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elena-14-01-66 [18.8K]

4 years ago

7

Rewrite x2 − 8x + 13 = 0 in the form (x − a)2 = b, where a and b are integers, to determine the a and b values. a = 4 and b = 3

a = 3 and b = 2 a = 2 and b = 1 a = 1 and b = 4

2 answers:

Lyrx [107]4 years ago

6 0

Answer:

A

Step-by-step explanation:

neonofarm [45]4 years ago

3 0

Answer:

A

Step-by-step explanation:

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3 years ago

What are the zeros of the quadratic function f(x) = 6x ^ 2 - 24x + 1 ?

ZanzabumX [31]

Answer:

x = $\frac{12+\sqrt{138} }{6}$ and x = $\frac{12-\sqrt{138} }{6}$

Step-by-step explanation:

Let's use the quadratic formula, which states that for a quadratic of the form ax² + bc + c, the zeroes are: $x=\frac{-b+\sqrt{b^2-4ac} }{2a}$ or $x=\frac{-b-\sqrt{b^2-4ac} }{2a}$ .

Here, a = 6, b = -24, and c = 1. Plug these in:

$x=\frac{-b+\sqrt{b^2-4ac} }{2a}$

$x=\frac{-(-24)+\sqrt{(-24)^2-4*6*1} }{2*6}=\frac{24+\sqrt{552} }{12}=\frac{24+2\sqrt{138} }{12} =\frac{12+\sqrt{138} }{6}$

AND

$x=\frac{-b-\sqrt{b^2-4ac} }{2a}$

$x=\frac{-(-24)-\sqrt{(-24)^2-4*6*1} }{2*6}=\frac{24-\sqrt{552} }{12}=\frac{24-2\sqrt{138} }{12} =\frac{12-\sqrt{138} }{6}$

Thus, the zeroes are: x = $\frac{12+\sqrt{138} }{6}$ and x = $\frac{12-\sqrt{138} }{6}$ .

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3 years ago

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