Question

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

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

between -1&0,0&1 and 1&2

b. between -2&-1, 0&1 and 1&2

d. between -2&1, 0&1 and 1&2

a.

c.

Please select the best answer from the choices provided

A

C

Oo

Answer 1

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

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

a) between -1&0, 0&1 and 2&3

Explanation:

Solving a quadratic equation:

Given a second order polynomial expressed by the following equation:

`ax^(2) + bx + c, a\\eq0`.

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) = (-b + √(\bigtriangleup))/(2*a)`

`x_(2) = (-b - √(\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) = (-(-3) + √(5))/(2*1) = 2.62`

`x_(2) = (-(-3) - √(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