Find the solution set.
8x^2 - 2x - 3 = 0
8x^2 - 2x - 3 = 0....factor the polynomial
(4x -3)(2x + 1) = 0...set each factor equal to zero and solve
4x - 3 = 0 2x + 1 = 0
4x = 3 2x = -1
x = 3/4 x = -1/2
{-1/2, 3/4}
Its too easy.
=> 8x^2 - 4x + 6x - 3 = 0
=> 4x(2x - 1) + 3(2x - 1)
=> (4x + 3) (2x - 1)
=> 4x + 3 = 0 or 2x - 1 = 0
=> x = -3/4 or x = 1/2
I hope it is helpful.
If u need any help in maths, contact me @ [email protected]
You can also use the quadratic formula noting that
a = 8
b = -2
c = -3
and your answers will be
x = 3/4 AND
x = -1/2
(2x+1)(4x-3)
x = -1/2 or 3/4
8x^2-6x+4x-3=0
2x(4x-3)+1(4x-3)=0
(2x+1)(4x-3)=0
2x+1=0 or 4x-3=0
x=-1/2 or 3/4
Solution set(-1/2, 3/4)
this is in the form ax^2+bx+c
so use the formula -b+((b^2-4ac)^(1/2)/2a) and
-b-((b^2-4ac)^(1/2)/2a)... substitute the values accordingly and u 'll get the two roots as
2 +((92)^(1/2))/6 and 2 -((92)^(1/2))/6
x={-.5, .75}
Comments
8x^2 - 2x - 3 = 0....factor the polynomial
(4x -3)(2x + 1) = 0...set each factor equal to zero and solve
4x - 3 = 0 2x + 1 = 0
4x = 3 2x = -1
x = 3/4 x = -1/2
{-1/2, 3/4}
Its too easy.
8x^2 - 2x - 3 = 0
=> 8x^2 - 4x + 6x - 3 = 0
=> 4x(2x - 1) + 3(2x - 1)
=> (4x + 3) (2x - 1)
=> 4x + 3 = 0 or 2x - 1 = 0
=> x = -3/4 or x = 1/2
I hope it is helpful.
If u need any help in maths, contact me @ [email protected]
You can also use the quadratic formula noting that
a = 8
b = -2
c = -3
and your answers will be
x = 3/4 AND
x = -1/2
(2x+1)(4x-3)
x = -1/2 or 3/4
8x^2-6x+4x-3=0
2x(4x-3)+1(4x-3)=0
(2x+1)(4x-3)=0
2x+1=0 or 4x-3=0
x=-1/2 or 3/4
Solution set(-1/2, 3/4)
this is in the form ax^2+bx+c
so use the formula -b+((b^2-4ac)^(1/2)/2a) and
-b-((b^2-4ac)^(1/2)/2a)... substitute the values accordingly and u 'll get the two roots as
2 +((92)^(1/2))/6 and 2 -((92)^(1/2))/6
x={-.5, .75}