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dolphi86
7 days ago
8

Which are the solutions of x2 = 19x + 1?

Mathematics
2 answers:
AnnZ [3.9K]7 days ago
6 0

Answer:

(\frac{19-\sqrt{365}} {2},\frac{19+\sqrt{365}} {2})

Step-by-step explanation:

We have

x^2=19x+1

To recall, the formula for solving a quadratic equation shaped like

ax^{2} +bx+c=0

is expressed as

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

For this problem, we have

-x^{2}-19x-1=0  

Thus,

a=1\\b=-19\\c=-1

Substitute into the formula:

x=\frac{-(-19)\pm\sqrt{-19^{2}-4(1)(-1)}} {2(1)}

x=\frac{19\pm\sqrt{365}} {2}

x=\frac{19+\sqrt{365}} {2}

x=\frac{19-\sqrt{365}} {2}

(\frac{19-\sqrt{365}} {2},\frac{19+\sqrt{365}} {2})

Consequently, the solutions are:

StartFraction 19 minus StartRoot 365 EndRoot Over 2 EndFraction, StartFraction 19 + StartRoot 365 EndRoot Over 2 EndFraction

Zina [3.9K]7 days ago
3 0

Answer:

The correct choice is B.

Step-by-step explanation:

I completed the test.

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