Answer:
Proof below.
Step-by-step explanation:
Quadratic Formula
[tex]x=\dfrac{-b \pm \sqrt{b^2-4ac} }{2a}\quad\textsf{when }\:ax^2+bx+c=0[/tex]
Given quadratic equation:
[tex]5x^2-6x-2=0[/tex]
Define the variables:
Substitute the defined variables into the quadratic formula and solve for x:
[tex]\implies x=\dfrac{-(-6) \pm \sqrt{(-6)^2-4(5)(-2)}}{2(5)}[/tex]
[tex]\implies x=\dfrac{6 \pm \sqrt{36+40}}{10}[/tex]
[tex]\implies x=\dfrac{6 \pm \sqrt{76}}{10}[/tex]
[tex]\implies x=\dfrac{6 \pm \sqrt{4 \cdot 19}}{10}[/tex]
[tex]\implies x=\dfrac{6 \pm \sqrt{4}\sqrt{19}}{10}[/tex]
[tex]\implies x=\dfrac{6 \pm2\sqrt{19}}{10}[/tex]
[tex]\implies x=\dfrac{3 \pm \sqrt{19}}{5}[/tex]
Therefore, the exact solutions to the given quadratic equation are:
[tex]x=\dfrac{3 + \sqrt{19}}{5} \:\textsf{ and }\:x=\dfrac{3 - \sqrt{19}}{5}[/tex]
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