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Chapter 5: Problem 57

Solve. Let \(g(x)=2 x^{2}+5 x .\) Find \(a\) such that \(g(a)=12\)

Short Answer

Expert verified
a = \frac{3}{2} or a = -4

Step by step solution

01

Substitute the function value

Replace the function value in the equation with the given value. In this case, we want to find the value of \(a\) such that \(g(a) = 12\). Start with the function \(g(x) = 2x^2 + 5x\). The equation becomes: \(2a^2 + 5a = 12\)
02

Rearrange the equation

Rearrange the equation to set it equal to zero by moving 12 to the left side: \(2a^2 + 5a - 12 = 0\)
03

Factor the quadratic equation

Factor the quadratic equation. In this case, factor \(2a^2 + 5a - 12\) into \( (2a - 3)(a + 4) = 0 \)
04

Solve for the roots

Set each factor equal to zero and solve for \(a\). \(2a - 3 = 0 \implies a = \frac{3}{2}\) and \(a + 4 = 0 \implies a = -4\)

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Key Concepts

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

function substitution
To solve the given equation, we start by substituting the function value into the equation. This process is known as function substitution.
It involves taking the given function, which in this case is \(g(x) = 2x^2 + 5x\), and replacing \(g(a)\) with the value we want to find, \(g(a) = 12\).
Hence, we rewrite it as: 2a^2 + 5a = 12.
This substitution helps to form a clear quadratic equation to work with.
rearranging equations
Next, we need to rearrange the equation to standard form, which helps in solving for the variable.
We do this by moving all terms to one side of the equation to set it equal to zero.
In our specific problem, we start with \(2a^2 + 5a - 12 = 0\).
This form is much easier to handle for further steps like factoring or using the quadratic formula.
factoring quadratics
Factoring quadratics is a technique to solve equations of the form \(ax^2 + bx + c = 0\) by expressing them as a product of binomials.
For the equation \(2a^2 + 5a - 12 = 0\), we factorize as follows:
Find two numbers that multiply to \(2a^2 \times (-12) = -24a^2\) and add to \(5a\). These numbers are \(8a\) and \(-3a\).
So we write: \(2a^2 + 8a - 3a - 12 = 0\)
Group the terms: \(2a(a+4) - 3(a+4) = 0\)
Factor out \(a + 4\), giving us: \((2a - 3)(a + 4) = 0\)
Now, our quadratic equation is factored, simplifying the finding of roots.
finding roots
Finally, to solve for the variable \(a\), we set each factor equal to zero. These factors are \((2a - 3)\) and \((a + 4)\).
Solving \(2a - 3 = 0\) gives: \(a = \frac{3}{2}\)
Solving \(a + 4 = 0\) gives: \(a = -4\).

Therefore, the roots of the original quadratic equation are \(a = \frac{3}{2}\) and \(a = -4\).
These are the values of \(a\) that make \(g(a) = 12\), as required.

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Most popular questions from this chapter

For \(P(x)\) and \(Q(x)\) as given, find the following. $$ \begin{aligned} &P(x)=13 x^{5}-22 x^{4}-36 x^{3}+40 x^{2}-16 x+75\\\ &Q(x)=42 x^{5}-37 x^{4}+50 x^{3}-28 x^{2}+34 x+100 \end{aligned} $$ $$ 2[Q(x)]-3[P(x)] $$

Factor completely. $$ -10 t^{3}+15 t $$

Find the domain of the function \(f\) given by each of the following. $$f(x)=\frac{3}{x^{2}-3 x-4}$$

Factor completely. $$ (1-x)^{3}-(x-1)^{6} $$

Can the number of solutions of a quadratic equation exceed two? Why or why not?
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