how do you solve for the roots:?

(2x - 5)(2x + 5) = 0
x = +/- 5/2

(3a - 4)(a+ 1) = 0
a = 4/3, -1

1)
4x^2 - 25 = 0
(2x)^2 - 5^2 = 0
(2x + 5)(2x - 5) = 0

2x + 5 = 0
2x = -5
x = -5/2 (-2.5)

2x - 5 = 0
2x = 5
x = 5/2 (2.5)

= = = = = = = =

2)
3a^2 - a - 4 = 0
3a^2 + 3a - 4a - 4 = 0
(3a^2 + 3a) - (4a + 4) = 0
3a(a + 1) - 4(a + 1) = 0
(a + 1)(3a - 4) = 0

a + 1 = 0
a = -1

3a - 4 = 0
3a = 4
a = 4/3

∴ a = -1 , 4/3

(2x-5)(2x+5)=0x = +/- 5/2

(3a-4)(a+1)=0
a = -1, 4/3

4x^2 - 25 = 0

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

x = 5/2
x = -5/2

----------------------------------

3a^2 - a - 4 = 0

(3a - 4)(a + 1) = 0

a = 4/3
a = -1

The first one is a difference of squares so:
4x^2 - 25 = ( 2x - 5 ) ( 2x + 5 ) = 0
then You can set 2x-5 = 0 and 2x + 5 = 0 to get the two values of x.

The second equation can be also factored to (3a - 4)(a + 1)
and using the same thing get the roots of the second equation.

the first one can be 2.5 or -2.5
the second one can be -1 or 1 1/3

4x^2 - 25 = 0

4x^2 = 25

x^2 = 25/4

x = ± 5/2 {square root both sides}


3a^2 - 1a - 4 = 0

ac = 3(-4) = -12

Factors of -12 that add to -1 are +3 & -4

3a^2 + 3a - 4a - 4 = 0 {rewrite for grouping}

3a(a + 1) - 4(a + 1) = 0

(3a - 4)(a + 1) = 0

a = 4/3, -1

WOW! All the Top Contributors of Mathematics have answered! All of them are correct!!!

4x^2-25=0 is the difference of squares, so
(2x+5)(2x-5)=0 set each of the factors equal to zero and you get
x= -5/2 and x= 5/2

3a^2 - a - 4 = 0 you just factor the polynomial
(3a-4)(a+1) = 0 and do the same as above, set each of the factors equal to zero, so
a = 4/3 and a = -1