solve for x in the equation $x^{2}+11x+\frac{121}{4}=\frac{125}{4}$. \n$x = - 11pm\frac{25}{2}$\n$x=-\frac{11…

Explanation:

Step1: Recognize the left - hand side as a perfect square

The left - hand side of the equation $x^{2}+11x+\frac{121}{4}$ is in the form of $(a + b)^2=a^{2}+2ab + b^{2}$, where $a = x$ and $b=\frac{11}{2}$ since $2\times x\times\frac{11}{2}=11x$ and $(\frac{11}{2})^{2}=\frac{121}{4}$. So the equation can be rewritten as $(x + \frac{11}{2})^{2}=\frac{125}{4}$.

Step2: Take the square root of both sides

Taking the square root of both sides gives $x+\frac{11}{2}=\pm\sqrt{\frac{125}{4}}$.

Step3: Simplify the square - root term

We know that $\sqrt{\frac{125}{4}}=\frac{\sqrt{125}}{\sqrt{4}}=\frac{5\sqrt{5}}{2}$ (since $125 = 25\times5$ and $\sqrt{125}=\sqrt{25\times5}=5\sqrt{5}$, $\sqrt{4} = 2$).

Step4: Solve for x

Subtract $\frac{11}{2}$ from both sides: $x=-\frac{11}{2}\pm\frac{5\sqrt{5}}{2}$.

Answer:

$x=-\frac{11}{2}\pm\frac{5\sqrt{5}}{2}$