Using Quadratic Equations To Solve Word Problems

Student difficulties in solving symbolic problems were mainly associated with arithmetic and algebraic manipulation errors.In the word problems, however, students had difficulties comprehending the context and were therefore unable to formulate the equation to be solved.The amount of effort you invest in practicing solving word problems will be directional proportional to your mastery of them.

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A second method of solving quadratic equations involves the use of the following formula: , b is the numeral that goes in front of x, and c is the numeral with no variable next to it (a.k.a., “the constant”).

When using the quadratic formula, you should be aware of three possibilities.

Author/s: Makbule Gozde Didis, Ayhan Kursat Erbas DOI: 10.12738/estp.2015.4.2743 Year: 2015 Vol: 15 Number: 4 Abstract This study attempts to investigate the performance of tenth-grade students in solving quadratic equations with one unknown, using symbolic equation and word-problem representations.

The participants were 217 tenth-grade students, from three different public high schools.

Based on this can you find a discrepancy between your approach and mine?

Baring errors does mine match your expected result?The second fact is that it takes the second pedestrian one more minute than the first to cover 1 km, so you have $$\frac1 1=\frac1.$$ Solve the two equations for $v_1$ and $v_2$ and then compute

Baring errors does mine match your expected result?

The second fact is that it takes the second pedestrian one more minute than the first to cover 1 km, so you have $$\frac1 1=\frac1.$$ Solve the two equations for $v_1$ and $v_2$ and then compute $1/v_1$ and $1/v_2=1/v_1 1$, or substitute $t_1=1/v_1$ and $t_2=1/v_2$ into the two equations and solve for the times directly.

There will be two solutions to this system of equations, but one of them doesn’t make physical sense for this problem, so that one will be rejected. So I will reproduce what you have with slightly different notation.

Many quadratic equations cannot be solved by factoring.

This is generally true when the roots, or answers, are not rational numbers.

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Baring errors does mine match your expected result?The second fact is that it takes the second pedestrian one more minute than the first to cover 1 km, so you have $$\frac1 1=\frac1.$$ Solve the two equations for $v_1$ and $v_2$ and then compute $1/v_1$ and $1/v_2=1/v_1 1$, or substitute $t_1=1/v_1$ and $t_2=1/v_2$ into the two equations and solve for the times directly.There will be two solutions to this system of equations, but one of them doesn’t make physical sense for this problem, so that one will be rejected. So I will reproduce what you have with slightly different notation.Many quadratic equations cannot be solved by factoring.This is generally true when the roots, or answers, are not rational numbers.I am answering because I do not have comment ability yet. Here is the problem, are you trying to get time or rate? d1 d2 = 76km d1 = v1*T d2 = v2*T T = 6h = 360min v1 = 1km/t v2 = 1km/(t-1min) Putting together...1/t 1/(t-1min) = 76/360 This leads to 19*t^2 - 180*t 71 = 0 (steps suppressed) The answer is ~9.06 min, the other is less than half a min.Of course you want to ensure you have a solid understanding of solving quadratic equations before watching this lesson.Like all word problems in math, there is no one single procedure you can use to solve a problem.Data was collected through an open-ended questionnaire comprising eight symbolic equations and four word problems; furthermore, semi-structured interviews were conducted with sixteen of the students.In the data analysis, the percentage of the students’ correct, incorrect, blank, and incomplete responses was determined to obtain an overview of student performance in solving symbolic equations and word problems.

/v_1$ and

Baring errors does mine match your expected result?

The second fact is that it takes the second pedestrian one more minute than the first to cover 1 km, so you have $$\frac1 1=\frac1.$$ Solve the two equations for $v_1$ and $v_2$ and then compute $1/v_1$ and $1/v_2=1/v_1 1$, or substitute $t_1=1/v_1$ and $t_2=1/v_2$ into the two equations and solve for the times directly.

There will be two solutions to this system of equations, but one of them doesn’t make physical sense for this problem, so that one will be rejected. So I will reproduce what you have with slightly different notation.

Many quadratic equations cannot be solved by factoring.

This is generally true when the roots, or answers, are not rational numbers.

||

Baring errors does mine match your expected result?The second fact is that it takes the second pedestrian one more minute than the first to cover 1 km, so you have $$\frac1 1=\frac1.$$ Solve the two equations for $v_1$ and $v_2$ and then compute $1/v_1$ and $1/v_2=1/v_1 1$, or substitute $t_1=1/v_1$ and $t_2=1/v_2$ into the two equations and solve for the times directly.There will be two solutions to this system of equations, but one of them doesn’t make physical sense for this problem, so that one will be rejected. So I will reproduce what you have with slightly different notation.Many quadratic equations cannot be solved by factoring.This is generally true when the roots, or answers, are not rational numbers.I am answering because I do not have comment ability yet. Here is the problem, are you trying to get time or rate? d1 d2 = 76km d1 = v1*T d2 = v2*T T = 6h = 360min v1 = 1km/t v2 = 1km/(t-1min) Putting together...1/t 1/(t-1min) = 76/360 This leads to 19*t^2 - 180*t 71 = 0 (steps suppressed) The answer is ~9.06 min, the other is less than half a min.Of course you want to ensure you have a solid understanding of solving quadratic equations before watching this lesson.Like all word problems in math, there is no one single procedure you can use to solve a problem.Data was collected through an open-ended questionnaire comprising eight symbolic equations and four word problems; furthermore, semi-structured interviews were conducted with sixteen of the students.In the data analysis, the percentage of the students’ correct, incorrect, blank, and incomplete responses was determined to obtain an overview of student performance in solving symbolic equations and word problems.

/v_2=1/v_1 1$, or substitute $t_1=1/v_1$ and $t_2=1/v_2$ into the two equations and solve for the times directly.There will be two solutions to this system of equations, but one of them doesn’t make physical sense for this problem, so that one will be rejected. So I will reproduce what you have with slightly different notation.Many quadratic equations cannot be solved by factoring.This is generally true when the roots, or answers, are not rational numbers.I am answering because I do not have comment ability yet. Here is the problem, are you trying to get time or rate? d1 d2 = 76km d1 = v1*T d2 = v2*T T = 6h = 360min v1 = 1km/t v2 = 1km/(t-1min) Putting together...1/t 1/(t-1min) = 76/360 This leads to 19*t^2 - 180*t 71 = 0 (steps suppressed) The answer is ~9.06 min, the other is less than half a min.Of course you want to ensure you have a solid understanding of solving quadratic equations before watching this lesson.Like all word problems in math, there is no one single procedure you can use to solve a problem.Data was collected through an open-ended questionnaire comprising eight symbolic equations and four word problems; furthermore, semi-structured interviews were conducted with sixteen of the students.In the data analysis, the percentage of the students’ correct, incorrect, blank, and incomplete responses was determined to obtain an overview of student performance in solving symbolic equations and word problems.

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