting

in front of each term
![\frac{d}{dx}[y cos x]= \frac{d}{dx}[5x^2]+ \frac{d}{dx}[ 3y^2]](https://tex.z-dn.net/?f=%20%5Cfrac%7Bd%7D%7Bdx%7D%5By%20cos%20x%5D%3D%20%5Cfrac%7Bd%7D%7Bdx%7D%5B5x%5E2%5D%2B%20%5Cfrac%7Bd%7D%7Bdx%7D%5B%203y%5E2%5D)
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Step 2:
Deal with the terms in 'x' and the constant terms
![\frac{d}{dx}[ycosx]= 10x+ \frac{d}{dx} [3y^2]](https://tex.z-dn.net/?f=%20%5Cfrac%7Bd%7D%7Bdx%7D%5Bycosx%5D%3D%2010x%2B%20%5Cfrac%7Bd%7D%7Bdx%7D%20%5B3y%5E2%5D%20%20)
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Step 3:
Use the chain rule for the terms in 'y'
![\frac{d}{dx}[ycosx]=10x+6y \frac{dy}{dx}](https://tex.z-dn.net/?f=%20%5Cfrac%7Bd%7D%7Bdx%7D%5Bycosx%5D%3D10x%2B6y%20%5Cfrac%7Bdy%7D%7Bdx%7D%20%20)
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Step 4:
Use the product rule on the term in 'x' and 'y'


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Step 5:
Rearrange to make

the subject


![[cos(x) - 6y] \frac{dy}{dx}=10x + y sin(y)](https://tex.z-dn.net/?f=%5Bcos%28x%29%20-%206y%5D%20%20%5Cfrac%7Bdy%7D%7Bdx%7D%3D10x%20%2B%20y%20sin%28y%29%20)

⇒ Final Answer
<span>(3x−2)(2x+3)=0
</span><span>(3x−2)=0; x = 2/3
</span><span>(2x+3)=0; x = -3/2
answer
x = -3/2 and x = 2/3</span>