Question

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

Answer 1

Answer:

c) $x=\frac{y^2\pm\sqrt{y^4-4a}}{2}$

d) $x= \frac{-3}{p-q^2u}$

Step-by-step explanation:

c) $y= \frac{\sqrt{x^2 + a} }{x}$

Solving the question:

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

d) $\sqrt{\frac{px+3}{ux}}=q$

Solving to find value of x

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