Natalie C.

asked • 04/08/15

Determine the solution set

Determine the solution set for 10x^2+x-3=0 A.1/3,3/5 b. -3/5,1/2 C.-1/2,3/5 D.-1/2,-3/5 D.-3,1

Lori C.

There are several ways to answer this.   1) Put 10x^2+x-3 into a graphing calculator.  Look at the graph to see where it crosses the x axis.  If you use the calculate function, you will get the answers -0.6 and 0.5.  Those answers in fraction form are -6/10 =-3/5 and 5/1- = 1/2.   2) Factor using the AC method.  Multiply the a by the c (10 * -3) = -30.  List the factors of -30 and see which set of factors add up to 1 (The coefficient of x).  Your factors will be 6 and -5.  They multiply to give you -30 and add to give you 1.  Replace the x in the formula with 6x - 5x.  You now have 10x2+ 6x -5x -3 = 0.  Next, factor by grouping.  Group the first 2 terms and factor them, 2x(5x+3).  Then group the last 2 terms and factor them, -1(5x+3).  Notice that the (5x+3) is in both expressions.  Factor it out and you will be left with the 2x and the -1.  So, your equation now is (5x+3)(2x-1) = 0.  Set each factor equal to zero, and solve...  5x+ 3= 0 implies that 5x = -3 and x = -3/5.  The second factor gives 2x-1=0 and that implies that 2x = 1 and x = 1/2.  Good - that's the same answer we got before!   3) Use the Quadratic Formula  a=10, b=1, and c=-3.... so x = ⌈ -1  ± √((1)2-4(10)(-4))⌉/2(10).  That is really hard to follow if you are not used to writing math...  Let's try to simplify...  x=(-1± √121)/20.  Breaking that down, x=(-1+11)/20=1/2  OR x=(-1-11)/20 = -12/20 = -0.6.   There are some other methods, but surely one of these will work!  Good luck!
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04/08/15

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