Respuesta :

dalendrk
[tex]x^2=-5x-3\\x^2+5x+3=0\\\\a=1;\ b=5;\ c=3\\\Delta=b^2-4ac\\\\\Delta=5^2-4\cdot1\cdot3=25-12=13 \ \textgreater \ 0\\\\therefore\\x_1=\dfrac{-b-\sqrt\Delta}{2a}\ and\ x_2=\dfrac{-b+\sqrt\Delta}{2a}\\\\x_1=\dfrac{-5-\sqrt{13}}{2\cdot1}=\boxed{\dfrac{-5-\sqrt{13}}{2}}\\\\x_2=\dfrac{-5+\sqrt{13}}{2\cdot1}=\boxed{\dfrac{-5+\sqrt{13}}{2}}[/tex]

Answer

Find out the  solutions of the quadratic equation.

x² = –5x – 3

To prove

As the quadratic equation be.

x² +  5x + 3 = 0

Discriminant Formula

[tex]x = \frac{-b\pm\sqrt{b^{2} -4ac} }{2a}[/tex]

As a = 1 , b  = 5 , c = 3

Put in the formula

[tex]x = \frac{-5\pm\sqrt{5^{2} -4\times 1\times 3} }{2\times 1}[/tex]

[tex]x = \frac{-5\pm\sqrt{25 -12} }{2\times 1}[/tex]

[tex]x = \frac{-5\pm\sqrt{13} }{2\times 1}[/tex]

Thus the solutions of x² = –5x – 3 are

[tex]x = \frac{-5\ +\sqrt{13} }{2}[/tex]

and

[tex]x = \frac{-5\ -\sqrt{13} }{2}[/tex]







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