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if the quadratic formula is used to find the roots of the equation x^2-6x-19=0, what is the correct roots

Sagot :

D1eformetalidn
[tex] x^{2} -6x-19=0 \\ \\discriminant= b^{2}-4ac=> \\ 36-(-76)=76+36=112 \\ \\ -b- \sqrt{disc.}/2a=>6- \sqrt{112}/2 \\ =>6- 4\sqrt{7}/2 \\ =3- 2\sqrt{7} \\ \\ -b +\sqrt{disc}/2a=6+4 \sqrt{7} /2 \\ =3+ 2\sqrt{7} \\ \\ Solution\: set:(3- 2\sqrt{7} ,3+ 2\sqrt{7} ) \\ \\ \framebox[1.1\width]{Good Luck!!} \par[/tex]
Konrad509idn
[tex]x^2-6x-19=0\\ x^2-6x+9-28=0\\ (x-3)^2=28\\ x-3=\sqrt{28} \vee x-3=-\sqrt{28}\\ x=3+2\sqrt7 \vee x=3-2\sqrt7[/tex]
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