Question

Guys help mee please with subject of the formula
these 2 questions ​

Answer 1

c)
`x=(y^2\pm√(y^4-4a))/(2)`

d)
`x= (-3)/(p-q^2u)`

Explanation:

c)
`y= (√(x^2 + a) )/(x)`

Solving the question:

`y= (√(x^2+a))/(x)\\Taking\,\,square\,\,on\,\,both\,\,sides\\(y)^2= ((√(x^2+a))/(x))^2\\y^2= (x^2+a)/(x)\\y^2.x = x^2+a\\x^2 +a - y^2x =0\\Rearranging\\x^2 -y^2x +a =0\\Solving \,\,using\,\,quadratic\,\,equation\,\,\\x=(-b\pm√(b^2-4ac))/(2a)\\where\,\, a= 1, b= -y^2 and c= a\\x=(-(-y^2)\pm√((-y^2)^2-4(1)(a)))/(2(1))\\x=(y^2\pm√(y^4-4a))/(2)`

d)
`\sqrt{(px+3)/(ux)}=q`

Solving to find value of x

`\sqrt{(px+3)/(ux)}=q\\ Taking\,\, square\,\, on\,\, both\,\, sides\,\,\\(\sqrt{(px+3)/(ux)})^2=q^2\\(px+3)/(ux) = q^2\\px+3 = q^2.ux\\px = q^2.ux -3\\px - q^2.ux = -3\\x(p-q^2u) = -3\\x= (-3)/(p-q^2u)`