Question

Which of the functions below have exactly
two distinct real zeros?
I. (v) = -12 + 14x – 45
II. 1.x) = 4x2 + 4
III. |(x) = 212 + 12x + 18
IV. /.x) = 3.x2 - 4x - 4
A. I and IV
C. II and III
B. I, II, and III
D. III and IV

Answer 1

Answer:

A. I and IV

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:

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

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

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

A quadratic equation has two distinct real zeros if:

$\bigtriangleup > 0$

So

I: $f(x) = -x^{2} + 14x - 45$

$a = -1, b = 14, c = -45$

$\bigtriangleup = 14^{2} - 4*(-1)*(-45) = 16$

So function I has two distinct real zeros.

II: $f(x) = 4x^{2} + 4$

$a = 4, b = 0, c = 4$

$\bigtriangleup = 0^{2} - 4*4*4 = -64$

Negative, so II has no real zeros.

III: $f(x) = 2x^{2} + 12x + 18$

$a = 2, b = 12, c = 18$

$\bigtriangleup = 12^{2} - 4*2*18 = 0$

So III has one real zero with double multiplicity, that is, two equal zeros.

IV: $f(x) = 3x^{2} -4x - 4$

$a = 3, b = -4, c = -4$

$\bigtriangleup = (-4)^{2} - 4*3*(-4) = 64$

So function IV has two distinct real zeros.

So the correct answer is:

A. I and IV