Quadratic equation to be solved without quadratic formula

[quadratic]

An interesting quadratic equation that have to be solved without using quadratic formula.
[tex](5+2y)^2-(1-y)^2=0[/tex]

[Guest]

To solve this equation rewrite it as shown below:

(5+2y)^2 - (1-y)^2 = 0

(5+2y)^2 = (1-y)^2

Now take square root on both sides, and we get rid of squares to get the following

5 + 2y = 1 - y

Now collect all the terms with "y" on one side and constants on the other

2y + y = 1 - 5

3y = - 4

y = - 4/3

[Guest]

I wonder why
[tex](5+2y)^2 = (1-y)^2[/tex] <=> 5+2y = 1-y
isn't it
|5 + 2y| = |1 - y|
?

[amit28it]

An interesting quadratic equation that have to be solved without using quadratic formula.
[tex](5+2y)^2-(1-y)^2=0[/tex]

(5 + 2y)^2 = (1 - y)^2
(5 + 2y) = (1 - y)
3y = -4
y = -4/3

[zain123]

Solve the following system of equations using the elimination method by equating coefficients;

11x - 5y + 61 = 0
3x - 20y - 2 = 0


_____________________________
aliiiiiii

[Guest]

[tex]\sqrt{(5+2y)^2}=\sqrt{(1-y)^2} \implies \left| 5+2y \right| = \left| 1-y \right| \implies 5+2y = 1-y \\ 5+2y \text{ or } -(1-y) \implies y =-\frac 43 \text{ or } y=-6[/tex]

[Guest]

The righthand side of [tex]0 = (5+2y)^2 - (1-y)^2[/tex] is the difference of squares, therefore it factors into
[tex]0 = [(5+2y) + (1-y)][(5+2y) - (1-y)] = (6 + y)(4 + 3y) \implies 6 + y = 0 \text{ or } 4 + 3y = 0 \implies y = -6 \text{ or } y = -\frac 43[/tex]