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Chapter 1: Problem 8

Solve each equation in Exercises \(1-14\) by factoring. $$4 x^{2}-13 x=-3$$

Short Answer

Expert verified
The solutions to the equation are \(x = 1\) and \(x = \frac{3}{4}\).

Step by step solution

01

Arrange in Standard Form

To arrange the equation in standard form, we add \(3\) to both sides to set the equation to zero. So we will have: \(4x^2 - 13x + 3 = 0\).
02

Factorize the Equation

The equation is now in the form \(ax^2+bx+c = 0\) and can be factorized. We are looking for two numbers that multiply to \(4 \cdot 3 = 12\) and add up to \(-13\). These numbers are \(-3\) and \(-4\). So the factored form of the equation will be: \(4(x - 1)(x - \frac{3}{4}) = 0\).
03

Solve for x

Having factored the equation, we set each factor equal to zero and solve for \(x\). From \(x-1 = 0\), we get \(x = 1\). From \(x - \frac{3}{4} = 0\), we get \(x = \frac{3}{4}\). Therefore, the solutions to the equation are \(x = 1\) and \(x = \frac{3}{4}\).

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Most popular questions from this chapter

In Exercises \(29-44,\) perform the indicated operations and write the result in standard form. $$\frac{-12+\sqrt{-28}}{32}$$

Solve each quadratic inequality in Exercises \(1-28\) and graph the solution set on a real number line. Express each solution set in interval notation. $$ x^{2}+2 x<0 $$

In Exercises \(1-8,\) add or subtract as indicated and write the result in standard form. $$(-2+6 i)+(4-i)$$

Solve each rational inequality in Exercises \(29-48,\) and graph the solution set on a real number line. Express each solution set in interval notation. $$ \frac{3 x+5}{6-2 x} \geq 0 $$

In Exercises \(29-44,\) perform the indicated operations and write the result in standard form. $$\sqrt{-8}(\sqrt{-3}-\sqrt{5})$$
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