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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?
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I usually use prime factorisation while dealing with big numbers such as 672 here. You can really quickly multiply prime factors to create possible factors adhering to sum of roots a.k.a. b.
For e.g. 672 = 2*2*2*2*2*3*7 = 2*2*2*3*2*2*7 = 24*28.
But then it all comes down to individual preference and comfort for a particular method.
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?
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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.
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 \((t + 2)^{2} - 676 = 0 \Rightarrow (t+2)^{2} = 26^{2} \Rightarrow t+2 = \pm 26 \Rightarrow t = 24 \textrm{ or } -28\)
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