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Chapter 0: Problem 80

Solve each quadratic equation using the quadratic formula. $$5 x^{2}+x-2=0$$

Short Answer

Expert verified
The solutions for the equation are \(x_1 = (sqrt(41) - 1) / 10\) and \(x_2 = -(sqrt(41) + 1) / 10\).

Step by step solution

01

Identify a, b, and c

From the given quadratic equation, identify the values of a, b, and c is the first thing that needs to be done. In this case, from the equation \(5x^2 + x - 2 = 0\), it can be noticed that \(a = 5\), \(b = 1\), and \(c = -2\).
02

Substitute a, b and c into the Quadratic formula

Plug the values of a, b, and c into the quadratic formula \(x = [-b 卤 sqrt(b^2 - 4ac)] / (2a)\). Plugging in our values gives: \(x = [-1 卤 sqrt((1)^2 - 4*5*(-2))] / (2*5)\).
03

Simplify the expression under the square root

The next step involves simplification. Start by calculating the expression under the square root: \(1^2 - 4*5*(-2) = 1 + 40 = 41\). This gives: \(x = [-1 卤 sqrt(41)] / 10\).
04

Separate into two different equations

Next, split the equation into two to account for the positive and negative square root: \(x = [-1 + sqrt(41)] / 10\) and \(x = [-1 - sqrt(41)] / 10\).
05

Simplify the answer

Lastly, simplify the two fractions to get the two possible solutions for x. This gives: \(x_1 = (sqrt(41) - 1) / 10\) and \(x_2 = -(sqrt(41) + 1) / 10\).

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Most popular questions from this chapter

In more U.S. marriages, spouses have different faiths. The bar graph shows the percentage of households with an interfaith marriage in 1988 and \(2012 .\) Also shown is the percentage of households in which a person of faith is married to someone with no religion. The formula $$ I=\frac{1}{4} x+26 $$ models the percentage of U.S. households with an interfaith marriage, \(I, x\) years after \(1988 .\) The formula $$ N=\frac{1}{4} x+6 $$ models the percentage of U.S households in which a person of faith is married to someone with no religion, \(N, x\) years after \(1988 .\) Use these models to solve Exercises \(107-108\). The formula for converting Celsius temperature, \(C,\) to Fahrenheit temperature, \(F\), is $$ F=\frac{9}{5} C+32 $$ If Fahrenheit temperature ranges from \(41^{\circ}\) to \(50^{\circ},\) inclusive, what is the range for Celsius temperature? Use interval notation to express this range.

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Determine whether each statement makes sense or does not make sense, and explain your reasoning. I subtracted \(\frac{3 x-5}{x-1}\) from \(\frac{x-3}{x-1}\) and obtained a constant.

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What is the discriminant and what information does it provide about a quadratic equation?
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