What is the solution set of the quadratic equation 8x2 +2x +1 =0?

1 answer:

4 0

$\displaystyle\\
 8x^2 +2x +1 =0\\\\
x_{12}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}=\frac{-2\pm \sqrt{2^2-4\cdot 8 \cdot 1}}{2\cdot 8}=\\\\
=\frac{-2\pm \sqrt{4-32}}{16}=\frac{-2\pm \sqrt{-28}}{16}=\frac{-2\pm \sqrt{4\cdot 7\cdot(-1)}}{16}=\\\\
=\frac{-2\pm 2i\sqrt{7}}{16}=\frac{-1\pm i\sqrt{7}}{8}= -\frac{1}{8}\pm \frac{\sqrt{7}}{8}i\\\\
x_1 = \boxed{-\frac{1}{8}+ \frac{\sqrt{7}}{8}i}\\\\
x_2 = \boxed{-\frac{1}{8}- \frac{\sqrt{7}}{8}i}\\\\
x_1,~x_2~\in \mathbb{C}$

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Roman55 [17]

$(x^3+x^2+x+2) divide by (x^2-1)$

We use long division to divide

There is no x term in x^2 -1 so we put 0x

x + 1

----------------------------

x^2+0x-1 x^3+ x^2 + x+ 2

x^3+0x^2-x

-----------------------------(subtract the bottom from top)

x^2 +2x + 2

x^2 +0x - 1

--------------------------------(subtract the bottom from top)

2x + 3

-----------------------------------------

Quotient : x+1

Remainder : 2x+3

5 0

3 years ago

Which of the following is the product of the rational expressions shown

Alexus [3.1K]

Answer: x = 0; x=10

Step-by-step explanation:

$\frac{x+2}{x-4} *\frac{3x}{x+4}\\ \\x+2(x+4) = x-4(3x)\\x^{2} +2x+ 4x + 8 = 3x^{2} -12x\\2x+4x+12x =3x^{2} - x^{2} \\20x= 2x^{2} \\10x = x^2\\x=0\\x=10$