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(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]

Try it!

(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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2.7 | Solving Quadratics

Solving through factored form

Factored form has been explained in the previous section, as well as how to achieve it. Therefore, this is a valid method of solving for quadratic roots, given “easy” numbers (integers).

Solving through vertex form

Factored form has also been explained to find the vertex. However, this form can also be used to find the roots of the function. Taking the previous equation as an example, the roots can be determined as follows:

\begin{align}
f(x) = 2(x + 2)^2 - 13 &= 0\\
2(x + 2)^2 &= 13\\
(x + 2)^2 &= {13 \over 2}\\
x + 2 &= ± \sqrt{{13 \over 2}}\\
x &= -2 ± \sqrt{{13 \over 2}}\\
\end{align}

This way, you can also find the values of \(x\) i.e. the roots of the function, even if the numbers are non-integer values.

Solving through quadratic formula, and the discriminant

Given a quadratic in standard form \(ax^2 + bx + c\), simply using a formula based on the coefficients \(a, b, \text{and} c\) suffices:

\[x = {-b ± \sqrt{b^2 - 4ac}\over 2a}\]

An important tip regarding the quadratic formula: the terms inside of the square root, namely \(b^2 - 4ac\) is called the discriminant, and is denoted by the symbol delta \(\Delta\). This should not be confused with delta referring to “change in”, as this is denoted by a \(d\) in the calculus section (topic 5) of the curriculum.

The discriminant \(\Delta\) indicates the number of roots that a quadratic function has. Since quadratics have a degree of 2 \(x^2\), the maximum number of roots a quadratic can have is 2, but what if it has only 1 root, or even 0? This can be cleared up by the discriminant.

If the discriminant is positive (\(\Delta > 0\)), then the quadratic has 2 roots.

If the discriminant is equal to 0 (\(\Delta = 0\)), then the quadratic has 1 root.

If the discriminant is negative (\(\Delta < 0\)), then the quadratic has 0 roots (aka complex roots for HL students).

For example, taking the same function \(f(x) = 2x^2 + 8x - 5\) again, we can determine how many roots it has, without wasting precious exam time actually calculating the roots, using the discriminant:

\begin{align}
\Delta &= b^2 - 4ac\\
\Delta &= (8)^2 - 4(2)(-5)\\
\Delta &= 64 - (-40)\\
\Delta &= 64 + 40\\
\Delta &> 0\\
\end{align}

Hence, since \(\Delta > 0\), the function \(f(x)\) has 2 roots.

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