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Chapter 5: Problem 20

Solve each equation. $$ 2 t^{2}+8 t=0 $$

Short Answer

Expert verified
The solutions are t = 0 and t = -4.

Step by step solution

01

Factor out the common factor

The given equation is 2t^2 + 8t = 0. First, factor out the common factor, which is 2t, from each term in the equation. This gives 2t(t + 4) = 0.
02

Set each factor equal to 0

The equation is now factored into 2t(t + 4) = 0. Set each factor equal to zero: 2t = 0 and t + 4 = 0.
03

Solve each equation

Now solve each of these equations separately. For 2t = 0, divide both sides by 2 to get t = 0. For t + 4 = 0, subtract 4 from both sides to get t = -4.

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Key Concepts

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

factoring quadratics
Factoring quadratics is like breaking a number into its prime factors. When you factor a quadratic equation, you're trying to express it as a product of simpler expressions. For the quadratic equation 2t^2 + 8t = 0, you can see that there's a common term in both parts: 2t. So, you factor it out. When you do this, the equation becomes 2t(t + 4) = 0. Now, instead of solving a quadratic equation directly, we have a simpler equation to work with. This method makes it much easier to find the values of t that make the equation true.
zero product property
The zero product property is a fundamental principle in algebra. It states that if the product of two numbers is zero, then at least one of the numbers must be zero. This principle is crucial when solving quadratics by factoring. In our factored equation, 2t(t + 4) = 0, it means either 2t = 0 or t + 4 = 0. Because if either factor is zero, the entire product is zero. So, to solve the equation, you set each factor to zero and solve for t. This gives you two simpler equations to work with.
solving linear equations
Solving linear equations is a key skill in algebra. Once you have factored the quadratic equation and applied the zero product property, you are left with linear equations to solve. For 2t = 0, you simply divide both sides by 2 to find t = 0. For t + 4 = 0, you subtract 4 from both sides to find t = -4. Each step involves straightforward operations that isolate the variable t and provide the solutions to the original quadratic equation. This method breaks down a seemingly complex problem into easier, manageable steps.

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Most popular questions from this chapter

Factor each trinomial. \(m^{2}(m-p)^{2}-m p(m-p)^{2}-12 p^{2}(m-p)^{2}\)

Skills In some cases, the method of factoring by grouping can be combined with the methods of special factoring discussed in this section. Consider this example. $$ \begin{array}{l} 8 x^{3}+4 x^{2}+27 y^{3}-9 y^{2} \\ \quad=\left(8 x^{3}+27 y^{3}\right)+\left(4 x^{2}-9 y^{2}\right) \\ \quad=(2 x+3 y)\left(4 x^{2}-6 x y+9 y^{2}\right)+(2 x+3 y)(2 x-3 y) \\ \quad=(2 x+3 y)\left[\left(4 x^{2}-6 x y+9 y^{2}\right)+(2 x-3 y)\right] \\ \quad=(2 x+3 y)\left(4 x^{2}-6 x y+9 y^{2}+2 x-3 y\right) \end{array} $$ Use the method just described to factor each polynomial. $$ 125 p^{3}+25 p^{2}+8 q^{3}-4 q^{2} $$

Solve each equation. $$ (x+8)(x-2)=-21 $$

Solve each equation. $$ (2 x+1)(x-3)=6 x+3 $$

Factor each trinomial. \(10(k+1)^{2}-7(k+1)+1\)
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