Question

The topic is solving radical equations, but im just confused on whether to squafe it or not$2 \sqrt{n} = n - 3$

Answer 1

a We are given the following radical equation

$2\sqrt[]{n}=n-3$

To solve this equation we need to square both sides of the equation

$\begin{gathered} (2\sqrt[]{n})^2=(n-3)^2 \\ 4n=(n-3)^2 \end{gathered}$

Apply the squares formula on right-hand side of the equation

$(a-b)^2=a^2+b^2-2ab$

So the equation will become

$\begin{gathered} 4n=n^2+3^2-2(n)(3) \\ 4n=n^2+9-6n \\ 0=n^2+9-6n-4n \\ 0=n^2+9-10n \\ n^2-10n+9=0 \end{gathered}$

So we are left with a quadratic equation.

The standard form of a quadratic equation is given by

$ax^2+bx+c=0$

Comparing the standard equation with our equation we get the following coefficients

a = 1

b = -10

c = 9

Now recall that quadratic formula is given by