Question

Determine between which consecutive integers the real zeros of f(x) = x2 – 3x + 1 are located.

Answer 1

Answer:

-1&0, 0&1 and 2&3

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

Answer 2

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:

$ax^{2} + bx + c, a\neq0$.

This polynomial has roots $x_{1}, x_{2}$ such that $ax^{2} + bx + c = a(x - x_{1})*(x - x_{2})$, given by the following formulas:

$x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}$

$x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}$

$\bigtriangleup = b^{2} - 4ac$

In this question:

$f(x) = x^{2} - 3x + 1$

So

$\bigtriangleup = (-3)^{2} - 4*1*1 = 5$

$x_{1} = \frac{-(-3) + \sqrt{5}}{2*1} = 2.62$

$x_{2} = \frac{-(-3) - \sqrt{5}}{2*1} = 0.38$

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