Solve these equations....?

Question

2x^2 – 5x = 12



2x(x + 3) = x + 25

Answer 1

2x² – 5x = 12
d = -5² - 4.2.-12
d = 25 + 96
d = 121

x = (5 +/- \/121) : 2.2
x' = (5 + 11) : 4
x' = 16 : 4 = 4
x" = (5 - 11) : 4
x" = -6 : 4 = -3/2
Sol: (There are x in R | x' = 4 and x" = -3/2)
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2x(x + 3) = x + 25
2x² + 6x = x + 25
2x² + 6x - x - 25 = 0
2x² + 5x - 25 = 0
d = 5² - 4.2.-25
d = 25 + 200
d = 225

x = (-5 +/- \/225) : 2.2
x' = (-5 + 15) : 4
x' = 10: 4 = 2,5
x" = (-5 -15) : 4
x" = -20 : 4 = -5
Sol: {There are x in R | x' = 2,5 and x" = -5}
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Answer 2

Answer 1:
2x^2 - 5x = 12
2x^2 - 5x - 12 = 0
2x^2 - 8x + 3x - 12 = 0
2x(x - 4) + 3(x - 4) = 0
(x - 4)(2x + 3) = 0

Either x - 4 = 0
or 2x + 3 = 0

If x - 4 = 0, x = 4
If 2x + 3 = 0, x = -3/2

There are two possible solutions, and they are:
x = 4 (or) -3/2


Answer 2:
2x(x + 3) = x + 25
2x^2 + 6x = x + 25
2x^2 + 5x = 25
2x^2 + 5x - 25 = 0
2x^2 - 5x + 10x - 25 = 0
x(2x - 5) + 5(2x - 5) = 0
(2x - 5)(x + 5) = 0

The two possible solutions are x = 5/2 (or) - 5

Answer 3

First one:

2x^2 - 5x - 12 = 0

Find two numbers that add to -5 and multiply to 2 x -12 = -24
These are: -8, 3

2x^2 - 8x + 3x - 12 = 0
2x ( x-4) + 3(x-4) = 0
(2x + 3)(x-4) = 0
x = 4, -1.5

Second one:

2x^2 + 6x = x + 25
2x^2 + 5x - 25 = 0

Then find two numbers that add to 5 and multiply to -25 x 2 = -50

The numbers are.... 10, 5

2x^2 + 10x - 5x - 25 = 0
2x(x+5) - 5 (x+5) = 0
(2x - 5)(x+5) = 0
x = 2.5, -5

Answer 4

For first equation:

1. rearrange the equation

2x^2-5x-12=0

2. use the quadratic formula (check wikipedia.com coz its difficult to show u the formula here)

x1 = (5+11)/4 = 4
x2 = (5-11)/4 = -3/2

For second equation:

1. expand and rearrange the equation

2x^2+5x-25=0

2. Use the quadratic formula. The result will be

x1 = (-5+15)/4 = 5/2
x2 = (-5-15)/4 = -5

Hope this helps u!

Answer 5

2x^2 - 5x = 12
2x^2 - 5x - 12 = 0
(2x + 3)(x - 4) = 0
x = {4, -3/2}

2x(x + 3) = x + 25
2x^2 + 6x = x + 25
2x^2 + 5x = 25
2x^2 + 5x - 25 = 0
(2x -5)(x +5) = 0
x = {-5, 5/2}

Answer 6

must use formula one.
x= (-b+/-(b^2 - 4ac)^0.5)/ 2a
whereby in this case, a=2, b= - 5, c = -12
the second one you can try out.

2x^2 - 5x = 12
2x^2 - 5x - 12 = 0
x = 5 +/- (25- 4(2)(-12))^0.5 over 2(2)
= 5 +/- (25+96)^0.5 over 4
= 5+/- 11 over 4
therefore x = (5+11)/4
=4

or x= (5-11)/4
=6/4
=1.5

second one answer is x = 2.5 or -5

Answer 7

2x² - 5x = 12

2x² - 5x - 12 = 12 - 12

2x² - 5x - 12 = 0

2x² - 8x + 3x - 12 = 0

2x(x - 4) + 3(x - 4)

(2x + 3)(x - 4)

- - - - - - -

2x(x + 3) = x + 25

2x² + 6x = x + 25

2x² + 6x - x = x + 25 - x

2x² + 5x = 25

2x² + 5x - 25 = 25 - 25

2x² + 5x - 25 = 0

2x² + 10x - 5x - 25

2x(x + 5) - 5(x + 5)

(2x - 5)(x + 5)

- - - - - - - - - -s-

Answer 8

the first one x=4
the second one is x=2.5

2x^2 – 5x = 12
2(4)^2 - 5(4) = 12
2(16) - 20 =12
32 - 20 = 12
12=12

2x(x+3)=x + 25
2(2.5) (2.5 + 3) = 2.5 +25
5(5.5) = 27.5
27.5 = 27.5

Answer 9

CASE I eq.1

Given: 2x²-5x=12

Using Quadratic Formula, we have,

ax²+bx +c = 0

x= (-b +√b²-4ac)/2a

solutions:

=[5+ √(5)²-4(2)(-12)]/2(2)

=[5+√25-4(-24)]/4

=[5+√25+96]/4

=[5+√121]/4

=(5+11)/4

=16/4

= 4........................ans

To check this,

2x²-5x=12

substitute the value of x.

2(4)²-5(4) = 12

32-20=12

12=12

or you may check this way,

2x²-5x-12=0

2(4)²-5(4)-12=0

32-20-12=0

0=0
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CASE II

Given eq. 2x(x+3) = x+25

solutions:

2x² +6x = x+25

2x²+6x-x = 25

2x² + 5x -25 = 0

Using the Quadratic Equation Formula:

ax²+bx+c = 0

x= [-b+√b²-4ac]/2a

x= [-5+√(5)²-4(2)(-25)]/2(2)

x=[-5+√25+200]/4

x=[-5+√225]/4

=(-5+15)/4

=10/4

=2.5 .......................answer

to prove it's correct!,

2x² + 5x-25 = 0

2(2.5)² +5(2.5) - 25 = 0

2(6.25) +12.5 -25 = 0

12.5 + 12.5 -25 =0

25-25 =0

0=0

Answer 10

2x^2-5x=12
2x^2-5x-12=0
(x-4)(x+1.5)
x=4, x=-1.5 (solved)


2x(x+3) = x+25
2x^x+6x = x+25
2x^x+6x-x-25=0
2x^x+5x-25=0
(x-2.5)(x+5)=0
x=2.5, x=-5 (solved)