Date: Mon, 29 Mar 1999 18:36:28 -0600 Sender: Ken
#1 a #2 t(t - 5) = 5 (t - 5) #3. (x + 4)(x - 3) = 8 Thanks. Hi Ken, I presume that you wish to solve these equations. (I.e. determine what value(s) of x will make the equations true.) - a
^{2}- 10a = 2a - 36 This is a quadratic equation since the highest power of x is the 2nd power. The usual strategy for solving a quadratic equation is to 1st get a zero on one side. a^{2}- 12a + 36 = 0. Then you can factorthis equation as (a - 6)(a - 6) = 0. Since the product of two binomials is 0 then at least one of them has to be 0. So a - 6 = 0 and a = 6. - t(t - 5) = 5(t - 5).
Again, moving both terms to the left side of the equation, t(t - 5) - 5(t - 5) = 0. One of the first steps in factoring is to look for a common factor. Here (t - 5) is a common factor so the equation becomes (t - 5)(t - 5) = 0. Thus, as in 1., t = 5. - (x + 4)(x - 3) = 8.
Notice that this is also a quadratic equation and there are, at most, 2 values of x which satisfy the equation. You can proceed as in 1.(move the 8 to the left side, expand and then factor the quadratic) but this is a problem where "guess-and-check" might work. If we can find 2 values of x by inspection we are finished. We see that the product of the 2 binomials on the Left Side is 8. The integer factors of 8 are 1,8,2,4,-1,-8,-2 and-4 so we can try these values for (x + 4).x + 4 = 1: Gives x = 5 and thus x - 3 = 2 so (x + 4)(x - 3) = 2 x + 4 = 8: Gives x = 4 and thus x - 3 = 1 so (x + 4)(x - 3) = 8 x + 4 = 2: Gives x = 6 and thus x - 3 = 3 so (x + 4)(x - 3) = 6 x + 4 = 4: Gives x = 0 and thus x - 3 = -3 so (x + 4)(x - 3) = -12 x + 4 = -1: Gives x = -5 and thus x - 3 = -5 so (x + 4)(x - 3) = 8
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