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REY [17]
3 years ago
15

What are the possible values of x in 8x2 + 4x = -1?

Mathematics
1 answer:
Westkost [7]3 years ago
4 0

Answer:

The possible values of x are:

x= \frac{-1+\sqrt{3}}{4} \,\,or\,\, x= 0.18

and

x= \frac{-1-\sqrt{3}}{4} \,\, or \,\, x= -0.68

Step-by-step explanation:

8x^2+4x+1=0

Using Quadratic formula to solve this equation:

x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\a = 8 \,\, b = 4\,\, c = -1\\Putting \,\, values \,\, in \,\, the\,\, equation\\x= \frac{-4\pm\sqrt{(4)^2-4(8)(-1)}}{2(8)}\\x= \frac{-4\pm\sqrt{16+32}}{16}\\x= \frac{-4\pm\sqrt{48}}{16}\\x= \frac{-4+ \sqrt{48}}{16} \,\, and \,\, x= \frac{-4- \sqrt{48}}{16}\\x= \frac{-1+ \sqrt{3}}{4} \,\, and \,\, x= \frac{-1- \sqrt{3}}{4}

The possible values of x are:

x= \frac{-1+\sqrt{3}}{4} \,\,or\,\, x= 0.18

and

x= \frac{-1-\sqrt{3}}{4} \,\, or \,\, x= -0.68

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