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Chapter 8: Problem 75

For each equation under the given condition, (a) find \(k\) and (b) find the other solution. Suppose that \(f(x)=a x^{2}+b x+c,\) with \(f(-3)=0\) \(f\left(\frac{1}{2}\right)=0,\) and \(f(0)=-12 .\) Find \(a, b,\) and \(c\)

Short Answer

Expert verified
a = 8, b = 20, c = -12

Step by step solution

01

- Set up the equations using given conditions

We know that the quadratic function is given by: \[f(x) = ax^2 + bx + c \] We are given three conditions: 1. \(f(-3)=0\) 2. \(f\left( \frac{1}{2} \right)=0\) 3. \(f(0)=-12\) Using these, we can set up the following system of equations: \[a(-3)^2 + b(-3) + c = 0\] \[a\left( \frac{1}{2} \right)^2 + b\left( \frac{1}{2} \right) + c = 0\] \[c = -12\]
02

- Substitute c in the first equation

Using \(c = -12\) in the first equation: \[a(9) - 3b - 12 = 0\] Simplifies to: \[9a - 3b = 12\] Divide by 3: \[3a - b = 4\]
03

- Substitute c in the second equation

Using \(c = -12\) in the second equation: \[a\left( \frac{1}{4} \right) + \frac{b}{2} - 12 = 0\] Multiply everything by 4 to clear the fraction: \[a + 2b - 48 = 0\] Rearranges to: \[a + 2b = 48\]
04

- Solve the system of linear equations

Now we have two simultaneous equations: \[3a - b = 4\] \[a + 2b = 48\] From the first equation, express \(b\) in terms of \(a\): \[b = 3a - 4\] Substitute this into the second equation: \[a + 2(3a - 4) = 48\] Simplify and solve for \(a\): \[a + 6a - 8 = 48\] \[7a - 8 = 48\] \[7a = 56\] \[a = 8\]
05

- Find b using the value of a

Substitute \(a = 8\) back into \(b = 3a - 4\): \[b = 3(8) - 4\] \[b = 24 - 4\] \[b = 20\]
06

- Verify values and write the final solution

We have found: \[a = 8\] \[b = 20\] \[c = -12\] Verify by substituting back into the original equations to ensure they hold true.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

quadratic function
A quadratic function is a type of polynomial function with the form:
  • %5C%5Bf(x) %3D ax%5E2 %2B bx %2B c%5C%5D
      • Here, a, b, and c are coefficients where a eq 0 short
    . The ax^2 term is what makes it quadratic.
    In this problem, we are given
    • three conditions: f(-3)=0, f(1/2)=0, and f(0)=-12.

      • We'll use these to create a system of equations. Our goal is to find a, b, and c once we have set up our system.
system of linear equations
To solve for the coefficients, we must use the conditions given in the problem to create linear equations.
  • First equation using f(-3)=0:
    • %5C%5B9a - 3b %2B c %3D 0%5C%5D
  • Second equation using f(1/2)=0:
    • %5C%5Ba%5Cleft(%5Cfrac%7B1%7D%7B4%7D%5Cright%29 %2B %5Cfrac%7Bb%7D%7B2%7D %2B c %3D 0%5C%5D
  • Third equation using f(0)=-12:
    • %5C%5Bc %3D -12%5C%5D

    Next, we substitute c=-12 into the first and second equations and simplify:
    • From f(-3)=0: %5C%5B9a - 3b %3D 12%5C%5D
    • From f(1/2)=0: %5C%5Ba %2B 2b %3D 48%5C%5D
This gives us a simplified system to solve.
solving equations step-by-step
Now, let's solve the system of linear equations step-by-step.
  • First, we have the equations:
    • %5C%5B3a - b %3D 4%5C%5D
    • %5C%5Ba %2B 2b %3D 48%5C%5D
    Next, we solve for one variable in terms of the other. From %5C%5B3a - b %3D 4%5C%5DIseolate b and find %5C%5Bb %3D 3a - 4%5C%5D.
    • Substitute this into the second equation:
    • %5C%5Ba %2B 2(3a - 4) %3D 48%5C%5D

        • We get %5C%5Ba %2B 6a - 8 %3D 48%5C%5D. Simplify to find a:
      • %5C%5B7a - 8 %3D 48%5C%5D
      • %5C%5B7a %3D 56%5C%5D
      • %5C%5Ba %3D 8%5C%5D
        • .
        Now substitute a = 8 back into b = 3a - 4: %5C%5Bb %3D 20%5C%5D.
      Finally, we check the solution by verifying:
      • %5Ca %3D 8 %5C, b %3D 20 %5C, c %3D -12%5C5B
        • tiThe original equations, these hold true.

      Following these steps ensures we have a complete solution for the coefficients of the quadratic function.

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