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

For $T$ to equal zero, the numerator must equal zero while the denominator is not zero. Thus, we set:

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

Squaring both sides leads to:

$2 - 5b = 0$

Solving for $b$ gives:

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

It's essential to ensure the denominator ($3b$) is not zero, which it is not for $b = \frac{2}{5}$.

To find the values of $k$ for which the equation has real roots, we first rewrite it in standard form:

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

The discriminant ($\Delta$) of a quadratic equation $ax^2 + bx + c = 0$ must be greater than or equal to zero for real roots:

$\Delta = b^2 - 4ac$

Here, $a = k$, $b = -2$, and $c = 35$. Therefore:

$\Delta = (-2)^2 - 4(k)(35)$

$\Delta = 4 - 140k$

For real roots, we need:

$4 - 140k \geq 0$

Solving this inequality provides:

$140k \leq 4$ $k \leq \frac{4}{140}$ $k \leq \frac{1}{35}$

Additionally, since the expression must be non-negative, we also check:

$4 + 140k \geq 0$

Which gives:

$k \geq -\frac{1}{35}$

Thus, the solutions for $k$ are:

$-\frac{1}{35} \leq k \leq \frac{1}{35}$