Given the equation
$$\sqrt{18 - 7 x} = x$$
$$\sqrt{18 - 7 x} = x$$
We raise the equation sides to 2-th degree
$$18 - 7 x = x^{2}$$
$$18 - 7 x = x^{2}$$
Transfer the right side of the equation left part with negative sign
$$- x^{2} - 7 x + 18 = 0$$
This equation is of the form
a*x^2 + b*x + c = 0
A quadratic equation can be solved
using the discriminant.
The roots of the quadratic equation:
$$x_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$x_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = -1$$
$$b = -7$$
$$c = 18$$
, then
D = b^2 - 4 * a * c =
(-7)^2 - 4 * (-1) * (18) = 121
Because D > 0, then the equation has two roots.
x1 = (-b + sqrt(D)) / (2*a)
x2 = (-b - sqrt(D)) / (2*a)
or
$$x_{1} = -9$$
Simplify$$x_{2} = 2$$
SimplifyBecause
$$\sqrt{18 - 7 x} = x$$
and
$$\sqrt{18 - 7 x} \geq 0$$
then
$$x \geq 0$$
or
$$0 \leq x$$
$$x < \infty$$
The final answer:
$$x_{2} = 2$$