solve the following quad. equation?

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Since A=lw and the width is (3x-4) and the length or height is (x+1), and the overall area is 10cm^2, then setup would be:

10 = (3x-4)(x+1)

10 = 3x^2-x-4

3x^2-x-4-10 = 0

3x^2-x-14 = 0

Solving for x we have

(3x-7)(x+2) = 0

3x-7=0
x=7/3

x+2=0
x=-2

Obviously we reject a negative because you can't have a negative length or height, etc.

So the height is: x+1 or 7/3+1 = 10/3cm

And the width is: 3x-4 or 3(7/3)-4 = 3cm

Have a good day!

area of parallelogram=base x corresponding altitude

(x+1)(3x-4)=10
3x^2-4x+3x-4=10
3x^2-x-4=10
3x^2-x-14=0
3x^2+6x-7x-14=0
3x(x+2)-7(x+2)=0
(3x-7)(x+2)=0
x=7/3 or -2
x=7/3 since the side can not be in negative.

Please rate me if i am correct.
:))

(x+1)(3x-4) = 10
3x²-x-4 = 10
3x²-x-14 = 0

   ax²+bx+c=0
where
   a=3
   b=-1
   c=-14

x = [-b ± √(b²-4ac)] / (2a)
   = [-(-1) ± √((-1)² - 4(3)(-14))] / 2(3)
   = [1 ± √(169)] / 6
   = [1 ± 13] / 6
   = 0.167 ± 2.167

x₁ = 0.167 + 2.167 = 2⅓
x₂ = 0.167 - 2.167 = -2 invalid length

height = 3⅓
width = 3

a) 4x2 -14x -30 = 0 8 - 14x -30 = 0 combine like words.. 8-30 is -22.. so 14x = -22 divide the two aspects by skill of 14 to get x by skill of itself x = -a million.fifty seven i do no longer know approximately B because of the fact does x2 propose x circumstances 2 or x squared?

(x + 1)(3x - 4) = 10
3x² - x - 14 = 0
(3x - 7)(x + 2) = 0
x = 7/3 cm

(x + 1)(3x - 4) = 10
x*3x + 1*3x - x*4 - 1*4 - 10 = 0
3x^2 + 3x - 4x - 4 - 10 = 0
3x^2 - x - 14 = 0
3x^2 + 6x - 7x - 14 = 0
(3x^3 + 6x) - (7x + 14) = 0
3x(x + 2) - 7(x + 2) = 0
(x + 2)(3x - 7) = 0

x + 2 = 0
x = -2

3x - 7 = 0
3x = 7
x = 7/3

∴ x = -2 , 7/3

height = h, width = w

the area of a parallelogram is hw:

hw = 10
w = 10 / h -- divide each side by h

w = 3x - 4
10 / h = 3x - 4 -- substitute w = 10 / h
h = 10 / (3x - 4) -- multiply each side by h and divide by (3x - 4)

h = x + 1

since h = 10 / (3x - 4) AND h = x + 1 we can say

x + 1 = 10 / (3x - 4)
(x + 1)(3x -4) = 10 -- multiply each side by (3x - 4)
3x^2 - x - 4 = 10 -- cross multiply
3x^2 - x = 14 -- add 4 to each side
(3x^2 - x + 1/12) = 14 + 1/12 -- prepare to make 3x^2 - x into (Ax - B)^2 form by adding 1/12 to each side
[√3x - 1/(2√3)]^2 = 169/12 -- convert to (Ax - B)^2 and combine constants
√3x - 1/(2√3) = 13/(2√3) -- take the square root of each side
√3x = 13/(2√3) + 1/(2√3) -- add 1/(2√3) to each side
√3x = 7/√3 -- combine constants and cancel 2/2
x = 7/3 -- divide each side by √3 and make √3 * √3 = 3

Multiply the width by the height, that is the area of a parallelogram. So
(3x-4)(x+1) = 10

Expand and rearrange to get a quadratic, the positive root is the answer.





3x^2 -x -14 = 0, x = 7/3

(x+1)(3x-4)=10
3x^2-x-4=10
3x^2-x-14=0
(3x-7)(x+2)=0, x=-2( not possible, can't have a negative height),so
3x=7, x=7/3
check
7/3+1=10/3
3(7/3)-4=3
10/3*3=10

x+1*3x-4=3x^2+x-4=10^2
=-1+- /1+48/6
x1=1
x2=-4/3