Answer:
a) between -1&0, 0&1 and 2&3
Step-by-step explanation:
Solving a quadratic equation:
Given a second order polynomial expressed by the following equation:
[tex]ax^{2} + bx + c, a\neq0[/tex].
This polynomial has roots [tex]x_{1}, x_{2}[/tex] such that [tex]ax^{2} + bx + c = a(x - x_{1})*(x - x_{2})[/tex], given by the following formulas:
[tex]x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}[/tex]
[tex]x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}[/tex]
[tex]\bigtriangleup = b^{2} - 4ac[/tex]
In this question:
[tex]f(x) = x^{2} - 3x + 1[/tex]
So
[tex]\bigtriangleup = (-3)^{2} - 4*1*1 = 5[/tex]
[tex]x_{1} = \frac{-(-3) + \sqrt{5}}{2*1} = 2.62[/tex]
[tex]x_{2} = \frac{-(-3) - \sqrt{5}}{2*1} = 0.38[/tex]
2.62 is between 2 and 3.
0.38 is between 0 and 1.
So the correct answer is:
a) between -1&0, 0&1 and 2&3
Answer:
Step-by-step explanation:
Given the quadratic expression f(x) = x2 – 3x + 1, the real zero of the expression will be the roots of the equation and this can be gotten as shown below
f(x) = x² – 3x + 1
On factorizing the equation using the general fomula when f(x) = 0;
x = -b±√b²-4ac/2a
from the equation above, a = 1, b = -3, c = 1
x = -(-3)±√(-3)²-4(1)(1)/2(1)
x = 3±√9-4/2
x = 3±√5/2
x = (3+√5)/2 or x = (3-√5)/2
x = 2.62 and 0.38
since 2.62 is between 2&3 and 0.38 is between 0&1 and -1&0, the required answer will be -1&0, 0&1 and 2&3
