(x)(x+1)=30?

Question

measuring area. x=L, x+1=W

Answer 1

(x)(x+1) = 30
x² + x = 30
x² + x - 30 = 0
Using quadratic formula: [-b±√(b² - 4ac)] / 2a
[-b±√(b² - 4ac)] / 2a
[-1±√(1² - 4(1)(-30)] / 2(1)
x = (-1 ± 11)/ 2
x = -6 or +5

Using positive value for length :
x = L = 5
x + 1 = 5 + 1 = 6
Area = 5*6 = 30 Units².

Answer 2

5 times 6 is 30

Answer 3

This is a quadratic equation
Therefore,

x squared + x = 30

x squared + x - 30 = 0

After doing some guessing here, you find that

(x+6) (x-5) = 0

Therefore,
(x+6) = 0 OR (x-5) = 0

and

x = -6 OR x=5

Since you are measuring length, this number should be positive. Thus, the length, x, =5.

And since the width is x+1, the width is 5+1 which is 6.

Answer 4

(x)(x+1)=30
x^2 + x = 30
x^2 + x - 30 = 0
(x - 5)(x + 6) = 0
x = 5 and x = -6 (rejected as length is always positive)

=> x = 5

To check:
Length (x) = 5
Width (x + 1) = 5 + 1
= 6

=> 6 * 5 = 30

Answer 5

(x)(x+1) = 30
x^2 + x - 30 = 0
(x-5)(x+6) = 0
Thus, x = 5 or x = -6. But because x states length, it can not be negative. Then, choose x = 5

Then we find the length is 5 and the width is 6

Answer 6

(x)(x+1) =30
x^2 +x = 30
x^2 +x-30=0
x^2 +6x -5x-30=0
x(x +6) - 5(x + 6) =0
(x+6) (x-5) =0
x=-6 or 5
i take x = 5 since length can not be a negative
L = 5
w = x+1 = 5+1 =6

Answer 7

(x)(x+1)=30
x^2 + x - 30 = 0
(x + 6)(x - 5) = 0
x = -6 OR x = 5

length cannot be negative
therefore x = 5

L = x = 5
W = x + 1 = 5 + 1 = 6

L = 5
W = 6

Answer 8

man x=5,
which means ur L=5, W=5+1=6

Answer 9

run it out

x^2 +x= 30
x^2 +x - 30=0

x+6 x-5

x=5

Answer 10

x ² + x - 30 = 0
( x + 6 ) ( x - 5 ) = 0
x = - 6 , x = 5