Question

I need help solving this it’s a radical equation I would appreciate the help:0

Answer 1

The given expression is

`\sqrt[]{6x\text{ }}+\text{ 3 = }\sqrt[]{9x\text{ + 19}}`

We would square both sides of the expression. It becomes

`\begin{gathered} (\sqrt[]{6x}+3)^2\text{ = (}\sqrt[]{9x+19)^2} \\ (\sqrt[]{6x}\text{ + 3)(}\sqrt[]{6x}\text{ + 3) = 9x + 19} \\ On\text{ the left, we would multiply each term in both parentheses. } \\ \sqrt[]{6x}*\sqrt[]{6x}\text{ + 3}\sqrt[]{6x}\text{ + 3}\sqrt[]{6x}\text{ + 3}*3\text{ = 9x + 19} \\ 6x\text{ + 6}\sqrt[]{6x}\text{ + 9 = 9x + 19} \\ \text{Collecting like terms, we have} \\ 6\sqrt[]{6x}\text{ = 9x - 6x + 19 - 9} \\ 6\sqrt[]{6x}\text{ = 3x + 10} \\ \text{There is still a radical on the left hand side. We would square both sides} \\ of\text{ the equation.} \\ (6\sqrt[]{6x)^2}=(3x+10)^2 \\ 6^2*6x\text{ = (3x + 10)(3x + 10)} \\ 216x=9x^2\text{ + 30x + 30x + 100} \\ 216x=9x^2+60x\text{ + 100} \\ 9x^2\text{ + 60x - 216x + 100 = 0} \\ 9x^2-\text{ 156x + 100 = 0} \end{gathered}`

We have a quadratic equation. The general form of a quadratic equation is expressed as

ax^2 + bx + c = 0

By comparing our equation with the quadratic equation, we have

a = 9, b = - 156, c = - 100

We would solve for x by applying the general formula for solving quadratic equations which is expressed as

`\begin{gathered} x\text{ =}\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}\text{ } \\ \text{substituting the given values, } \\ x\text{ = }\frac{-\text{ - 156 }\pm\sqrt[]{(-156)^2^{}-\text{ 4}*9\text{ x }100}}{2*9} \\ x\text{ = }\frac{156\text{ }\pm\sqrt[]{24336-\text{ 3600}}}{18} \\ x\text{ = }\frac{156\text{ }\pm\sqrt[]{20736}}{18} \\ x\text{ = }\frac{156\text{ + 144}}{18}\text{ or x = }\frac{156\text{ - 144}}{18} \\ x\text{ = }(300)/(18)\text{ or x = }(12)/(18) \end{gathered}`

Dividing the numerators and denominators by 6, we have

x = 50/3 or x = 2/3