Respuesta :
You can use the formula for finding the roots of a quadratic equation here.
The solution of given equation is: x = -1
What are the solutions of standard form of quadratic equation?
The standard form is [tex]ax^2 + bx + c = 0[/tex]
The solutions are:
[tex]x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}[/tex]
How to find the solution of given quadratic equation?
To find the solutions, firstly we've to simplify it.
[tex]\sqrt{1-3x} = x + 3\\ \text{Taking square on both sides}\\ (\sqrt{1-3x})^2 = (x+3)^2\\ 1-3x = x^2 + 9 + 6x\\ x^2 + 9x + 8 = 0[/tex]
Know that we took square.
Also know that [tex](\sqrt{1-3x})^2 = (-\sqrt{1-3x})^2 = 1-3x\\ [/tex] This is the reason that both roots we will obtain won't belong to our quadratic equation we obtained. Thus, we will have to check which root is correct solution of the obtained quadratic equation.
Thus, we have a = 1, b = 9, c= 8, and thus,
[tex]x = \dfrac{-9 \pm \sqrt{81 - 32}}{2}\\\\ x = \dfrac{-9 \pm \sqrt{49}}{2}\\\\ x = \dfrac{-9 \pm 7}{2}\\\\ x = \dfrac{-9-7}{2} , \: x = \dfrac{-9+7}{2}\\\\ x = -8, \: x = -1[/tex]
Putting x = -8 doesn't satisfy the equation. But x = -1 does.
Thus, the solution to the given equation is x = -1
Learn more about solutions of a quadratic equation here:
https://brainly.com/question/3358603
