Question 1
The equation $k x^2 + 8x + 3k = 0$ where $k$ is a constant, has real unequal roots.
Find the set of values of $k$ giving your answer in an exact simplified form. (5)
Step 1: Condition for real and unequal roots
For a quadratic equation $ax^2 + bx + c = 0$, the roots are real and unequal if and only if the discriminant $\Delta$ satisfies:
$ \Delta > 0. $
The discriminant is given by:
$ \Delta = b^2 - 4ac. $
Step 2: Identifying coefficients
In the given equation $kx^2 + 8x + 3k = 0$, the coefficients are:
$ \begin{aligned} a &= k, \\ b &= 8, \\ c &= 3k. \end{aligned} $
Step 3: Compute the discriminant
Substitute the coefficients into the formula for the discriminant:
$ \begin{aligned} \Delta &= b^2 - 4ac \\ &= 8^2 - 4(k)(3k) \\ &= 64 - 12k^2. \end{aligned} $
Step 4: Solve the inequality $\Delta > 0$
For the roots to be real and unequal, we require $\Delta > 0$:
$ \begin{aligned} 64 - 12k^2 &> 0 \\ 12k^2 &< 64 \\ k^2 &< \frac{64}{12} \\ k^2 &< \frac{16}{3}. \end{aligned} $
Step 5: Solve for $k$
Taking the square root on both sides, we obtain:
$ -\sqrt{\frac{16}{3}} < k < \sqrt{\frac{16}{3}}. $
$ -\frac{4\sqrt{3}}{3} < k < \frac{4\sqrt{3}}{3}. $
Step 6: Final answer
The set of values of $k$ for which the equation has real and unequal roots is:
$ k \in \left(-\frac{4\sqrt{3}}{3}, \frac{4\sqrt{3}}{3}\right). $
Question 2
In triangle $ABC$, $AB = 3x$ cm, $BC = 5x$ cm and $\angle ABC = 110^{\circ}$.
(a) Find, in degrees to one decimal place, the size of $\angle BCA$. (4)
The area of triangle $ABC$ is $24$ cm$^2$.
(b) Find, to 3 significant figures, the value of $x$. (3)
Step 1: Understanding the problem
We are given a triangle $ABC$ where:
- $AB = 3x\,\mathrm{cm}$,
- $BC = 5x\,\mathrm{cm}$,
- $\angle ABC = 110^{\circ}$.
Additionally, the area of $\triangle ABC$ is given as $24\,\mathrm{cm}^2$. The goal is to:
- Find the size of $\angle BCA$ (in degrees, to one decimal place).
- Determine the value of $x$ (to 3 significant figures).
Step 2: Using the Sine Rule to find $\angle BCA$
To find $\angle BCA$, we first determine the length of $AC$ using the Law of Cosines:
$ AC^2 = AB^2 + BC^2 - 2 \cdot AB \cdot BC \cdot \cos(\angle ABC). $
Substituting the known values:
$ \begin{aligned} AC^2 &= (3x)^2 + (5x)^2 - 2 \cdot (3x) \cdot (5x) \cdot \cos(110^{\circ}) \\ &= 9x^2 + 25x^2 - 30x^2 \cdot \cos(110^{\circ}). \end{aligned} $
Simplify to:
$ AC^2 = 34x^2 - 30x^2 \cdot \cos(110^{\circ}). $
$ AC = \sqrt{34x^2 - 30x^2 \cdot \cos(110^{\circ})}. $
Next, we apply the Sine Rule:
$ \frac{AB}{\sin(\angle BCA)} = \frac{AC}{\sin(\angle ABC)}. $
Let $\angle BCA = \theta$. Substituting the known values:
$ \frac{3x}{\sin(\theta)} = \frac{\sqrt{34x^2 - 30x^2 \cdot \cos(110^{\circ})}}{\sin(110^{\circ})}. $
Rearranging for $\sin(\theta)$:
$ \sin(\theta) = \frac{3x \cdot \sin(110^{\circ})} {\sqrt{34x^2 - 30x^2 \cdot \cos(110^{\circ})}}. $
Using numerical values:
- $\cos(110^{\circ}) \approx -0.3420$,
- $\sin(110^{\circ}) \approx 0.9397$.
Substitute these into the equation for $\sin(\theta)$:
$ \sin(\theta) = \frac{3x \cdot 0.9397} {\sqrt{34x^2 - 30x^2 \cdot (-0.3420)}}. $
Simplify the denominator:
$ \sqrt{34x^2 - 30x^2 \cdot (-0.3420)} = \sqrt{34x^2 + 10.26x^2} = \sqrt{44.26x^2} = \sqrt{44.26} \cdot x \approx 6.65x. $
Thus:
$ \sin(\theta) = \frac{3x \cdot 0.9397}{6.65x} = \frac{2.8191}{6.65} \approx 0.4239. $
Now, find $\theta$:
$ \theta = \arcsin(0.4239) \approx 25.1^{\circ}. $
Step 3: Using the area formula to find $x$
The area of $\triangle ABC$ can also be expressed using the formula:
$ \text{Area} = \frac{1}{2} \cdot AB \cdot BC \cdot \sin(\angle ABC). $
Substituting the known values:
$ \begin{aligned} 24 &= \frac{1}{2} \cdot (3x) \cdot (5x) \cdot \sin(110^{\circ}) \\ 24 &= \frac{1}{2} \cdot 15x^2 \cdot 0.9397. \end{aligned} $
Simplify to:
$ 15x^2 \cdot 0.9397 = 48. $
Divide through by $15 \cdot 0.9397$:
$ x^2 = \frac{48}{15 \cdot 0.9397} \approx \frac{48}{14.0955} \approx 3.406. $
Take the square root to find $x$:
$ x = \sqrt{3.406} \approx 1.85. $
Step 4: Final answers
- The size of $\angle BCA$ is $25.1^{\circ}$ (to one decimal place).
- The value of $x$ is $1.85$ (to 3 significant figures).
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