Respuesta :
Given
[tex]x^2+11x+\frac{121}{4}=\frac{125}{4} \\ \\ \Rightarrow x^2+11x+\frac{121}{4}-\frac{125}{4}=0 \\ \\ \Rightarrow x^2+11x-1=0 \\ \\ \Rightarrow x= \frac{-11\pm\sqrt{11^2-4(1)(-1)}}{2(1)} \\ \\ = \frac{-11\pm\sqrt{121+4}}{2} = \frac{-11\pm\sqrt{125}}{2} \\ \\ = \frac{-11\pm5\sqrt{5}}{2} [/tex]
[tex]x^2+11x+\frac{121}{4}=\frac{125}{4} \\ \\ \Rightarrow x^2+11x+\frac{121}{4}-\frac{125}{4}=0 \\ \\ \Rightarrow x^2+11x-1=0 \\ \\ \Rightarrow x= \frac{-11\pm\sqrt{11^2-4(1)(-1)}}{2(1)} \\ \\ = \frac{-11\pm\sqrt{121+4}}{2} = \frac{-11\pm\sqrt{125}}{2} \\ \\ = \frac{-11\pm5\sqrt{5}}{2} [/tex]
Answer:
The value of x is [tex]-\dfrac{11}{2}\pm\dfrac{5\sqrt{5}}{2}i[/tex]
Step-by-step explanation:
Given the equation
[tex]x^2+11x+\dfrac{121}{4}+\dfrac{125}{4}[/tex]
For making perfect square root
Firstly, we will half of middle term then add and subtract the square of half of the middle term in the equation
[tex]x^2+11x+\dfrac{121}{4}+\dfrac{125}{4}+(\dfrac{11}{2})^2-(\dfrac{11}{2})^2[/tex]
[tex](x+\dfrac{11}{2})^2+\dfrac{121}{4}+\dfrac{125}{4}-\dfrac{121}{4}[/tex]
Now, the like terms will be the cancel.
[tex](x+\dfrac{11}{2})^2+\dfrac{125}{4}[/tex]
[tex](x+\dfrac{11}{2})^2=-\dfrac{125}{4}[/tex]
[tex](x+\dfrac{11}{2})=\sqrt{\dfrac{125}{4}}[/tex]
[tex](x+\dfrac{11}{2})=\dfrac{5\sqrt{5}}{2}i[/tex]
[tex]x= -\dfrac{11}{2}\pm\dfrac{5\sqrt{5}}{2}i[/tex]
Hence, The value of x is [tex]-\dfrac{11}{2}\pm\dfrac{5\sqrt{5}}{2}i[/tex]
