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Cez005
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BrushMyQuant
How about using the quadratic formula?

https://en.wikipedia.org/wiki/Quadratic_formula
Cez005
Hi all,

Factoring (t^2)+4t-672=0 was required for a OG question. Any tips on how to approach this other than trial and error using the answer choices?


Thanks, that's not a bad suggestion but it can still be long-winded to find the root. Any tips on finding the root of a large number like 2704?

Would you just narrow down the range of possible numbers. E.g. 50^2 is 2500 and 60^2 is 3600, so the sqrt must be between 50 and 60. The only integers that. when squared, result in a units digit of 4, are 2 and 8. Trying 52^2 gives the answer.
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I prefer long division
https://www.math-only-math.com/square-ro ... ethod.html

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Cez005
Hi all,

Factoring (t^2)+4t-672=0 was required for a OG question. Any tips on how to approach this other than trial and error using the answer choices?

Solving and Factoring Quadratics:
https://www.purplemath.com/modules/solvquad.htm
https://www.purplemath.com/modules/factquad.htm

Hope it helps.
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Hi all,

Factoring (t^2)+4t-672=0 was required for a OG question. Any tips on how to approach this other than trial and error using the answer choices?

Hi Cez005,

There are mainly three ways to solve the quadratic equation problems.
1. Using formula
2. Middle term factorization method, and
3. Method of completing Square

I'll explain the Method of completing Square.

Any quadratic equation of the form \(ax^{2} + bx + c = 0\) can be transformed into \(a(x+d)^{2} + e = 0\),
where \(d = \frac{b}{2a} \textrm{ and } e = c - \frac{b^{2}}{2a}\)

In the given example, we have \(a = 1, b = 4 \textrm{ and }c = -672\)
\(d = 4/2 = 2, e = -672 - 16/4 = -676\)

Hence, we have \((t + 2)^{2} - 676 = 0 \Rightarrow (t+2)^{2} = 26^{2} \Rightarrow t+2 = \pm 26 \Rightarrow t = 24 \textrm{ or } -28\)

For more detail please visit completing-square.html

Please remember squares of 1 to 30. what-to-memorize-in-gmat-quant-for-the-gmat.

Hope this helps.
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