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# Super hard algebra, prove your math skills!?

Topic: Super hard algebra, prove your math skills!?
June 16, 2019 / By Chrystal
Question: Solve the verbal problem with a quadratic equation 1. The length of a rectangle is 3 inches more than twice its width, and its area is 65 square inches. Find the dimensions. length = 2w+3, width = w. 2. The width of a rectangle is two-thirds of its length, and its area is 216 square meters. Find the dimensions. length = L, width = 2/3L

Avis | 7 days ago
1) Because area = lw, this gives the equation 'w(2w + 3) = 65'. Distributing gives 2w^2 + 3w - 65 = 0 The Quadratic Formula is {-b ± sqrt(b^2 - 4ac)}/2a. a = coefficient of w^2, 2 b = coefficient of w, 3 c = the constant, -65 Now you have everything you need to do this on your own. 2) L(2/3L) = 216. 2/3L^2 - 216 = 0. Now repeat the steps shown in problem 1, knowing that b will be 0 because there is no x.
👍 182 | 👎 7
Originally Answered: Super hard!math problem need help!?
Hi, The first boy takes 1/3 of the coins and leaves 2/3 of the coins. The second boy takes 1/3 of the remaining 2/3, which is 1/3 x 2/3 =2/9 of the original amount. The 1/3 the first boy took plus the 2/9 the second boy took adds up to 5/9 of the original number of coins, meaning only 4/9 of the original coins are there when the third boy gets up. The third boy takes 1/3 of 4/9 which is 4/27 of the original coins. Adding this 4/27 onto the 5/9 the original 2 boys took means they have taken 19/27 of all of the coins. That means the remaining 216 coins the leprechaun fins are only 8/27 of his original amount. You can find that original amount by setting up a proportion and cross-multiplying it to solve. 8_____216 === = ===== 27_____x underlines are for spacing purposes 8x = 27(216) 8x = 5,832 x = 729 This is the amount the leprechaun had originally. Boy 1 had 1/3 of 729 or 243 coins Boy 2 had 2/9 of 729 or 162 coins Boy 3 had 4/27 of 729 or 108 coins 216 + 243 + 162 + 108 = 729, the original amount I hope that helps!

Abigayle
1) area=65=(2w+3)*w=2w2+3w implies 2w2+3w-65=0 (2w + 13)(w - 5)=0 implies 2w+13=0 implies w= - 13/2 impossible or w-5 =0 implies w=5 inches and L = 13 inches 2) 216 = L*2/3L=2/3L2 implies L2 = 3*216/2=324 IMPLIES L= 18 and w =12
👍 70 | 👎 0

Stu
1) w*(2w+3)=65 ------ 2w^2 +3w-65=0 you will need to do the Pythagorean theorem for this one. I don't really feel like doing it but it is just plugging in numbers. 2) 2/3L*L=216 ------- 2/3L^2=216 ----- L^2=324-------- L=18 meters^2
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Pancras
area = width*length area = 65 in^2 = (2w + 3)*w 2w^2 + 3w = 65 w + (3/2)w = 65/2 (w + 3/4)^2 = 529/16 w = 5.75 - 0.75 = 5 in therefore length = 2(5) + 3 = 13 area = 65 = width*length = 5*13 The second one is done the same way
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Livy
1. 65=(2w+3)w 65=2w^2+3w 0=2w^2+3w-65 0=(w-5)(2w+13) w=5 2w=-13, w= -13/2 (width can't be negative so this value is not the answer. Therefore, w=5 in l=2w+3 l=2(5)+3 l=13 in 2. w=2/3L, A=lw 216=L(2/3L) 216=2/3L^2 2/3L^2-216=0 3/2(2/3L^2-216)=3/2(0) L^2-324=0 L^2=324 L=√324 L=18 m w=2/3L w=2/3(18) w=12m
👍 52 | 👎 -21

Jashub
1. 2w^2 + 3w -65 = 0 (w-5)(2w+13) = 0 w = 5 l = 13 2. 2L^2/3 = 216 L^2 = 3*216/2 = 144 L = 18 W = 12
👍 46 | 👎 -28

Originally Answered: super hard math problem. I can't solve it, need help?
AC : y = x + a, so A(-a,0) y = x^2 = x + a x^2 - x - a = 0 (x - 1/2)^2 = a + 0.25 x = 0.5 +- √(a+0.25) B with the -ve x, so B(0.5 - √[a+0.25],Yb) and C(0.5 + √[a+0.25],Yc) B is the middle /center point of AC, so 2(0.5 - √[a+0.25]) = -a + [0.5 + √(a+0.25)] 0.5 + a = 3√[a+0.25] (a^2 + a + 1/4) = 9(a + 1/4) a^2 - 8a - 2 = 0 (a - 4)^2 = 2 + 16 a = 4 +- 3√2 since a > 0, a = 4 + 3√2 y-intercept = a = 3√2 + 4 = 8.24264

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