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The Revised GRE– What’s New on the Math Front?

Concept-wise, I would say the math section is business as usual, save for two things: coordinate geometry and the quadratic equation.

While parabolas are tested on the current GRE, they are oddball questions, geometry outliers in the usually slew of right triangle and circle problems. In the revised GRE book, an entire section is devoted to understanding not only parabolas but also absolute value graphs (they are represented in a V-shape that can be placed in a horizontal or a vertical position. Such focus leads me to think that we’ll have a greater proportion of these questions compared to before (a good way to practice these is via SAT prep books—the SAT has many parabola questions.)

As for the quadratic equation, it is only employed to help us derive the intersection of a parabola and a line. (See page 240). Still, now that ETS is requiring us to know the quadratic equation, it is fair game for simple polynomial expressions. For instance try the problem below:

x^2 + x + 6 = 0

What are the two values of x that satisfy the equation above? If you haven’t learned the quadratic equation then definitely add it to your arsenal.



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4 Responses to The Revised GRE– What’s New on the Math Front?

  1. sonali August 13, 2016 at 1:30 am #

    can you provide solution to the above polynomial?

    • Magoosh Test Prep Expert
      Magoosh Test Prep Expert August 15, 2016 at 3:18 am #

      Hi Sonali 🙂

      Happy to help 🙂 It turns out that the polynomial does not have any rational roots, and we can use the quadratic formula to see this:

      x^2 + x + 6 = 0
      a = 1
      b = 1
      c = 6

      -b/(2a) +/- sqrt(b^2 – 4*a*c)/(2a)
      -1/2 +/- sqrt(1^2 – 4*1*6)/2
      -1/2 +/- sqrt(1-24)/2
      -1/2 +/- sqrt(-23)/2

      Since there is a negative number under the radical, there are no real roots. The quantity (b^2 – 4ac) is called the determinant and must be positive for a quadratic equation to have real roots (x intercepts).

      We can confirm that the equation x^2 + x + 6 = 0 does not have real roots by graphing the equation. When we do so, we see that the curve does not intersect the x-axis and so there are no x-intercepts.

      Hope this helps 🙂

  2. Nick July 27, 2011 at 4:16 pm #

    just wanted to point out the typo in the above quadratic formula: “-b” should be a numerator not a stand-alone symbol.

    • Margarette Jung
      Margarette July 27, 2011 at 4:55 pm #

      Thanks, Nick– fixed the formatting on the equation so it should look correct now.

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