Question

The perimeter of a triangle is $15{{x}^{2}}-23x+9$ and two of its sides are $5{{x}^{2}}+8x-1$ and $6{{x}^{2}}-9x+4$. Find the third side.

Answer 1

Hint:
In this problem we have the perimeter of the triangle in quadratic equation and also the two sides of the triangle in the quadratic form. We assume the third side of the triangle as a quadratic equation. Now we will obtain the perimeter of the triangle having three sides and equate it to the given perimeter to find the third side of the triangle.

Formula Used:
We will use the perimeter of the triangle having sides $a$, $b$, $c$ which is given by $p=a+b+c$.

Complete Step by Step Solution:
Given that,
First side of the triangle is $5{{x}^{2}}+8x-1$.
Second side of the triangle is $6{{x}^{2}}-9x+4$.
Perimeter of the triangle is $15{{x}^{2}}-23x+9$.
Let the third side of the triangle is $a{{x}^{2}}+bx+c$.
Now the perimeter of the triangle having sides $5{{x}^{2}}+8x-1$, $6{{x}^{2}}-9x+4$, $a{{x}^{2}}+bx+c$ is given by
$p=\left( 5{{x}^{2}}+8x-1 \right)+\left( 6{{x}^{2}}-9x+4 \right)+\left( a{{x}^{2}}+bx+c \right)$
Using distribution law of multiplication in the above equation, then we will get
$p=5{{x}^{2}}+8x-1+6{{x}^{2}}-9x+4+a{{x}^{2}}+bx+c$
Rearranging the terms in the above equation them we will get
$p=5{{x}^{2}}+6{{x}^{2}}+a{{x}^{2}}+8x-9x+bx-1+4+c$
Taking ${{x}^{2}}$ common from the terms $5{{x}^{2}}+6{{x}^{2}}+a{{x}^{2}}$ and $x$ from $8x-9x+bx$ in the above equation, then we will get
$\begin{align}
  & p=\left( 5+6+a \right){{x}^{2}}+\left( 8-9+b \right)x-1+4+c \\
 & \Rightarrow p=\left( a+11 \right){{x}^{2}}+\left( b-1 \right)x+\left( c+3 \right) \\
\end{align}$
But in the problem we have the perimeter of the triangle as $15{{x}^{2}}-23x+9$.
$\therefore \left( a+11 \right){{x}^{2}}+\left( b-1 \right)x+\left( c+3 \right)=15{{x}^{2}}-23x+9\cdot \cdot \cdot \cdot \left( \text{i} \right)$
Equating the coefficient of ${{x}^{2}}$ term in the above equation, then we will get
$a+11=15$
Subtracting $11$ from the both sides of the above equation, then we will get
$\begin{align}
  & a+11-11=15-11 \\
 & \Rightarrow a=4 \\
\end{align}$
Equating the coefficient of $x$ terms in equation $\left( \text{i} \right)$, then we will get
$b-1=-23$
Adding one on both sides of the above equation, then we will get
$\begin{align}
  & b-1+1=-23+1 \\
 & \Rightarrow b=-22 \\
\end{align}$
Equating constants in the equation $\left( \text{i} \right)$, then we will get
$c+3=9$
Subtracting $3$ from the both sides of the above equation, then we will get
$\begin{align}
  & c+3-3=9-3 \\
 & \Rightarrow c=6 \\
\end{align}$
We have the values $a=4$, $b=-22$, $c=6$, then the third side of the triangle is

$a{{x}^{2}}+bx+c=4{{x}^{2}}-22x+6$.

Note:
We can directly find the third side from the perimeter formula
$\begin{align}
  & p=a+b+c \\
 & \Rightarrow c=p-\left( a+b \right) \\
\end{align}$
Substituting the given values, then we will get
$\begin{align}
  & c=15{{x}^{2}}-23x+9-\left( 5{{x}^{2}}+8x-1+6{{x}^{2}}-9x+4 \right) \\
 & \Rightarrow c=15{{x}^{2}}-23x+9-\left( 11{{x}^{2}}-x+3 \right) \\
 & \Rightarrow c=15{{x}^{2}}-11{{x}^{2}}-23x+x+9-3 \\
 & \Rightarrow c=4{{x}^{2}}-22x+6 \\
\end{align}$
From both the methods we got the same answer.