Given: $$T = \frac{\sqrt{2 - 5b}}{3b}$$ Determine the numerical value of $b$ for which $T$ is: 2.1.1 Undefined 2.1.2 Equal to zero 2.2 Determine the value(s) of $k$ for which the equation $k^2 - 35 - 2x$ has real roots. - NSC Technical Mathematics - Question 2 - 2024 - Paper 1

To determine when $T$ equals zero, we first note that a fraction is zero when its numerator is zero (as long as the denominator is not zero). The numerator is:

$\sqrt{2 - 5b}$

Setting this equal to zero:

$\sqrt{2 - 5b} = 0$

Squaring both sides yields:

$2 - 5b = 0$

Now, solving for $b$:

$5b = 2$

$b = \frac{2}{5}$

Therefore, the numerical value of $b$ for which $T$ is equal to zero is: $b = \frac{2}{5}$.

To find the values of $k$ for which the quadratic equation $k^2 - 35 - 2x = 0$ has real roots, we first rewrite it in standard form:

$k^2 - 2x - 35 = 0$

Next, we calculate the discriminant ($\Delta$) using the formula:

$\Delta = b^2 - 4ac,$

where $a = 1$, $b = -2x$, and $c = -35$. Thus, we obtain:

$\Delta = (-2x)^2 - 4(1)(-35)$

This simplifies to:

$\Delta = 4x^2 + 140$

For the quadratic to have real roots, we set the discriminant greater than or equal to zero:

$4x^2 + 140 \geq 0$

Since $4x^2$ is always non-negative, this inequality holds for all real values of $x$.

Next, to reason about $k$, we also rearrange the equation:

$k^2 - 35 - 2x = 0$
leading to

$k^2 = 35 + 2x$

We need to ensure that this is non-negative:

$k^2 \geq 0$. This indicates:

$35 + 2x \geq 0$

Finally, solving for $k$ gives:

1. For $x = 0$:
   • $k^2 = 35 ightarrow k = \pm \sqrt{35}$
2. For $x < -\frac{35}{2}$:
   • The left-hand side can be negative leading to no real roots.

So, the values of $k$ for which the equation has real roots is:

$k = \pm \sqrt{35}$