This section consists of 20 questions of 1 mark each.

If the sum of zeroes of the polynomial [tex]$p(x)=2x^2 - k \sqrt{2} x + 1[tex]$[/tex] is [tex]$[/tex]\sqrt{2}[tex]$[/tex], then the value of [tex]$[/tex]k$[/tex] is:



Answer :

To determine the value of \( k \) for the polynomial \( p(x) = 2x^2 - k\sqrt{2} x + 1 \) given that the sum of its zeroes is \( \sqrt{2} \), let's follow these steps:

1. Recall the general properties of a polynomial:
For any quadratic polynomial \( ax^2 + bx + c \), the sum of the zeroes (roots) is given by the formula:
[tex]\[ \text{Sum of zeroes} = -\frac{b}{a} \][/tex]

2. Identify the coefficients:
In the polynomial \( p(x) = 2x^2 - k\sqrt{2} x + 1 \), the coefficients are:
[tex]\[ a = 2, \quad b = -k\sqrt{2}, \quad c = 1 \][/tex]

3. Set up the relationship using the sum of the zeroes:
We are given that the sum of the zeroes is \( \sqrt{2} \). According to the formula for the sum of the zeroes, we have:
[tex]\[ -\frac{b}{a} = \sqrt{2} \][/tex]

4. Substitute the coefficients into the formula:
Substituting \( b = -k\sqrt{2} \) and \( a = 2 \) into the formula, we get:
[tex]\[ -\frac{-k\sqrt{2}}{2} = \sqrt{2} \][/tex]

5. Simplify the equation:
[tex]\[ \frac{k\sqrt{2}}{2} = \sqrt{2} \][/tex]

6. Solve for \( k \):
To isolate \( k \), multiply both sides by 2:
[tex]\[ k\sqrt{2} = 2\sqrt{2} \][/tex]

Now, divide both sides by \( \sqrt{2} \):
[tex]\[ k = \frac{2\sqrt{2}}{\sqrt{2}} \][/tex]

Simplify the right-hand side:
[tex]\[ k = 2 \][/tex]

Therefore, the value of [tex]\( k \)[/tex] is [tex]\( \boxed{2} \)[/tex].