Find the solution to the system of equations. Round to the nearest tenth if necessary.
[tex]
\begin{array}{l}
f(x)=3 x+6 \\
g(x)=x^2+4 x+1
\end{array}
[/tex]
Answer (1)
Set the two functions equal to each other: 3 x + 6 = x 2 + 4 x + 1 .
Rearrange the equation into standard quadratic form: x 2 + x − 5 = 0 .
Apply the quadratic formula: x = 2 a − b ± b 2 − 4 a c .
Calculate the two solutions and round to the nearest tenth: x ≈ 1.8 and x ≈ − 2.8 .
Explanation
Understanding the Problem We are given two functions, f ( x ) = 3 x + 6 and g ( x ) = x 2 + 4 x + 1 , and we need to find the solutions to the equation f ( x ) = g ( x ) . This means we need to find the values of x for which the two functions have the same value.
Setting the Functions Equal To find the solutions, we set the two functions equal to each other:
3 x + 6 = x 2 + 4 x + 1
Now, we rearrange the equation to get a quadratic equation in the standard form a x 2 + b x + c = 0 .
Rearranging the Equation Subtract 3 x and 6 from both sides of the equation:
0 = x 2 + 4 x + 1 − 3 x − 6
Simplify the equation:
0 = x 2 + x − 5
Applying the Quadratic Formula Now we have a quadratic equation x 2 + x − 5 = 0 . We can use the quadratic formula to solve for x . The quadratic formula is given by:
x = 2 a − b ± b 2 − 4 a c
In our equation, a = 1 , b = 1 , and c = − 5 .
Simplifying the Expression Substitute the values of a , b , and c into the quadratic formula:
x = 2 ( 1 ) − 1 ± 1 2 − 4 ( 1 ) ( − 5 )
Simplify the expression:
x = 2 − 1 ± 1 + 20 = 2 − 1 ± 21
Calculating the Roots Now we have two possible values for x :
x 1 = 2 − 1 + 21 and x 2 = 2 − 1 − 21
We need to approximate these values to the nearest tenth.
Approximating the Solutions Using a calculator, we find the approximate values:
x 1 = 2 − 1 + 21 ≈ 2 − 1 + 4.58 ≈ 2 3.58 ≈ 1.79 ≈ 1.8
x 2 = 2 − 1 − 21 ≈ 2 − 1 − 4.58 ≈ 2 − 5.58 ≈ − 2.79 ≈ − 2.8
Final Answer Therefore, the solutions to the system of equations, rounded to the nearest tenth, are x ≈ 1.8 and x ≈ − 2.8 .
Examples
Understanding systems of equations is crucial in many real-world applications. For instance, in business, you might have one function representing the cost of producing items and another representing the revenue generated from selling those items. Finding the points where these functions intersect (solving the system of equations) helps determine the break-even points, where costs equal revenue. This analysis is vital for making informed decisions about production levels and pricing strategies. Similarly, in physics, you might use systems of equations to model the motion of objects under different forces, finding when and where their paths intersect.
