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06 Feb 2020, 12:34
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Well I came across a new way to solve quadratic equations. It works on equations which are hard to factor. I read about it on New York Times. in this parabola:
y = x2 – 4x – 5
The two solutions when y = 0 are the symmetrical points r and s, where the parabola crosses the x-axis.
The midpoint, or average, of r and s is the axis of symmetry of the parabola. We want r + s = –b, which happens when the average of r and s is –b ÷ 2. In this example: 4 ÷ 2 = 2.
The two solutions to the quadratic equation will be the axis of symmetry plus or minus an unknown amount, which we’ll call u. In this example:
r = 2 – u and s = 2 + u
To find u, we want the product of r and s to be equal to c, which in this example is –5. Rewriting r and s in terms of u:
r × s = –5
(2 – u) × (2 + u) = –5
Solving that yields 22 – u2 = –5 or u2 = 9, so u = 3 works.
The two solutions to this quadratic equation are 2 – u and 2 + u, or –1 and 5. In other words, this parabola intersects the x-axis when x = –1 and x = 5.
y = x2 – 4x – 5
The two solutions when y = 0 are the symmetrical points r and s, where the parabola crosses the x-axis.
The midpoint, or average, of r and s is the axis of symmetry of the parabola. We want r + s = –b, which happens when the average of r and s is –b ÷ 2. In this example: 4 ÷ 2 = 2.
The two solutions to the quadratic equation will be the axis of symmetry plus or minus an unknown amount, which we’ll call u. In this example:
r = 2 – u and s = 2 + u
To find u, we want the product of r and s to be equal to c, which in this example is –5. Rewriting r and s in terms of u:
r × s = –5
(2 – u) × (2 + u) = –5
Solving that yields 22 – u2 = –5 or u2 = 9, so u = 3 works.
The two solutions to this quadratic equation are 2 – u and 2 + u, or –1 and 5. In other words, this parabola intersects the x-axis when x = –1 and x = 5.
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06 Feb 2020, 12:45
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On Second read i notice u^2 =9 so u=3 or -3 as GMAT sees it, but know that u here is the distance so u got to have a positive root.
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Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
