Given the surface area is the rectangular object is 78, what is x = ? 

Greetings. The surface areas of the sides of the rectangular figure is 2(3X^2+X^2+2X+3X^2+6X)= 2(7X^2+8X)=14X^2+16X. Now, we are told that the total areas is equal to 78 square units. Therefore, 14X^2+16X=78 and 14X^2+16X-78=0. That is 7X^2+8X-39=0 after dividing throughout by 2. Now, we will factor the expression to get (X+3)(7X-13)=0. Setting each factor to 0, we have X=-3 or X=13/7 units. The total surface areas works out to 78.48, (43.23+20.83+14.42). This problem is quite involved in that it requires the knowledge of algebraic and geometric calculations.

Rectangular object, drawn with dimensions x+2, 3*x, and x, has surface area of 78. What is x? 2*[(x+2)*(3*x) + (x+2)*x + (3*x)*x] = 78, x > 0 3*x^2+6*x + x^2+2*x + 3*x^2 = 39 7*x^2 + 8*x - 39 = 0 (7*x - 13)*(x + 3) = 0 x = 13/7, -3 x = 13/7

A hint ,be careful volume is different from surface area you use different formulas With the above information you can figure out volume

I did 14x^2 +16x-78=0. Then I divided everything by 2 because why not? I got 7x^2+8x-39=0 which equals to 7x^2+21x-13x-39=0. That gave me x=13/7 and -3. For obvious reasons -3 will not work so the only answer is 13/7. I was so happy when I pressed forward the video and saw our answers were the same!

7x^2+8x-39=0 Is as far as I got. Not bad considring I graduated HS in 1970. I could not recall how to solve for x in this quadradic equation

three sides x x+2 3x the shape has 6 faces 3 same sized pairs face1= x(x+2) face2= x(3x) face3= 3x(x+2) 2(face1+face2+face3)=78 2(x(x+2)+x(3x)+3x(x+2))=78 (x(x+2)+x(3x)+3x(x+2))=39 x^2+2x+3x^2+3x^2+6x=39 x^2+3x^2+3x^2+2x+6x=39 7x^2+8x=39 in quadratic form 7x^2+8x-39=0 (-b+/-sqrt(b^2-4ac))/2a a=7 b=8 c=-39 (-8+/-sqrt(8^2-4(7)(-39)))/2(7) (-8+/-sqrt(64+1092))/14 (-8+/-sqrt(1156))/14 (-8+/-34)/14 tossing negative results x= 26/14 = 13/7 verify x= 13/7 = 1.857 2(x(x+2)+x(3x)+3x(x+2))=78 2((1.857(1.857+2) +(1.857)(3)(1.857) +3(1.857)(1.857+2)) =?78 2((1.857)(3.857) +(1.857)(3)(1.857) +(3)(1.857)(3.857)) =?78 (7.162)+(10.345)+(21.487) =? 39 38.994 =❤ 39 E X T R A C R E D I T: complete the square: 7x^2+8x-39=0 7x^2+8x=39 (7/7)x^2+(8/7)x=(39/7) x^2+(8/7)x=(39/7) to get 8/7x with two factors yields: (x+(4/7))×(x+(4/7)) = x^2+(8/7)x+(4/7)^2 so (4/7)^2 must be added to both sides, thus x^2+(8/7)x+(4/7)^2 =(39/7)+(4/7)^2 so that (x+(4/7))^2=(39/7)+(4/7)^2 (x+(4/7))^2=(39/7)+(16/49) =((39×7)/(7×7))+16/49 =(273+16)/49 =(289/49) so that (x+(4/7))^2 =(289/49) sqrt((x+(4/7))^2) =+/-sqrt(289/49) x+(4/7)=+/-(17/7) x.1=(17/7)-(4/7) =13/7 verified previously x.2=(-17/7)-(4/7) =-21/7 =-3 EXTRA EXTRA CREDIT Rational Root Theorm 7x^2+8x-39=0 p= 39 q= 7 possible factors 39=> 1,3,13,39 7=> 1,7 possible roots (p/q)=> +/-1, +/-1/7, +/-3, +/-3/7, +/-13, +/-13/7, +/-39, +/-39/17 and our root is in there: +13/7 and -21/7 (=-3) is too I guess I'll not try them all as I'm just trying to reinforce how it is employed as learning math is repeat, repeat, repeat. I know how to expand am equation in x.

{360*6}=({720 ÷9}=8o 8/10 (x+8x-10)

What softwares do you use?

13/7

Doh! Did it correctly, but forgot to use the 2a at the end.

I came up with 1.86 as the approximate value of "x".

×= 1.858 approximately , plug inn ✌

Also rectangular object? Really? the proper name is cuboid.

'pretty big numbers' to attempt factorising? You really have trouble with those big numbers? If so then your maths skills are quite poor. What big numbers? 7 is a prime number so that takes care of that and the factors of 39 are 1, 3, 13, and 39, so not really a lot of choice to get a set of brackets out of. Over here it would be expected to solve using brackets. Only those who as I said have poor maths skills would use a far more complicated method of using the formula which takes one to far bigger numbers and the good possibly of making mistakes with multiplying negatives and having to know the square root of a big number which most students will not know, cannot find and they would grind to a halt there and then. Are you really advocating such a method? if so you're not doing your students any good at all and poor teaching in my view. Anyway a decent teacher would show all methods to show why factorising is a better option.