Respuesta :

alinakincsem

Answer:

[tex](z-\frac{3}{2} )^2-\frac{29}{4}[/tex]

Step-by-step explanation:

We are given the following quadratic equation by completing the square:

[tex]z^2 - 3z - 5 = 0[/tex]

Rewriting the equation in the form [tex]x^2+2ax+a^2[/tex] to get:

[tex]z^2 - 3z - 5+(-\frac{3}{2} )^2-(-\frac{3}{2} )^2[/tex]

[tex]z^2-3z+(-\frac{3}{2} )^2=(z-\frac{3}{2} )^2[/tex]

Completing the square to get:

[tex] ( z - \frac{ 3 } { 2 } )^ 2 - 5 - ( - \frac { 3 } { 2 } ) ^ 2[/tex]

[tex](z-\frac{3}{2} )^2-\frac{29}{4}[/tex]

luisejr77

Answer: [tex]z_1=4.19\\\\z_2=-1.19[/tex]

Step-by-step explanation:

Add 5 to both sides of the equation:

[tex]z^2 - 3z - 5 +5= 0+5\\\\z^2 - 3z = 5[/tex]

Divide the coefficient of [tex]z[/tex] by two and square it:

[tex](\frac{b}{2})^2= (\frac{3}{2})^2[/tex]

Add it to both sides of the equation:

[tex]z^{2} -3z+ (\frac{3}{2})^2=5+ (\frac{3}{2})^2[/tex]

Then, simplifying:

[tex](z- \frac{3}{2})^2=\frac{29}{4}[/tex]

Apply square root to both sides and solve for "z":

[tex]\sqrt{(z- \frac{3}{2})^2}=\±\sqrt{\frac{29}{4} }\\\\z=\±\sqrt{\frac{29}{4}}+ \frac{3}{2}\\\\z_1=4.19\\\\z_2=-1.19[/tex]

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