Example 1
Completion requirements
| Example 1 |
Solve the given system algebraically. Verify graphically.
\(\left\{ \begin{array}{l}
y = 4x^2 + 32x + 63 \\
y = 4\left( {x + 4} \right)^2 - 1
\end{array} \right.\)
The equation \(4x^2 + 32x + 63 = 4x^2 + 32x + 63\) can be further simplified to
\(\begin{align}
32x + 63 &= 32x + 63 \\
63 &= 63 \\
0 &= 0 \end{align}\)
Any other equations where the left side is identical to the right side are examples of true statements, showing the system has an infinite number of solutions.
Graphical verification:
\(\left\{ \begin{array}{l}
y = 4x^2 + 32x + 63 \\
y = 4\left( {x + 4} \right)^2 - 1
\end{array} \right.\)
The \(y\) is isolated in the first equation, so substitute its equivalent into the second equation to solve by substitution.
\(\begin{align}
y &= 4\left( {x + 4} \right)^2 - 1 \\
4x^2 + 32x + 63 &= 4\left( {x + 4} \right)^2 - 1 \\
4x^2 + 32x + 63 &= 4\left( {x^2 + 8x + 16} \right) - 1 \\
4x^2 + 32x + 63 &= 4x^2 + 32x + 64 - 1 \\
4x^2 + 32x + 63 &= 4x^2 + 32x + 63 \\
\end{align}\)
The two sides of the equation are identical, so this is a true statement. This system has an infinite number of real solutions.
\(\begin{align}
y &= 4\left( {x + 4} \right)^2 - 1 \\
4x^2 + 32x + 63 &= 4\left( {x + 4} \right)^2 - 1 \\
4x^2 + 32x + 63 &= 4\left( {x^2 + 8x + 16} \right) - 1 \\
4x^2 + 32x + 63 &= 4x^2 + 32x + 64 - 1 \\
4x^2 + 32x + 63 &= 4x^2 + 32x + 63 \\
\end{align}\)
The two sides of the equation are identical, so this is a true statement. This system has an infinite number of real solutions.
The equation \(4x^2 + 32x + 63 = 4x^2 + 32x + 63\) can be further simplified to
\(\begin{align}
32x + 63 &= 32x + 63 \\
63 &= 63 \\
0 &= 0 \end{align}\)
Any other equations where the left side is identical to the right side are examples of true statements, showing the system has an infinite number of solutions.
Graphical verification: