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Marrrta [24]
2 years ago
11

Helppp!!!!!

Mathematics
2 answers:
gladu [14]2 years ago
8 0

Answer:

fyi b's answer has imaginary numbers in it...

Imaginary: 1 +\frac{\sqrt{2i} }{2 }    

Imaginary: 1 -  \frac{\sqrt{2i} }{2 }  

Step-by-step explanation:

2x^{2} - 4x -3 = 0

\sqrt{-4^{2} -4(2)(3)} = \sqrt{-8} ... the negative root will produce imaginary solutions

Gelneren [198K]2 years ago
6 0

9514 1404 393

Answer:

  1a. -4, 3/4

  1b. 1-0.71i, 1+0.71i

Step-by-step explanation:

The directions tell you to use the quadratic formula. Factoring may get you the solution somewhat more easily, but does not comply with the directions.

The quadratic formula tells you ...

  \text{The solution to }ax^2+bx+c=0\text{ is given by }\\\\x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}

__

1a. a=4, b=13, c=-12

  x=\dfrac{-13\pm\sqrt{13^2-4(4)(-12)}}{2(4)}=\dfrac{-13\pm\sqrt{361}}{8}=\dfrac{-13\pm19}{8}\\\\x=\left\{-4,\dfrac{3}{4}\right\}

__

1b. After adding 3 to both sides, a=2, b=-4, c=3

  x=\dfrac{-(-4)\pm\sqrt{(-4)^2-4(2)(3)}}{2(2)}=\dfrac{4\pm\sqrt{-8}}{4}=1\pm\dfrac{\sqrt{2}}{2}i\\\\x=\left\{1-\dfrac{\sqrt{2}}{2}i,1+\dfrac{\sqrt{2}}{2}i\right\}\approx\{1-0.71i,1+0.71i\}

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