Answer:
[tex]x=\frac{-5(+/-)\sqrt{5}} {2}[/tex]
x = StartFraction negative 5 plus or minus StartRoot 5 EndRoot Over 2 EndFraction
Step-by-step explanation:
we know that
The formula to solve a quadratic equation of the form
[tex]ax^{2} +bx+c=0[/tex]
is equal to
[tex]x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}[/tex]
in this problem we have
[tex]f(x)=x^{2} +5x+5[/tex]
equate the function to zero
[tex]x^{2} +5x+5=0[/tex]
so
[tex]a=1\\b=5\\c=5[/tex]
substitute in the formula
[tex]x=\frac{-5(+/-)\sqrt{5^{2}-4(1)(5)}} {2(1)}[/tex]
[tex]x=\frac{-5(+/-)\sqrt{5}} {2}[/tex]
therefore
x = StartFraction negative 5 plus or minus StartRoot 5 EndRoot Over 2 EndFraction
The zeros of the function f(x) are :
Discriminant of quadratic equation ( ax² + bx + c = 0 ) could be calculated by using :
From the value of Discriminant , we know how many solutions the equation has by condition :
D < 0 → No Real Roots
D = 0 → One Real Root
D > 0 → Two Real Roots
An axis of symmetry of quadratic equation y = ax² + bx + c is :
[tex]\large {\boxed {x = \frac{-b}{2a} } }[/tex]
Let us now tackle the problem!
Given:
[tex]f(x) = x^2 + 5x + 5[/tex]
The zeros of the quadratic function could be calculated when f(x) = 0 :
[tex]f(x) = x^2 + 5x + 5[/tex]
[tex]0 = x^2 + 5x + 5[/tex]
[tex]0 = (x + \frac{5}{2})^2 - (\frac{5}{2})^2 + 5[/tex]
[tex](x + \frac{5}{2})^2 = (\frac{5}{2})^2 - 5[/tex]
[tex](x + \frac{5}{2})^2 = \frac{25}{4} - 5[/tex]
[tex](x + \frac{5}{2})^2 = \frac{5}{4}[/tex]
[tex]x + \frac{5}{2} = \pm \sqrt{\frac{5}{4}}[/tex]
[tex]x + \frac{5}{2} = \pm \frac{\sqrt{5}}{2}[/tex]
[tex]x = - \frac{5}{2} \pm \frac{\sqrt{5}}{2}[/tex]
[tex]x = \boxed{ \frac{-5 \pm \sqrt{5}}{2} }[/tex]
[tex]\texttt{ }[/tex]
Grade: High School
Subject: Mathematics
Chapter: Quadratic Equations
Keywords: Quadratic , Equation , Discriminant , Real , Number
