Question

Sq root of 13t+ 4 – 2 sq root of t + 2 +1 = 0

Answer 1

`\begin{gathered} \sqrt{3t\text{ + 4 }}\text{ - 2}\sqrt{t\text{ + 2 }}+\text{ 1 = 0} \\ \sqrt{3t\text{ + 4 }}=\text{ 2}\sqrt{t\text{ + 2}}\text{ - 1} \\ (√(3t+4))^2=\text{ (2}\sqrt{t\text{ + 2}}-1)^2 \\ \text{ 3t + 4 = 4(t + 2) - 4}\sqrt{t\text{ + 2 }}\text{ + 1} \\ 3t\text{ + 4 = 4t + 8 +1 - 4}\sqrt{t\text{ + 2}} \\ 3t\text{ + 4 - 4t - 9 = -4}\sqrt{t\text{ + 2}} \\ \text{ -t - 5 = -4}\sqrt{t\text{ + 2}} \\ \text{ (-t - 5)}^2\text{ = (-4}\sqrt{t\text{ + 2}})^(^2) \\ \text{ t}^2\text{ + 10t + 25 = 16(t + 2)} \\ \text{ t}^2\text{ + 10t + 25 = 16t + 32} \\ \text{ t}^2\text{ +10t - 16t + 25 - 32 = 0} \\ \text{ t}^2\text{ -6t + 7 = 0} \\ \text{ (t -7)(t + 1) = 0} \\ \text{ t1 = 7 t2 = -1} \end{gathered}`