A quadratic equation is given by x2 + (2k+2)x + 2k+5 = 0. Part (a) asks for values of k where the roots are equal. Part (b) asks for a new quadratic equation with integral coefficients for k > 0, whos

Substitute the values of $\alpha+\beta$ and $\alpha\beta$:

Factor out $\alpha$ from the first term and $\beta$ from the second term:

Substitute the values of $\alpha+\beta$ and $\alpha\beta$:

A quadratic equation with roots $r_1$ and $r_2$ is given by $x^2 - (r_1+r_2)x + r_1r_2 = 0$.

The numerator is zero when $2-x=0 \implies x=2$.
The denominator is zero when $x+1=0 \implies x=-1$.
These points divide the number line into three intervals: $(-\infty, -1)$, $(-1, 2)$, and $(2, \infty)$.

• For $x < -1$ (e.g., $x=-2$): $\frac{2-(-2)}{-2+1} = \frac{4}{-1} = -4$. This is $< 0$.
• For $-1 < x < 2$ (e.g., $x=0$): $\frac{2-0}{0+1} = \frac{2}{1} = 2$. This is $\ge 0$.
• For $x > 2$ (e.g., $x=3$): $\frac{2-3}{3+1} = \frac{-1}{4}$. This is $< 0$.

We need $\frac{2-x}{x+1} \ge 0$. This is satisfied when $-1 < x < 2$.
Since the inequality includes "equal to" ($\ge$), the value $x=2$ (where the numerator is zero) is included in the solution.
The value $x=-1$ (where the denominator is zero) makes the expression undefined, so it must be excluded.
Therefore, the range of values for $x$ is $-1 < x \le 2$.