Thank you Bonnie. It takes a long time and much effort to make these videos and it motivates me to do more when I get positive feedback. Please use the video any way you think it will help people teach or learn. Steven G 3/30/2014 | Steven G.
I am doing "solving quadratic equations by factoring" and "zero product property"... in all the examples of factoring, they do NOT have any with fractions - so i'm not sure what to do first. Can you help me to find the solution set for this : 4/5x^{2} = 2x - 4/5 I've tried this: 4/5x^{2 }- 2x +4/5 =0 but now i'm sure what to do next...do I factor or get rid of the fraction part? 5/25/2014 | Jason from Frederick, MD
Jason, The technique in the video works only when the binomial factors include integers. Here is how I solved the problem you listed. Multiply by 5/4 to get x^{2}- (5/2)x + 1 = 0 Use the quadratic formula to get x = 1/2, 2 (x - 1/2)(x - 2) = 0 Steven 5/26/2014 | Steven G.
(4/5)x^{2} - 2x + (4/5) = 0 The "common denominator" is 5. So multiply both sides by 5" 4x^{2} - 10x + 4 = 0 Since you can get the original equation back by dividing both sides by 5, both equations must have the same solution set. But this equation is better since it has no fractions in it. You may notice that 4, 10, and 4 have a common divisor of 2. If you want to use the quadratic formula, then let a = 4, b = -10 and c = 4. 10 ± √((-10)^{2} - 4(4)(4)) --------------------------- 2(4) 10 ± √(100 - 64) ------------------- 8 10 ± 6 ------- 8 2, 1/2 But it would really simplify the arithmetic if you divided both sides by 2 first. 2x^{2} - 5x + 2 = 0 Now you can let a = 2, b = -5 and c = 2: 5 ± √((-5)^{2} - 4(2)(2)) ------------------------- 2(2) 5 ± √(25 - 16) ----------------- 4 5 ± 3 ------ 4 2, 1/2 You can also factor 2x^{2} - 5x + 2 = 0 (x - 2)(2x - 1) = 0 x-2 = 0 or 2x - 1 = 0 x = 2, x = 1/2 Dec 31 | Steven G.
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