Respuesta :
Answer:
[tex]x=\dfrac{3+ \sqrt{29}}{2}, \quad \dfrac{3- \sqrt{29}}{2}[/tex]
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]x^2-3x-5=0[/tex]
Define the variables:
[tex]\implies a=1, \quad b=-3, \quad c=-5[/tex]
Substitute the defined variables into the quadratic formula and solve for x:
[tex]\implies x=\dfrac{-(-3) \pm \sqrt{(-3)^2-4(1)(-5)}}{2(1)}[/tex]
[tex]\implies x=\dfrac{3 \pm \sqrt{9+20}}{2}[/tex]
[tex]\implies x=\dfrac{3 \pm \sqrt{29}}{2}[/tex]
Therefore, the exact solutions to the given quadratic equation are:
[tex]x=\dfrac{3+ \sqrt{29}}{2}, \quad \dfrac{3- \sqrt{29}}{2}[/tex]
Learn more about the quadratic formula here:
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Answer:
[tex]\boxed {\frac{3+\sqrt{29}}{2}} \boxed{\frac{3-\sqrt{29}}{2}}[/tex]
Step-by-step explanation:
Quadratic Formula :
[tex]\boxed {\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}}[/tex]
We are given the equation x² - 3x - 5 = 0.
Here,
- a = 1
- b = -3
- c = -5
Solving :
- 3 ± √3² - 4(1)(-5) / 2(1)
- 3 ± √29 / 2
Hence, the solutions are :
[tex]\boxed {\frac{3+\sqrt{29}}{2}} \boxed{\frac{3-\sqrt{29}}{2}}[/tex]
