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2.3 - 2.6 | Forms of Quadratics

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2.3-6 | Forms of Quadratics

Standard form

As we have explored linear functions (polynomials to the first power), we will now look into quadratic functions (polynomials to the second power).

Quadratics are composed of 3 terms: \(x^2\), \(x\), and constant terms. In the standard form of quadratics, they are represented as follows:

\[f(x) = ax^2 + bx + c\]

where \(c\) is the y-intercept, and \(a\) indicates whether the quadratic points upwards or downwards. Specifically, if \(a\) is negative, then the parabola opens up downwards, but it opens upwards when \(a\) is positive. I.e. the function has a maximum (range) when \(a\) is negative, and the function has a minimum (range) when \(a\) is positive:

Another useful hint we can derive from the standard form of quadratics is the line of symmetry. It can be noticed that the parabolic shapes that quadratics generate are symmetrical, with a line of symmetry traveling through the center vertically. To determine the equation of the line of symmetry (which can also be useful in finding the vertex i.e. the minimum and maximum points of the function– since the vertices lie on the line of symmetry), we can make use of the coefficients as follows:

\[x = -{b \over 2a}\]

This provides the line of symmetry. To find the vertex, we can use this value for \(x\), and plug it in for \(f(x)\), such that the y-value of the function at \(x\) can be determined:

\[\text{Coordinates of Vertex} = (-{b \over 2a}, f(-{b \over 2a}))\]

Factored form

This form is key in identifying the roots of the quadratic. The factored form is represented as follows:

\[f(x) = a(x-p)(x-q)\]

This is special, because when the y-value aka \(f(x) = 0\), the \(x\) value provides the intersection with the x-axis, i.e. the roots.

Therefore, if the value of the function is 0, it means that \(x\) is a root. In factored form, if \(x=p\) or \(x=q\), it would cause either \(q-q=0\) or \(p-p=0\), i.e. the value of the function being 0, rendering \(x=p\) or \(x=q\) roots of the function \(f(x)\).

To get to this form, it is crucial to understand how to factor quadratics– going from standard form to factored form. Although there is indeed a formula (the quadratic formula), you cannot always just plug in numbers there, as paper 1 does not allow a calculator to be used, causing calculations to become extremely tedious. Therefore, factoring is highly important.

Going from standard form to factored form has a basic procedure which can be applied to any given quadratic \(ax^2 + bx + c\). However, it must be kept in mind that although all quadratics can be factored, not all of them can be factored with nice, integer numbers. Therefore if you cannot easily factor the quadratic, you have most likely either made a mistake, or it is a paper 2 or 3 question where you may use your calculator to calculate/graph roots.

  1. Multiply \(a\) and \(c\) to get \(ac\)

  2. Find 2 numbers that multiply to \(ac\) and add up to \(b\) simultaneously

  3. Letting these 2 numbers in step \(2\) be \(p\) and \(q\), separate \(bx\) to \(px + qx\)

  4. Factor like terms

  5. Factor like terms again

This may seem confusing, but it really is quite simple. It can easily be visualized through an example:

Solve \(x^2 - 5x + 6 = 0\)

In this case, \(a = 1\), \(b = -5\), \(c = 6\). Therefore:

\begin{align}
ac &=  1 \times 6\\
ac &=  6\\
\end{align}

Now we must find 2 numbers that multiply to \(6\), but add up to \(-5\).

This is possible with the numbers \(-2\) and \(-3\), because \((-2) \times (-3) = 6\), and \((-2) + (-3) = - 2 - 3 = - 5\).

Hence, we separate \(-5x\) into \(-2x-3x\):

\begin{align}
x^2 - 5x + 6 &= 0\\
x^2 - 2x - 3x + 6 &= 0\\
\end{align}

We can notice that the leftmost 2 terms have an \(x\) in common, and the right-most 2 terms have a 3 in common (instead of 3 we can factor negative 3, such that the minus corresponds in the same way as the leftmost terms):

\begin{align}
x^2 - 2x - 3x + 6 &= 0\\
x(x - 2) - 3(x - 2) &= 0\\
\end{align}

Now we can factor a second time, as we can notice that the term \((x-2)\) is common for both terms:

\begin{align}
x(x - 2) - 3(x - 2) &= 0\\
(x - 2)(x - 3) &= 0\\
\end{align}

Hence, we have achieved factored form.

We can also easily tell that the equation would be satisfied by \(x = 2\) and \(x = 3\), because then the equation would lead to \(0 = 0\) which is a true statement.

Therefore, \(x = 2\) and \(x = 3\) are roots of the function \(f(x) = x^2 - 5x + 6 = 0\).

Vertex Form

This is the 3rd form of quadratics which should be studied. Vertex form is represented as follows:

\[f(x) = a(x - h) + k\]

Through this form, the coordinates of the vertex (i.e. the minimum or maximum point of the quadratic) can be identified as \((h,k)\).

In the exam, you can be explicitly asked to use vertex form to determine the vertex.

Going from standard form to vertex form is also systematic, such that it just requires a number of basic steps given a quadratic \(ax^2 + bx + c\):

  1. Take out the Greatest Common Factor (GCF) of the first two terms \(a\) and \(b\), such that \(x^2\) no longer has a coefficient, and the equation resembles the form:

    \[p(x^2 + qx) + c)\]

  2. Take the number \(({q \over 2})^2\) (half of q, squared). Add and subtract this number simultaneously, such that the equation resembles the form:

    \[p(x^2 + qx + ({q \over 2})^2 - ({q \over 2})^2) + c)\]

  3. Multiply \(({q \over 2})^2\) by \(p\) to take it out of the brackets, such that the equation resembles the form:

    \[p(x^2 + qx + ({q \over 2})^2) + c - ({q \over 2})^2 \times p)\]

  4. After writing down \(p\) open-bracket, write down \(x\), the sign after \(x^2\), and \({q \over 2}\) close-bracket, all squared, and then the rest of the remaining terms, such that the equation resembles the form:

    \[p(x + ({q \over 2}))^2 + c - ({q \over 2})^2 \times p\]

  5. Hence, vertex form has been achieved, with the vertex being \((-{q \over 2}, (c - ({q \over 2})^2 \times p)\), where \(h = -{q \over 2}\) and \(k = c - (({q \over 2})^2 \times p)\)

This probably seems super confusing just because of the number of variables, but this is simply the generalized form. The best way to understand this would be through an example:

Given the function \(f(x) = 2x^2 + 8x - 5\), find the coordinates of the vertex through vertex form.

Taking the GCF of the coefficients of \(2x^2\) and \(8x\) is \(2\). Hence:

\begin{align}
& 2x^2 + 8x - 5\\
&= 2(x^2 + 4x) - 5\\
\end{align}

Taking half of \(4\) and squaring, leads to \(({4 \over 2})^2 = 2^2\). Adding and subtracting:

\begin{align}
&= 2(x^2 + 4x + 2^2 - 2^2) - 5\\
\end{align}

Taking the \(-2^2\) out of the brackets, we must multiply it by the coefficient on the bracket, which is \(2\):

\begin{align}
&= 2(x^2 + 4x + 2^2 - 2^2) - 5\\
&= 2(x^2 + 4x + 2^2) - 5 - (2 \times 2^2)\\
\end{align}

Writing down according to step 4 and simplifying:

\begin{align}
&= 2(x^2 + 4x + 2^2) - 5 - (2 \times 2^2)\\
&= 2(x + 2)^2 - 5 - (2 \times 2^2)\\
&= 2(x + 2)^2 - 5 - (2 \times 4)\\
&= 2(x + 2)^2 - 5 - 8\\
&= 2(x + 2)^2 - 13\\
\end{align}

Hence, vertex form has been achieved, with the coordinates of the vertex being \((-2,-13)\).

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Consider the triangle \(PQR\) such that \(PQ=12\) cm, \(QR=24\) cm, and \(∠QPR=73°\).

(a) Find the length of \(PR\).

(b) Find the area of triangle \(PQR\).

(a) Find the length of \(PR\).

{METHOD 1}

Use cosine rule, by strategically inputting values for the variables.

\[{c}^{2}={a}^{2}+{b}^{2}-2(a)(b)\cos(C)\]

As we are given \(∠QPR\), we can find a way to use it in the cosine rule directly, and place other values in accordingly to find :

\[{QR}^{2}={PQ}^{2}+{PR}^{2}-2(PQ)(PR)\cos(∠QPR)\]

Then simply substitute and simplify using GDC. Remember to change from radians to degrees for this case.

\begin{align*}
{24}^{2}&={12}^{2}+{PR}^{2}-2(12)(QR)\cos(73°)\\
576&=144+{PR}^{2}-24(0.29237)QR\\
144-576+{PR}^{2}-7.01692PR&=0\\
{PR}^{2}-7.01692PR-432&=0\\
\end{align*}

Solving the quadratic using GDC (by graphing to find roots or through polynomial solve function):

\[PR=24.578 \text{ or } PR=-17.570\]

Reject the negative value (\(PR=-17.570\)), as a geometric line in 2D space cannot have a negative length.

Therefore:

\[PR=\boxed{24.6 \text{ cm}}\]

{METHOD 2}

The sine rule permits us to find \(α\) in the following sketch (not to scale, and may not reflect the true nature of the triangle, but provides a good approximate geometrical representation):

Using sine rule, we can instead attempt to find another angle in the triangle; by finding \(α\), we can find \(θ\).

Therefore using GDC by changing from radians into degrees:

\begin{align*}
\frac{\sin(∠QPR)}{QR}&=\frac{\sin(α)}{PQ}\\
α&=\arcsin(\frac{PQ\sin(∠QPR)}{QR})\\
α&=\arcsin(\frac{12\sin(73°)}{24})\\
α&=28.5648°\\
\end{align*}

As the sum of angles in a triangle add up to 180°:

\begin{align*}
α+θ+∠QPR&=180°\\
θ&=180°-α-∠QPR\\
θ&=180°-28.5648°-73°\\
θ&=78.4352°\\
\end{align*}

Now, the cosine rule can be used directly. Inputting the following values and solving with GDC:

\begin{align*}
{c}^{2}&={a}^{2}+{b}^{2}-2(a)(b)\cos(C)\\
{PR}^{2}&={QR}^{2}+{PQ}^{2}-2(QR)(PQ)\cos(θ)\\
PR&=\sqrt{{24}^{2}+{12}^{2}-2(24)(12)\cos(78.4352°)}\\
PR&=\boxed{24.6 \text{ cm}}\\
\end{align*}

[3]

(b) Find the area of triangle \(PQR\)

Through part (a), all the information acquired can be used in 1 equation to find the area. Namely:

\[A=\frac{1}{2}ab\sin(C)\]

Any appropriate values can be used to gain the value, as there are multiple different values which can be substituted to gain the area.

\begin{align*}
A&=\frac{1}{2}ab\sin(C)\\
A&=\frac{1}{2}(PQ)(QR)\sin(θ)\\
A&=\frac{1}{2}(12)(24)\sin(78.4352°)\\
A&=\boxed{141 {\text{ cm}}^{2}}\\
\end{align*}

[2]

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The complex numbers \(p\) and \(q\) lie on a complex plane, within a circle of radius 6 units.

a) Determine the exact value of \(|p-q|\).

b) Determine the area of triangle \(\text{ROS}\).

a) Determine the exact value of \(|p-q|\).

METHOD 1

First, it must be understood what is meant by \(|p-q|\). Essentially, we must find the modulus (magnitude) of the complex number produced by \(p-q\). The best way to do this with the given information is geometrically; the trick is, treating this question as a geometry and vectors question.

Geometrically representing \(p-q\) using \(p\) and \(-q\) would look like this:

Looking at the diagram as a whole, it can be identified that both \(p\) and \(q\) are of the same magnitude, as they are encompassed in a circle of radius \(6\) from the origin \(\text{O]\). Hence, an isosceles triangle is formed:

Using the cosine rule here from the data booklet to find \(|p-q|\):

\begin{align}c^2 &= a^2 + b^2 - 2ab \cos C\\{(|p-q|)}^{2} &= 6^2 + 6^2 - 2 \times 6 \times 6 \times \cos ({\pi \over 3})\\\end{align}

In this case, \(\cos ({\pi \over 3})\) (which is the same as \(\cos (60°)\)) is \({1 \over 2}\), which should be learnt by heart.

Hence:

\begin{align}{(|p-q|)}^{2} &= 6^2 + 6^2 - 2 \times 6 \times 6 \times \cos ({\pi \over 3})\\{(|p-q|)}^{2} &= 6^2 + 6^2 - 2 \times 6^2 \times {1 \over 2}\\{(|p-q|)}^{2} &= 6^2 + 6^2 - 6^2\\|p-q| &= \sqrt{6^2}\\& \rightarrow \boxed{|p-q| = 6 \text{ units}}\\\end{align}

METHOD 2

The first few steps of this method are the same as the previous method. Getting to the point of identifying that an isosceles triangle is formed is crucial:

It is known that all interior angles in a triangle add up to 180°, or \(\pi\) radians:

\[\angle SOR + \angle OSR + \angle ROS = \pi\]

It is also known that in an isosceles triangle, isosceles angles \(\angle SOR\) and \(\angle OSR\) must be equal:

\[\angle SOR = \angle OSR\]

Hence:

\begin{align}\angle SOR + \angle OSR + \angle ROS &= \pi \\(\angle SOR + \angle OSR) + {\pi \over 3} &= \pi \\2(\angle SOR) &= \pi - {\pi \over 3} \\2(\angle SOR) &= {2 \pi \over 3} \\\angle SOR &= {\pi \over 3} \\\angle SOR &= {\pi \over 3} = \angle OSR \\\end{align}

As a result, it is clear that all angles within triangle \(ROS\) are equal to \({\pi \over 3}\), meaning that they are all the same.

If all angles are the same, it means that the triangle is an equilateral triangle, such that all sides must be the same.

Hence, 

\[OR = OS = \boxed{|p-q| = 6 \text{ units}}\]

b) Determine the area of triangle \(ROS\).

To determine the area, the easiest way is to use the formula provided in the data booklet:

\begin{align}\text{Area} &= {1 \over 2}ab \sin C\\\text{Area} &= {1 \over 2} \times 6 \times 6 \times \sin ({\pi \over 3})\\\end{align}

In this case, \(\sin ({\pi \over 3})\) (which is the same as \(\sin (60°)\)) is \({\sqrt{3} \over 2}\), which should be learnt by heart.

\begin{align}\text{Area} &= {1 \over 2} \times 36 \times {\sqrt{3} \over 2}\\\text{Area} &= {1 \over 4} \times 36 \times \sqrt{3}\\& \rightarrow \boxed{\text{Area} = 9 \sqrt{3} \text{ square units}}\\\end{align}

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A soldier operates a mortar, preparing to launch a shell. The trajectory of the shell depends on the initial angle \(\theta\) with the ground.

The horizontal and vertical trajectories are given by the functions \(x(t)=6t\sin\theta\) and \(y(t)=3{t}^{2}-6t\cos\theta-6\) respectively, with respect to time \(t\).

Diagram 1.png

Standing on a cliff, the mortar sits upon land which is at a non-zero height, but the ground surrounding the cliff is flat and at a height of \(0\).

  1. Find the time \(t\) at which the shell hits the ground in terms of \(\theta\).

  2. Determine the optimal angle \(\theta\), such that the horizontal range \(R\) of the shell shot by the mortar is greatest.

  3. Determine the greatest horizontal range \(R\) the shell shot by the mortar can reach.

(a) Find the time \(t\) at which the shell hits the ground in terms of \(\theta\).

The shell hits the ground when the vertical trajectory (height) is 0 (\(y(t)=0\)):

\[y(t)=0=3{t}^{2}-6t\cos\theta-6\]
[1]

Using the quadratic formula to solve for \(t\) in terms of \(\theta\):

\begin{align}
0&=3{t}^{2}-6t\cos\theta-6\\
t&={-b ± \sqrt{b^2-4ac}\over 2a}\\
t&={-(-6\cos\theta) ± \sqrt{{(-6\cos\theta)}^{2}-4(3)(-6)}\over 2(3)}\\
t&={6\cos\theta ± \sqrt{{36{\cos}^{2}\theta}+72}\over 6}\\
t&=\cos\theta ± {\sqrt{{36({\cos}^{2}\theta}+2)}\over 6}\\
t&=\cos\theta ± {6\sqrt{{{\cos}^{2}\theta}+2}\over 6}\\
t&=\cos\theta ± \sqrt{{{\cos}^{2}\theta}+2}\\
\end{align}
[1]

Because the time \(t\) cannot be negative, the negative option is rejected. Hence:

\[\boxed{t=\cos\theta + \sqrt{{{\cos}^{2}\theta}+2}}\]
[1]

(b) Determine the optimal angle \(\theta\), such that the horizontal range \(R\) of the shell shot by the mortar is greatest.

The horizontal distance/range is described by the given function \(x(t)=6t\sin\theta\).

From part (a), \(t\) is found in terms of \(\theta\) for when the shell reaches the ground, such that it can be substituted in for \(t\) for the function \(x(t)\):

\begin{align}
x(t)&=6t\sin\theta\\
x(\theta)&=6(\cos\theta + \sqrt{{{\cos}^{2}\theta}+2})\sin\theta=R\\
R&=6(\sin\theta\cos\theta + \sin\theta\sqrt{{{\cos}^{2}\theta}+2})\\
\end{align}
[1]

For the greatest horizontal range, \(R={x}_{\text{max}}(t)\). To find the maximum, the first derivative of the function must be at \(0\) (\({x}_{\text{max}}^\prime(t)={dR \over d\theta}=0\)). To do this, the product and the chain rule can be used:

\begin{align}

0&={dR \over d\theta}\\

0&={d \over d\theta}[6(\sin\theta\cos\theta + \sin\theta\sqrt{{{\cos}^{2}\theta}+2})]\\

0&=6\cdot {d \over d\theta}[\sin\theta\cos\theta + \sin\theta\sqrt{{{\cos}^{2}\theta}+2}]\\

0&={d \over d\theta}[\sin\theta\cos\theta + \sin\theta\sqrt{{{\cos}^{2}\theta}+2}]\\

0&=\sin\theta \cdot {d \over d\theta}(\cos\theta) + \cos\theta \cdot {d \over d\theta}(\sin\theta) + \sin\theta \cdot {d \over d\theta}(\sqrt{{{\cos}^{2}\theta}+2}) + \sqrt{{{\cos}^{2}\theta}+2} \cdot {d \over d\theta}(\sin\theta)\\

0&={-\sin}^{2}\theta + {\cos}^{2}\theta + {\sin\theta \over \sqrt{{{\cos}^{2}\theta}+2}} \cdot {d \over d\theta}({\cos}^{2}\theta+2) + \cos\theta\sqrt{{{\cos}^{2}\theta}+2}\\

0&={\cos}^{2}\theta - {\sin}^{2}\theta + {\sin\theta \over \sqrt{{{\cos}^{2}\theta}+2}} \cdot \cos\theta(-\sin\theta) + \cos\theta\sqrt{{{\cos}^{2}\theta}+2}\\

0&={\cos}^{2}\theta - {\sin}^{2}\theta - {{\sin}^{2}\theta\cos\theta \over \sqrt{{{\cos}^{2}\theta}+2}} + \cos\theta\sqrt{{{\cos}^{2}\theta}+2}\\

\end{align}
[4]

Bringing the terms with the square root to one side, and then squaring both sides of the equation, will ultimately cancel out the square roots:

\begin{align}

0&={\cos}^{2}\theta - {\sin}^{2}\theta - {{\sin}^{2}\theta\cos\theta \over \sqrt{{{\cos}^{2}\theta}+2}} + \cos\theta\sqrt{{{\cos}^{2}\theta}+2}\\

{\sin}^{2}\theta - {\cos}^{2}\theta &= \cos\theta\sqrt{{{\cos}^{2}\theta}+2} - {{\sin}^{2}\theta\cos\theta \over \sqrt{{{\cos}^{2}\theta}+2}}\\

{({\sin}^{2}\theta - {\cos}{2}\theta)}{2} &= {(\cos\theta\sqrt{{{\cos}^{2}\theta}+2} - {{\sin}^{2}\theta\cos\theta \over \sqrt{{{\cos}{2}\theta}+2}})}2\\

{\sin}^{4}\theta + {\cos}^{4}\theta -2{\sin}2\theta{\cos}{2}\theta &= {\cos}{2}\theta({\cos}{2}\theta+2) + {{\sin}{4}\theta{\cos}{2}\theta \over {\cos}^{2}\theta+2} - 2(\cos\theta\sqrt{{\cos}{2}\theta+2})({{\sin}{2}\theta\cos\theta \over \sqrt{{{\cos}^{2}\theta}+2}})\\

{\sin}^{4}\theta + {\cos}^{4}\theta -2{\sin}2\theta{\cos}{2}\theta &= {\cos}{2}\theta({\cos}{2}\theta+2) + {{\sin}{4}\theta{\cos}{2}\theta \over {\cos}^{2}\theta+2} - 2(\cos\theta\sqrt{{\cos}{2}\theta+2})({{\sin}{2}\theta\cos\theta \over \sqrt{{{\cos}^{2}\theta}+2}})\\

{\sin}^{4}\theta + {\cos}^{4}\theta - 2{\sin}2\theta{\cos}{2}\theta &= {\cos}^{4}\theta + 2{\cos}^{2}\theta + {{\sin}{4}\theta{\cos}{2}\theta \over {\cos}^{2}\theta+2} - 2{\sin}2\theta{\cos}{2}\theta\\

{\sin}^{4}\theta &= 2{\cos}^{2}\theta + {{\sin}{4}\theta{\cos}{2}\theta \over {\cos}^{2}\theta+2}\\

{\sin}{4}\theta({\cos}{2}\theta+2) &= 2{\cos}{2}\theta({\cos}{2}\theta+2) + {\sin}{4}\theta{\cos}{2}\theta\\

{\sin}{4}\theta{\cos}{2}\theta + 2{\sin}^{4}\theta &= 2{\cos}^{4}\theta + 4{\cos}^{2}\theta + {\sin}{4}\theta{\cos}{2}\theta\\

2{\sin}^{4}\theta &= 2{\cos}^{4}\theta + 4{\cos}^{2}\theta\\

{\sin}^{4}\theta &= {\cos}^{4}\theta + 2{\cos}^{2}\theta\\

\end{align}
[4]

Now that the simplified equation comprises of both \(\sin^2\) and \(\cos^2\), one or the other could be manipulated and substituted using the identity \({\sin}^{2}\theta + {\cos}^{2}\theta = 1\). It is easier to substitute in place of \(\sin\) to make the entire equation comprise of \(\cos\), because there are more \(\cos\). Thus manipulating the identity provides \({\sin}^{2}\theta = 1 - {\cos}^{2}\theta\):

\begin{align}

{\sin}^{4}\theta &= {\cos}^{4}\theta + 2{\cos}^{2}\theta\\

{({\sin}{2}\theta)}2 &= {\cos}^{4}\theta + 2{\cos}^{2}\theta\\

{(1 - {\cos}{2}\theta)}2 &= {\cos}^{4}\theta + 2{\cos}^{2}\theta\\

1 + {\cos}^{4}\theta - 2{\cos}^{2}\theta &= {\cos}^{4}\theta + 2{\cos}^{2}\theta\\

1 - 2{\cos}^{2}\theta &= 2{\cos}^{2}\theta\\

1 &= 4{\cos}^{2}\theta\\

{1 \over 4} &= {\cos}^{2}\theta\\

± \sqrt{{1 \over 4}} &= \cos\theta\\

\cos\theta &= ± {1 \over 2}\\

\end{align}
[3]

To find the values that \(\theta\) can take to produce the greatest range, either a unit circle, or a sketch of \(\cos\theta\) in combination with the known angles can be used:

Diagram 2.png

It should be known by heart that \(\cos(60°)={1 \over 2}\).

Any values other than \(60°\) are rejected, as \(\theta<90°\).

Hence:

\[\boxed{{\theta}_{\text{optimal}}=60°}\]
[1]

(c) Determine the greatest horizontal range \(R\) the shell shot by the mortar can reach.

From part (a) and (b), the function for the horizontal range \(x(t)\) is found in terms of \(\theta\), as well as the optimal angle \(\theta\). Therefore, substituting the value for \(theta\) in the horizontal range function in terms of \(\theta\), \(x(\theta)\), provides the greatest horizontal range \(R\):

(The values for \(\sin(60°)\) and \(\cos(60°)\) should be known by heart.)

\begin{align}
x(t)&=6t\sin\theta\\
x(\theta)&=6(\cos\theta + \sqrt{{{\cos}^{2}\theta}+2})\sin\theta=R\\
R&=6(\sin(60°)\cos(60°) + \sin(60°)\sqrt{{{\cos}^{2}(60°)}+2})\\
R&=6({\sqrt{3} \over 2} \cdot {1 \over 2} + {\sqrt{3} \over 2}\sqrt{{{({1 \over 2})}^{2}}+2})\\
&\rightarrow \boxed{R={27 \over 8} \text{units}}\\ \end{align}
[2]

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