*Of course you want to ensure you have a solid understanding of solving quadratic equations before watching this lesson.*

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.

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.

What is the maximum height the ball reaches, and how far (horizontally) from Audrey does is the ball at its maximum height?

How far does the ball travel before it hits the ground?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.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...Baring errors does mine match your expected result?Quadratic applications are very helpful in solving several types of word problems (other than the bouquet throwing problem), especially where optimization is involved.Solution: Note that in this problem, the \(x\)-axis is measuring the horizontal distance of the path of the ball, not the time, so when we draw the parabola, it’s a true indication of the trajectory or path of the ball.Note also that the equation given is in vertex form (if we add Since the quadratic is already in vertex form (\(y=a k\), where \((h,k)\) is the vertex), we can see that the vertex from \(0=-0.018 8\) is \((20,8)\).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.Based on this can you find a discrepancy between your approach and mine?In addition, the students’ written responses and interview data were qualitatively analyzed to determine the nature of the students’ difficulties in formulating and solving quadratic equations.The findings revealed that although students have difficulties in solving both symbolic quadratic equations and quadratic word problems, they performed better in the context of symbolic equations compared with word problems.

## Comments Solving Word Problems Using Quadratic Equations

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