Certainly! To solve the given equation:
[tex]\[ x - 3\sqrt{x} - 18 = 0 \][/tex]
we can use a substitution method. Let's set [tex]\( u = \sqrt{x} \)[/tex]. Then, we know that [tex]\( u^2 = x \)[/tex]. Substituting [tex]\( u \)[/tex] for [tex]\( \sqrt{x} \)[/tex], the equation becomes:
[tex]\[ u^2 - 3u - 18 = 0 \][/tex]
This is now a quadratic equation in terms of [tex]\( u \)[/tex]. The general form of a quadratic equation is:
[tex]\[ au^2 + bu + c = 0 \][/tex]
where, in our case, [tex]\( a = 1 \)[/tex], [tex]\( b = -3 \)[/tex], and [tex]\( c = -18 \)[/tex].
To solve this quadratic equation, we'll use the quadratic formula, which is:
[tex]\[ u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Plugging in our values ([tex]\( a = 1 \)[/tex], [tex]\( b = -3 \)[/tex], [tex]\( c = -18 \)[/tex]):
1. Calculate the discriminant:
[tex]\[ \text{Discriminant} = b^2 - 4ac \][/tex]
[tex]\[ \text{Discriminant} = (-3)^2 - 4(1)(-18) \][/tex]
[tex]\[ \text{Discriminant} = 9 + 72 \][/tex]
[tex]\[ \text{Discriminant} = 81 \][/tex]
2. Substitute the values into the quadratic formula:
[tex]\[ u = \frac{-(-3) \pm \sqrt{81}}{2 \cdot 1} \][/tex]
[tex]\[ u = \frac{3 \pm 9}{2} \][/tex]
3. Solve for [tex]\( u \)[/tex]:
[tex]\[ u_1 = \frac{3 + 9}{2} = \frac{12}{2} = 6 \][/tex]
[tex]\[ u_2 = \frac{3 - 9}{2} = \frac{-6}{2} = -3 \][/tex]
Since [tex]\( u = \sqrt{x} \)[/tex] and [tex]\( \sqrt{x} \)[/tex] cannot be a negative number, we discard the solution [tex]\( u = -3 \)[/tex], as it does not give a real value for [tex]\( x \)[/tex].
4. Now, using the valid solution [tex]\( u = 6 \)[/tex]:
[tex]\[ \sqrt{x} = 6 \][/tex]
Squaring both sides, we get:
[tex]\[ x = 6^2 = 36 \][/tex]
Therefore, the solution to the equation [tex]\( x - 3\sqrt{x} - 18 = 0 \)[/tex] is:
[tex]\[ x = 36 \][/tex]
Additionally, other value is also derived for [tex]\(\sqrt(x) = -3 \)[/tex], thus giving
[tex]\[
x = (-3)^2 = 9
\][/tex]
and the solutions are:
\[ x = 36 \text{ and }x = 9]
