please answer!! algebra.. quadratic formula by factoring.?

Question

If the length of each side of a square is increased by 5cm, the area is multiplied by 4. What is the length of the original side of the square..
[please show the step by step solving]

Answer 1

Let x = length of sides of original square.

We know that the formula of a square is x² ; therefore

A = x²

Increasing the length of a side of a square is interpreted as
x + 5, and doing so means the area is 4 times bigger. That is

4A = (x + 5)²

Two equations, two unknowns.

A = x²
4A = (x + 5)²

Substitute A = x² into the second equation, and solve for x.

4x² = (x + 5)²

Let's move everything to the left hand side and factor as a difference of squares; it saves a step.

4x² - (x + 5)² = 0

[2x - (x + 5)] [2x + (x + 5)] = 0

[2x - x - 5] [2x + x + 5] = 0

[x - 5] [3x + 5] = 0

This means

x - 5 = 0, or
3x + 5 = 0

x = 5

x = -5/3

But x = -5/3 is extraneous (we cannot have a negative length), so we reject it.

x = 5.

Answer 2

Let length be x and area be a.
Area of a square = x^2
a= x^2 -------- (1)
But x is increased by 5 and a is multiplied by 4
=> 4a = (x+5)^2 --------- (2)
Expand (x+5)^2
x^2 + 10x + 25
=> 4a = x^2 + 10x + 25 ------------(3)
In equation (1), multiply both the sides by 4
=>4a = 4* x^2 ---------------- (4)
Now, (4) - (3)
=>4a-4a = 4x^2 - (x^2 + 10x + 25)
3x^2 - 10x-25
Factor this.
You'll get the answer as 5

Answer 3

Let length = x cm
Area = x² cm²
New length = (x + 5) cm
New area = (x + 5)²
(x + 5)² = 4x²
x² + 10x + 25 = 4x²
3x² - 10x - 25 = 0
(3x + 5).(x - 5) = 0
x = 5 (taking +ve value for x)
Length of side of original square = 5 cm

Answer 4

let the length of each side be (x) and the area be (a)

a = x² --- (1)
4a = (x+5)² --- (2)

from (2):
4a = x² + 10x + 25 --- (3)

from (1):
4a = 4x² --- (4)

(4) - (3):
4a - 4a = 4x² - (x² + 10x + 25)
0 = 3x² - 10x - 25
3x² - 10x - 25 = 0
(3x + 5) (x - 5) = 0
3x + 5 = 0 ... or ... x - 5 = 0
3x = -5 ... or ... x = 5
x = -(5/3) <--- nonsensical answer (length of side cannot be negative)

therefore the original side of the square is 5cm.

Answer 5

Let the original length be X.

then,original a=x^2

(x+5)^2=4x^2

x^2+10x+25=4x^2

3x^2-10x-25=0

use cross method:

(3x+5)(x-5)=0

x= -3/5 , x=5

x>=0
so x=5

Answer 6

the two solutions (roots) of the quadratic equation are (5/8+sq. root of fifty seven/8) and (5/8-sq. root of fifty seven/8) that are (0.625+sqrt57/8) and ().625-sqrt57/8) i'm sorry, I misplaced my scientific Texas gadgets calculator that does sq. roots. basically placed, the quadratic equation is ax2+ bx+c=0. a=4 b=--5 and c=--2. The quadratic formulation is -b plus or minus the sq. root of (b sqaured - 4 a c ) throughout 2 a. because of the "plus and minus" there are 2 solutions. I e 2 factors the place the PARABOLA crosses the x-axis.

Answer 7

Let the side of the square be x units
The,
Area= x²
a=x²........................................1

When 5 is added to each side the new side becomes (x+5)units

But we know,
4a=(x+5)²................................2

From Equation 1 we know a x²
Therefore
4x²=(x+5)²
2x=x+5
x=5


Therefore the original side of the square is 5 units