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Chapter 2: Problem 11

Solve the quadratic equation by factoring. $$x^{2}-7 x+12=0$$

Short Answer

Expert verified
The solutions for the equation are \(x = 4\) and \(x = 3\).

Step by step solution

01

Factoring the quadratic equation

We want to factor the equation \(x^2 - 7x + 12 = 0\). We need to find two numbers that multiply to \(+12\) (the constant term) and add to \(-7\) (the coefficient of x). These numbers are -4 and -3. So the factored form would be \((x-4)(x-3) = 0\).
02

Setting each factor equal to zero

Now that we have factored the equation, we set each factor equal to zero and solve for x, i.e., \(x-4 = 0\) and \(x-3 = 0\).
03

Solving for x

Solving these two equations, we get: \(x = 4\) from \(x-4 = 0\), and \(x = 3\) from \(x-3 = 0\). So the solutions to the original quadratic equation are \(x = 4\) and \(x = 3\).

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Most popular questions from this chapter

This set of exercises will draw on the ideas presented in this section and your general math background. How many zeros, real and nonreal, does the function \(f(x)=x^{4}-1\) have? How many \(x\) -intercepts does the graph of \(f\) have?

Graph each quadratic function by finding a suitable viewing window with the help of the TABLE feature of a graphing utility. Also find the vertex of the associated parabola using the graphing utility. $$y_{1}(x)=0.4 x^{2}+20$$

In Exercises \(101-104,\) let \(f(t)=3 t+1\) and \(g(x)=x^{2}+4\). Evaluate \((g \circ g)\left(\frac{1}{2}\right)\)

This set of exercises will draw on the ideas presented in this section and your general math background. Why must we have \(a \neq 0\) in the definition of a quadratic function?

Can you write down an expression for a quadratic function whose \(x\) -intercepts are given by (2,0) and (3,0)\(?\) Is there more than one possible answer? Explain.
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