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Novay_Z [31]
2 years ago
7

What types of solutions does 6x^2 - 20x + 1 have?​

Mathematics
1 answer:
elena55 [62]2 years ago
3 0

Answer:

2 real solutions

Step-by-step explanation:

We can use the determinant, which says that for a quadratic of the form ax² + bx + c, we can determine what kind of solutions it has by looking at the determinant of the form:

b² - 4ac

If b² - 4ac > 0, then there are 2 real solutions. If b² - 4ac = 0, then there is 1 real solution. If b² - 4ac < 0, then there are 2 imaginary solutions.

Here, a = 6, b = -20, and c = 1. So, plug these into the determinant formula:

b² - 4ac

(-20)² - 4 * 6 * 1 = 400 - 24 = 376

Since 376 is clearly greater than 0, we know this quadratic has 2 real solutions.

<em>~ an aesthetics lover</em>

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Rewrite the equation of a circle given below in standard form x² + 2x + y² + 22y - 10 = 0
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