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marysya [2.9K]
3 years ago
13

Solve square root X^2+12x+43=0

Mathematics
2 answers:
AlekseyPX3 years ago
5 0
x^2+12x+43=0
 

x= \dfrac{-b+ \sqrt{b^2-4ac} }{2a}, \dfrac{-b- \sqrt{b^2-4ac} }{2a} 

x= \dfrac{-12+ \sqrt{12^2-4*43} }{2},\dfrac{-12- \sqrt{12^2-4*43} }{2} 

x= \dfrac{-12+2 \sqrt{7i} }{2}, \dfrac{-12-2 \sqrt{7i} }{2} 

x=-6+ \sqrt{7i},-6- \sqrt{7i}
Novay_Z [31]3 years ago
5 0
X²+12x+43=0

This a quadratic equation ax²+bx+c=0 (in our case a=1, b=12, & c=43)

to solve it, means to calculate the value of x which render this equation nil

In short we have to find the roots x' & x'' by applying the following formula:

x' = [-b+√(b² - 4ac)] / 2a  & x" =[-b - √(b² - 4ac)] / 2a 
If you plug the related values you will notice the amount inside the radical (√)
is negative. As you know such an amount should be positif, hence their is no roots & consequently no solution

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