Answer:
The solution to the equation are [tex]5+\frac{\sqrt{42} }{2\\} \ and \ 5-\frac{\sqrt{42} }{2\\}\\[/tex]
Both of his values are positive real numbers
Step-by-step explanation:
The general formula of a quadratic equation is expressed as [tex]ax^{2}+bx+c = 0\ where;\\x = -b\±\frac{\sqrt{b^{2}-4ac } }{2a}[/tex]
Given the expression 0 = x² – 5x – 4 which can be rewritten as shown below;
x² – 5x – 4 = 0
Comparing this to the general equation; a = 1, b = -5, c= -4
To get the solution to the quadratic equation, we will use the general formula above;
[tex]x = -b\±\frac{\sqrt{b^{2}-4ac } }{2a}\\x = -(-5)\±\frac{\sqrt{(-5)^{2}-4(1)(-4) } }{2(1)}\\\\x = 5\±\frac{\sqrt{25+16 } }{2}\\x =5\±\frac{\sqrt{41} }{2}\\x = 5+\frac{\sqrt{42} }{2}\ and \ 5-\sqrt{42} /2\\[/tex]
Both of his values are positive real numbers
