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# GR.11 Math question help please!?

Topic: GR.11 Math question help please!?
July 18, 2019 / By Andrew
Question: Can anyone help me solve this question? The length of a square is increase by 3 cm. The width of the square is increase by 1 cm. The area of the resultant rectangle is 34cm. To one decimal place, what is the side of the original square? Please help and show all your work!! Thank you!

## Best Answers: GR.11 Math question help please!?

Totty | 9 days ago
From the way I understand the problem, here's how you solve: Set up an equation for your situation. You want to know the square's original side length, and you have other data points that can be used to find it. We'll call the original side length x. We're onyl using one variable because the problem states that it's a square, so length and width are the same. If it was a rectangle, we would ahve to use x and y and this porblem would be a lot rmoe complicated. So we know that length times width gives us area, right? (l)(w)=a. In our case (square) we can start with (x)(x)=a. Let's start filling in data. We are given that a is 34cm^2. So lets add that in. (x)(x)=34 But, we also know that 34 is the area of the modified square, not the original. How was the square modified? length was increased by 3 and width was increase by 1. So (x+3)(x+1)=34. Now solve the equation for x in any way you'd like. I just grabbed my TI-89 and had it solve it for me, but we can do it by hand too. FOIL out and we get x^2+4x+3=34. Again, solve however you'd like from here. You could also graph f(x)=x^2+4x+3 and f(x)=34 and find where they intersect. However you choose to solve, you should get two answers, -7.92 and 3.92. Obviously, the side of a square can't be negative, so 3.92 is the answer. x ( the side of the square) was 3.92 cm before the length was increased by 3 and the width as increased by 1.
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Originally Answered: Math homework question, 8th grade math [( Write the product in simplest form )] Help? :]?
As derived from the theorem of Pascal's Triangle, there are 3 basic rules/laws of quadratic equations which are namely: 1. (a + b)(a - b) = (a² - b²) 2. (a + b)² = a² + 2ab + b² 3. (a - b)² = a² - 2ab + b² When all the above is not applicable, you use the FOIL (First, Outer, Inner, Last) method, the slow factorisation method, which currently applies to your given question. Solution: (t + 3)(t - 4) = t² - 4t + 3t - 12 = t² - t - 12 Multiply the first term of (t + 3), which is t, by the first term of (t - 4), which is t. You get t². Then multiply the outer term of (t + 3), which is t, by the outer term of (t - 4), which is -4. You get -4t. Next, multiply the inner term of (t + 3), which is 3, by the outer term of (t - 4), which is t. You get 3t. Lastly, multiply the last term of (t + 3), which is 3, by the last term of (t - 4), which is -4. You get -12. It's just following patterns. It's so hard to explain through writing. I hope you've understood though.
Originally Answered: Math homework question, 8th grade math [( Write the product in simplest form )] Help? :]?
( t + 3 ) ( t - 4 ) = t.t - 4.t + 3.t - 12 = t^2 - 4t + 3t - 12 = t^2 - t - 12 i hope i remember right.

Sabina
With a square, the length is equal to the width, so we'll call THAT original length/width 'x': (x + 3)(x + 1) = 34 x² + 4x + 3 = 34 x² + 4x - 31 = 0 x = 3.9 OR x = -7.9. Obviously, x (the original length/width) cannot be a negative number, so we are left with x = 3.9. So, the side of the original square was 3.9 cm. =======================================...
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Nessa
You might like to start by guessing the answer will be near 4 because 5*7 = 35 (x + 1)(x + 3) = 34 x^2 + 4x - 31 = 0 quadratic formula, positive solution x = sqrt(35) - 2 ~ 3.9 You can see that this works because [sqrt(35) - 1][sqrt(35) + 1] = 35 - 1 = 34 Regards - Ian
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Originally Answered: i have this riddle/question/math problem and i need help solving it because it is part of my math grade!?
10 children * 6 Eggs= 60 eggs total 60/4=15, fifteen is one fourth of 60. three fourths would be 45. 60-45=15. 15 eggs are left!

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