quadratic equations?

Question

The perimeter of a rectangle is 41 inches and its area is 91 sqare inches. Find the length of its shortest side.

and

The sum of two numbers is 6 and their product is 4. Find the larger of the two numbers.

I really am not sure where I am going... I need an explanation of how to solve the problem and answer would also be useful. Thanks.

-asian_al

Answer 1

Let
l = length of the rectangle
w = width of the rectangle
2l + 2w = perimeter of the rectangle
lw = area of the rectangle

Thus, the equations are:
2l + 2w = 41
lw = 91

From the first equation, we can solve for l:
l = (41 - 2w)/2

Substitute this value to the second equation. Solving for w means solving for the length of the shortest side.
(41 - 2w)/2 (w) = 91

Distribute w, then cross-multiply 2
41w - 2w² = 182

Transpose everything to the right
2w² - 41w + 182 = 0

You can see that it is very difficult to factor this equation, so you can simply use the quadratic formula:
x = [-b ± √(b² - 4ac)]/2a

Use this formula using a = 2, b = -41 and c = 182, and using a negative sign for the square root (since we are solving for the shortest side):
w = [41 - √(41² - 4 · 2 · 182)]/(2 · 2)

simplify
w = [41 - √(1681 - 1456)]/4

simplify
w = (41 - √225)/4

get the square root
w = (41 - 15)/4

Subtract
w = 26/4

Divide
w = 13/2 inches= 6.5 inches

Therefore, the length of the shortest side is 6 1/2 inches.

_________________________
Let
x = be the first number

Since their sum is 6, then
6 - x = the second number

Since their product is 4, then
x(6 - x) = 4

Distribute x
6x - x² = 4

Transpose everything to the right
x² - 6x + 4 = 0

Since it is again hard to factor out, then use the quadratic formula once again(with a = 1, b = -6 and c = 4). This time we are looking for the larger of the two, so use a positive sign now.
x = [6 + √(6² - 4 · 1 · 4)]/(2 · 1)

Simplify
x = [6 + √(36 - 16)]/2

subtract
x = (6 + √20)/2

We can simplify √20 = 2√5
x = (6 + 2√5)/2

Divide 2
x = 3 + √5

Therefore, the larger number is 3 + √5.

^_^

Answer 2

For Part I of your question, use the equations for the perimeter and area of a rectangle:
1. P = 2(L+w) = 41
A= L*w = 91
2. As you can see, there are two variables in each equation, so you need to solve for one of the variables in terms of the other. I used the Area equation (since it is simpler) and solved for w in terms of l): w = 91/L
3. Now that we are able to use the one variable "L", substitute the 91/L for the "w" in the perimeter equation and solve:
41 = 2(L + 91/L)
41 = 2L + 182/L ---> we now have to multiply both sides by L in the denominator to get it out of there
L(41) = L (2L + 182/L)
41L = 2L^ + 182 (^ means squared)
0 = 2L^ -41L + 182 (here is your quadratic)
*** This next step is tricky because you have to find the factors of 182 that when manipulated according to the terms of the equation will give you the middle term of -41. The two factors end up being 13 and 14:
(2L-13) (L-14) = 0
Set each term equal to 0 to determine the values of L:
2L -13 =0; L = 13/2
L-14 = 0; L = 14
Since both values of L are positive and can represent a length, either can be substitued back into the original equations to determine the width. . Again the easiest equation to use is the Area one: 91 = L*w
91 = 14 *w
91/14 = w
13/2 = w -----> we already obtained this second value for L in the above quadratic, so L= 14, w = 13/2 or 6.5 and the shortest side is 6.5.

6.5 is the answer for part I.


Part 2
We know that the sum of two numbers x and y is 6
We know that the product of the same two numbers xy is 4.

x + y = 6
xy = 4

Again to variables in each quation. Solve for one of them in either equation:
xy = 4
x= 4/y <---- take this term and substitute it for the x in the x + y equation:
4/y + y = 6 <----multiply both sides by the y in the denominator to get it out of there; this will form the quadratic:
4 + y^ = 6y
y^-6y+4 = 0
*** Again, this next step is tricky. We are looking for the factors of 4 which when manipulated according to the signs of the quadratic will give you - 6. The only factors of 4 that I know of are 1,4 and 2,2. There is no possible way that either set of factors will give you -6. Unless I am too sleepy to see otherwise, I am afraid I, personally, cannot continue to solve the equation. Are you sure you have written the problem correctly? I later read yup's answer and he used the correct procedure for determining the roots of squared terms. I apologize, I'd forgotten that step. It's been 15 years since I've taken algebra.

Answer 3

You need a system of equations.
The perimeter equation is P = 2x + 2y
The area equation is A = xy
x = shorter side
y = longer side
Because the x and the y are the same, you can use the two equations together to solve this.
41 = 2x + 2y
91 = xy

If we solve for x in the second equation, we get x = 91/y
We then plug this into the first equation, and solve for y
41 = 2(91/y) + 2y
41 - 2y = 182/y
41y - 2y^2 = 182
-2y^2 + 41y - 182 = 0

Solve for y using the quadratic equation.
-B plus or minus the square root of b^2 - 4ac, all over 2a
Then plug that back into the second equation (91 = xy)
and solve for x

do the same for the other one,
x + y = 6
xy = 4

x = 6 - y

(6-y)y = 4
-y^2 + 6y -4 =0

etc, etc, etc

Any help at all?

Answer 4

For each problem you must solve 2 equations simultaneously.

1: P = 2 L+ 2 W and A = L times W

41 = 2 L + 2 W and 91 = L times W Solve for W

2: let L= large # and S = small #

S + L = 6 and S times L = 4 Solve for L

Answer 5

Heres the first one. You know P=2l + 2w and A=lw, where l and w are length and width. so 41=2l+2w and 91=lw. What are the only two numbers that mutliply to get 91? 13 and 7.So now you know the shortest side is 7. If you put these two numbers into the perimeter equation they work too.
sorry i dont have time for the other problem at the moment.

Answer 6

One straightforward term that looks many times is "equations." for many, this would be a buzz word this is recognizable as a math term yet relatively comprehend-how the term would sometime get away human beings. the two maximum straightforward equations are linear equations and quadratic equations yet, of direction, in effortless terms giving an equation a acceptance does not inevitably clarify what that's. So, a clearer definition of what precisely the two maximum straightforward equations - linear and quadratic equations - is mandatory. A linear equation refers to a undeniable equation this is graphed on a at cutting-edge line. additionally, a linear equation possesses on the line one variable this is often stated as "X" and "X" will continuously be of a level this is a million at maximum. (this is, there are no exponents; yet once you're in seek of for exponents then wait and spot because of the fact we can get to them presently!) a straightforward occasion of a linear equation would be 1x + 2 = 3. needless to say, x would equivalent a million in this actual occasion and it relatively is discovered by utilising in effortless terms utilising slightly algebra on the equation to come to a decision X. 3 minus 2 equals a million. for this reason, X would desire to equivalent one as a million x a million equals one. And, nope, no longer all linear equations are that straightforward as they arrive as complicated as 6(x + 3) = 24 (x +0), however the straightforward ingredient of isolating x to locate the answer does not substitute. A quadratic equation is barely diverse from a linear equation in one admire: a selection of of of the figures is squared. (The word quadratic derives from the Latin word for squared) the straightforward form of a quadratic equation is ax2 + bx + c = 11. In this type of equation, if a = a million, b = 2 and c = 3 then X would desire to equivalent 2. all of us comprehend this because of the fact 2 squared is 4 and four x a million = 4. 2 x 2 = 4. As with a linear equation, there would be greater complicated variations of a quadratic equation yet only with the easy and complicated linear equations straightforward algebraic operations can yield the main appropriate answer.

Answer 7

2L+2w=41
LW=91

(41-2w)/2=L 91w+2w=41/2 w(91+2)=41/2 41/2times1/93
answer=41/186

s+l=6
sl=4
6-l=s (6-l)L=4 4L=6-L 4L+L= 6 L=6/5
6/5s=4 20=6s 20/6=s 10/3=s
answer: s= 10/3 L= 6/5

Answer 8

I'll solve 1 for you, you can then solve the 2nd
2(x+y)=41
xy=91
so 2(x+91/x)=41
2x^2+182=41x
Transposing, 2x^2-41x+182=0, or 2x^2-13x-28x+182=0
x(2x-13)-14(2x-13)=0
(2x-13)(x-14)=0
So, the shortest side is 6.5 inches. Incidentally the longer side is 14. BTW, a rectangle doesn't have a shortest side, just a shorter side

Answer 9

The perimeter of a rectangle is 41 inches and its area is 91 sqare inches. Find the length of its shortest side.


p=2l+2w=41
l+w=20.5
A=lw=91
l=20.5-w substitute
(20.5-w)w=91
-w^2+20.5w-91=0
w^2-20.5w+91=0
w=(20.5+/-√(20.5^2-4*91))/2
w=(20.5+/-√56.25)22
w=(20.5+/-7.5)/2
w=13/2=6.5
w=28/2=14
the sides are 6.5 & 14
w=6.5

and

The sum of two numbers is 6 and their product is 4. Find the larger of the two numbers.

xy=4
x+y=6
x=6-y
y(6-y)=4
-y^2+6y-4=0
y^2-6y+4=0
y=(6+/-√(36-16))/2
y=3+/-(1/2)√20
y=3+/-√5
larger number is 3+√5