Question

14. Set up and solve a quadratic equation to find

the value of x in the following parallelogram.
2
CE= 12, EB-x² + 4x.
C
A
E
D
B

help me pleaseeeee

Answer 1

x = 2

Explanation:

In a parallelogram

• The diagonals, bisect each other

Then

EB = CE ( substitute values )

x² + 4x = 12 ( subtract 12 from both sides )

x² + 4x - 12 = 0 ← in standard form

(x + 6)(x - 2) = 0 ← in factored form

equate each factor to zero and solve for x

x + 6 = 0 ⇒ x = - 6

x - 2 = 0 ⇒ x = 2

As a check

x = - 6

EB = x² + 4x = (- 6)² + 4(- 6) = 36 - 12 = 24 ≠ 12 ← not a solution

x = 2

EB = x² + 4x = 2² + 4(2) = 4 + 8 = 12 ← solution

the solution is then x = 2

Answer 2

`x=2`

Explanation:

CE and EB appear o be equal so we can set it up as such in an equation

`12=x^2+4x`

In a quadratic equation it is always equal to zero so we subtract 12 from both sides

`0=x^2+4x-12`

Now to find the x we fill it into the quadratic formula

`x=(-4+√(4^2-4(1)(-12)) )/(2(1))`

`x=(-4+√(16+48) )/(2)`

`x=(-4+√(64) )/(2) \\x=(-4+8)/(2) \\x=(4)/(2) \\x=2`

Now we do the same thing except we subtract what was under the square root

`(-4-8)/(2) \\(-12)/(2) \\-6`

But we can't use this because it will give us -12 as a side and we can't have a negative side so we will just use the 2

Hope this helps