Question

SOLVE for x in the equation x^2+11x+121/4+125/4

Answer 1

Given

$x^2+11x+\frac{121}{4}=\frac{125}{4} \\ \\ \Rightarrow x^2+11x+\frac{121}{4}-\frac{125}{4}=0 \\ \\ \Rightarrow x^2+11x-1=0 \\ \\ \Rightarrow x= \frac{-11\pm\sqrt{11^2-4(1)(-1)}}{2(1)} \\ \\ = \frac{-11\pm\sqrt{121+4}}{2} = \frac{-11\pm\sqrt{125}}{2} \\ \\ = \frac{-11\pm5\sqrt{5}}{2}$

Answer 2

Answer:

The value of x is $-\dfrac{11}{2}\pm\dfrac{5\sqrt{5}}{2}i$

Step-by-step explanation:

Given the equation

$x^2+11x+\dfrac{121}{4}+\dfrac{125}{4}$

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

$x^2+11x+\dfrac{121}{4}+\dfrac{125}{4}+(\dfrac{11}{2})^2-(\dfrac{11}{2})^2$

$(x+\dfrac{11}{2})^2+\dfrac{121}{4}+\dfrac{125}{4}-\dfrac{121}{4}$

Now, the like terms will be the cancel.

$(x+\dfrac{11}{2})^2+\dfrac{125}{4}$

$(x+\dfrac{11}{2})^2=-\dfrac{125}{4}$

$(x+\dfrac{11}{2})=\sqrt{\dfrac{125}{4}}$

$(x+\dfrac{11}{2})=\dfrac{5\sqrt{5}}{2}i$

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

Hence, The value of x is $-\dfrac{11}{2}\pm\dfrac{5\sqrt{5}}{2}i$