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Nookie1986 [14]
3 years ago
5

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

Mathematics
2 answers:
kondaur [170]3 years ago
8 0

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

NISA [10]3 years ago
5 0
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}
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