Answer:
c) [tex]x=\frac{y^2\pm\sqrt{y^4-4a}}{2}[/tex]
d) [tex]x= \frac{-3}{p-q^2u}[/tex]
Step-by-step explanation:
c) [tex]y= \frac{\sqrt{x^2 + a} }{x}[/tex]
Solving the question:
[tex]y= \frac{\sqrt{x^2+a}}{x}\\Taking\,\,square\,\,on\,\,both\,\,sides\\(y)^2= (\frac{\sqrt{x^2+a}}{x})^2\\y^2= \frac{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=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\where\,\, a= 1, b= -y^2 and c= a\\x=\frac{-(-y^2)\pm\sqrt{(-y^2)^2-4(1)(a)}}{2(1)}\\x=\frac{y^2\pm\sqrt{y^4-4a}}{2}[/tex]
d) [tex]\sqrt{\frac{px+3}{ux}}=q[/tex]
Solving to find value of x
[tex]\sqrt{\frac{px+3}{ux}}=q\\ Taking\,\, square\,\, on\,\, both\,\, sides\,\,\\(\sqrt{\frac{px+3}{ux}})^2=q^2\\\frac{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= \frac{-3}{p-q^2u}[/tex]
