Solving Quadratic Equations by completing the square: z^2 – 3z – 5 = 0

Question

Solving Quadratic Equations by completing the square:

z^2 – 3z – 5 = 0

in progress 0
Abigail 6 days 2021-09-14T22:31:50+00:00 2 Answers 0

Answers ( )

    0
    2021-09-14T22:32:51+00:00

    Answer:

    (z-\frac{3}{2} )^2-\frac{29}{4}

    Step-by-step explanation:

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

    z^2 - 3z - 5 = 0

    Rewriting the equation in the form x^2+2ax+a^2 to get:

    z^2 - 3z - 5+(-\frac{3}{2} )^2-(-\frac{3}{2} )^2

    z^2-3z+(-\frac{3}{2} )^2=(z-\frac{3}{2} )^2

    Completing the square to get:

     ( z - \frac{ 3 } { 2 } )^ 2 - 5 - ( - \frac { 3 } { 2 } ) ^ 2

    (z-\frac{3}{2} )^2-\frac{29}{4}

    0
    2021-09-14T22:33:16+00:00

    Answer: z_1=4.19\\\\z_2=-1.19

    Step-by-step explanation:

    Add 5 to both sides of the equation:

    z^2 - 3z - 5 +5= 0+5\\\\z^2 - 3z = 5

    Divide the coefficient of z by two and square it:

    (\frac{b}{2})^2= (\frac{3}{2})^2

    Add it to both sides of the equation:

    z^{2} -3z+ (\frac{3}{2})^2=5+ (\frac{3}{2})^2

    Then, simplifying:

    (z- \frac{3}{2})^2=\frac{29}{4}

    Apply square root to both sides and solve for “z”:

    \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

Leave an answer

Browse
Browse

27:3+15-4x7+3-1=? ( )