Question

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

Answer 1

Answer:

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

Step-by-step explanation:

we have

$x^2=19x+1$

we know that

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

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

in this problem we have

$-x^{2}-19x-1=0$  

so

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

substitute in the formula

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

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

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

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

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

therefore

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

Answer 2

Answer:

The answer is B

Step-by-step explanation:

I took the test