Excelsior! Academy Posted May 27, 2013 Share Posted May 27, 2013 Feeling very ignorant at the moment. Help please. Substitute the exponent 2 for the * and Greek letters for the Â£ and â‚¬. I didn't have the correct symbols on my iPad. Problem #1 X*+10x+Â£=(x+â‚¬)* Problem #2 In a right triangle, one leg is 3ft longer than the other. The hypotenuse is 6ft long. How long is the shorter leg? Simplify. The answer given is (-3+3 [ 7)/2 Substitute a square root sign for [. Problem #3 6x*-7x=2 Quote Link to comment Share on other sites More sharing options...

nmoira Posted May 27, 2013 Share Posted May 27, 2013 Feeling very ignorant at the moment. Help please. Substitute the exponent 2 for the * and Greek letters for the Â£ and â‚¬. I didn't have the correct symbols on my iPad. Problem #1 X*+10x+Â£=(x+â‚¬)* What are we supposed to do with this? We can't solve for x with three variables and only one equation. Quote Link to comment Share on other sites More sharing options...

nmoira Posted May 27, 2013 Share Posted May 27, 2013 Problem #2 In a right triangle, one leg is 3ft longer than the other. The hypotenuse is 6ft long. How long is the shorter leg? Simplify. Let the shortest side be x. a^2 + b^2 = c^2 (Pythagorean theorem) (x)^2 + (x+3)^2 = 36 x^2 + (x^2 + 6x +9) = 36 2x^2 + 6x - 27 = 0 Then either complete the square or use the quadratic formula, eliminating the negative solution. Sorry, there's no way I can type all that in. :) Quote Link to comment Share on other sites More sharing options...

Excelsior! Academy Posted May 27, 2013 Author Share Posted May 27, 2013 What are we supposed to do with this? We can't solve for x with three variables and only one equation. Thank you! My thoughts exactly. The answer given is Â£=25 and â‚¬=5 Quote Link to comment Share on other sites More sharing options...

Excelsior! Academy Posted May 27, 2013 Author Share Posted May 27, 2013 We tried the Pathagorean Theorem, but apparently didn't go far enough. Thank you! Quote Link to comment Share on other sites More sharing options...

nmoira Posted May 27, 2013 Share Posted May 27, 2013 6x*-7x=2 (My daughter pointed out there's a superscript button... months of using the new boards and I never noticed.) 6x^{2} - 7x - 2 = 0 This also needs you to either complete the square or use the quadratic formula. :) Quote Link to comment Share on other sites More sharing options...

Dana Posted May 27, 2013 Share Posted May 27, 2013 Typing on iPad, so ask if something isn't clear. These are all about solving quadratic a by completing the square. For the first problem, the goal is to have the student comfortable with the special products rules: (a + B)^2 and (a - B)^2 In both rules, when expanded, the first term is x^2 and the last term is +b^2. The difference is the sign of the middle term...either +2ab or -2ab. In your problem, you're trying to fill in the blanks so you'll have a perfect square.... x^2 + 10x + _ = (x + _ )^2 It's often easiest to start with the blank on the right. To get the middle term of 10x on the left, we have +2ab a is x, so 10x is 2(x)(B) This means b has to be 5. We now have x^2 + 10x + _ = (x + 5)^2 When we expand, the blank on the left is 5^2, so 25. You now have the true statement x^2 + 10x + 25 = (x + 5)^2 Quote Link to comment Share on other sites More sharing options...

regentrude Posted May 28, 2013 Share Posted May 28, 2013 Feeling very ignorant at the moment. Help please. Substitute the exponent 2 for the * and Greek letters for the Â£ and â‚¬. I didn't have the correct symbols on my iPad. Problem #1 X*+10x+Â£=(x+â‚¬)* Let me rewrite it with more standard symbol for square: x^2 +10x+a = (x+b.)^2 (I don't have your symbols, so will use a and b instead) If this relationship is to be true for ALL values of x, you do have enough information to solve for both unknowns by comparing coefficients in front of the different powers of x. To do that, expand the right side: x^2 +10x+a = x^2 + 2bx +b^2 For this to be true, 10=2b hence b=5 and a= b^2=25 because whatever is in front of x at the left must be in front of x on the right, and whatever is the constant term without any x on the left must equal the constant term without any x on the right. Quote Link to comment Share on other sites More sharing options...

Excelsior! Academy Posted May 28, 2013 Author Share Posted May 28, 2013 Thank you, thank you, thank you all. I love this board! Quote Link to comment Share on other sites More sharing options...

letsplaymath Posted May 28, 2013 Share Posted May 28, 2013 You might find this new site helpful: mini-courses about quadratic equations and how to understand them, from one of my most-favoritest math teachers on the internet.. G'day Math Quote Link to comment Share on other sites More sharing options...

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