Sure, let's solve these questions step-by-step:
### Question 13:
We need to find [tex]\( p(x) + p(-x) \)[/tex] where [tex]\( p(x) = x + 3 \)[/tex].
1. Calculate [tex]\( p(x) \)[/tex]:
Given [tex]\( p(x) = x + 3 \)[/tex],
[tex]\[
p(x) = x + 3
\][/tex]
2. Calculate [tex]\( p(-x) \)[/tex]:
Substitute [tex]\(-x\)[/tex] into the function [tex]\( p(x) \)[/tex]:
[tex]\[
p(-x) = -x + 3
\][/tex]
3. Add [tex]\( p(x) \)[/tex] and [tex]\( p(-x) \)[/tex]:
[tex]\[
p(x) + p(-x) = (x + 3) + (-x + 3) = x - x + 3 + 3 = 6
\][/tex]
Therefore, [tex]\( p(x) + p(-x) \)[/tex] is equal to [tex]\( 6 \)[/tex].
Answer: d) 6
### Question 14:
We need to determine one of the zeroes of the polynomial [tex]\( 2x^2 + 7x - 4 \)[/tex].
1. Identify the polynomial:
The polynomial is [tex]\( 2x^2 + 7x - 4 \)[/tex].
2. Find the roots of the polynomial:
To find the roots, we solve the equation [tex]\( 2x^2 + 7x - 4 = 0 \)[/tex].
The roots of a quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex] are found using the quadratic formula:
[tex]\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
where [tex]\( a = 2 \)[/tex], [tex]\( b = 7 \)[/tex], [tex]\( c = -4 \)[/tex].
3. Calculate the discriminant:
[tex]\[
\text{Discriminant} = b^2 - 4ac = 7^2 - 4 \cdot 2 \cdot (-4) = 49 + 32 = 81
\][/tex]
4. Calculate the roots:
[tex]\[
x = \frac{-7 \pm \sqrt{81}}{2 \cdot 2} = \frac{-7 \pm 9}{4}
\][/tex]
- For [tex]\( + \sqrt{81} \)[/tex]:
[tex]\[
x = \frac{-7 + 9}{4} = \frac{2}{4} = \frac{1}{2}
\][/tex]
- For [tex]\( - \sqrt{81} \)[/tex]:
[tex]\[
x = \frac{-7 - 9}{4} = \frac{-16}{4} = -4
\][/tex]
Thus, the roots of the polynomial [tex]\( 2x^2 + 7x - 4 \)[/tex] are [tex]\( \frac{1}{2} \)[/tex] and [tex]\( -4 \)[/tex].
One of the zeroes is [tex]\( \frac{1}{2} \)[/tex].
Answer: b) [tex]\(\frac{1}{2}\)[/tex]
