How do you solve this? 2x squared - 9x - 5 = 0

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neela | High School Teacher | (Level 3) Valedictorian

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To solve 2x^2-9x-5=0.

We shall solve this  making it as  X^2 - n^2 =0 or X^2 = n^2 , where X is is function of x and n is numeric.

2(x^2-4.5x)-5=0

2[x^2-2(4.5/2)x +(4.5/2)^2]- 2(4.5/2)^2-5=0. Here 2(4.5/2)^2 is added and subtracted with the purpose of making the expression to bring to the form of 2X^2 -n^2 form.

2(x-4.5/2)^2 =  2(4.5/2)^2+5 =15.125

(x-2.25)^2= 15.125/2 =7.5625 which is in X^2=n^2 form.

Take the square root on both sides :

x-2.25 = +2.75  or x-2.25 = -2.75

x=2.25+2.75  =  5 or

x=2.25-2.75 = -0.5

krishna-agrawala's profile pic

krishna-agrawala | College Teacher | (Level 3) Valedictorian

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In this solution we will represent x squared as x^2. Therefore the given equation becomes:

2x^2 - 9x - 5 = 0

Therefore: 2x^ - 10x + x - 5 = 0

Therefore: 2x(x - 5) + 1(x - 5) = 0

Therefore:(2x + 1)*(x - 5) = 0

Therefore:(2x + 1) = 0, or (x - 5) = 0

Therefore: x = -1/2 or x = 5 Answer.

malkaam's profile pic

malkaam | Student, Undergraduate | (Level 1) Valedictorian

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2x squared - 9x - 5 = 0

We can solve with the help of quadratic formula, as the question is in quadratic equation form i.e.

2x^2-9x-5=0

Where a=2

          b=-9

          c=-5

Input the above values into the quadratic formula:

x=[9+sqrt{(-9)^2-4(2)(-5)}]/4               x=[9-sqrt{(-9)^2-4(2)(-5)}]/4   

  =[9+sqrt(81+40)]/4                               =[9-sqrt(81+40)]/4 

  =[9+sqrt(121)]/4                                   =[9-sqrt(121)]/4

  =(9+11)/4                                             =(9-11)/4

  =5                                                        =-0.5

Solution Set= (5,-0.5)

Wiggin42's profile pic

Wiggin42 | Student, Undergraduate | (Level 2) Valedictorian

Posted on

 `2x^2 - 9x - 5`
``
`(2x + 1)(x - 5) = 0`
``
`2x + 1 = 0`
`x = -1/2`
``
`x - 5 = 0`
`x = 5`
` `
atyourservice's profile pic

atyourservice | Student, Grade 11 | (Level 3) Valedictorian

Posted on

`2x^2 - 9x - 5 = 0`

a=2 b=-9 c=-5

`cxxa ` and then factors of `axxc ` that equal b

`2xx-5=-10`

`-10xx1= -10 `       `-10+1=-9`

group:

`(2x^2 - 10x) ( x - 5) `

factor out:

`2x(x - 5) + 1(x - 5) = 0 `

`(2x + 1) (x - 5)`

set parentheses equal to 0

(2x + 1) = 0

(x - 5) = 0

x = -1/2

x = 5 

 
 
giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

First of all, let's put the second grade equation under the general form:

a*x^2+b*x+c=0, where a,b,c, are the quotients and x is the variable, or unknown.

Let's identify the quotients of our equation:

a=2, b=-9, c=-5

Now, we cand use the formula for finding the two solution of the equation:

x1=[-b+sqrt(b^2-4*a*c)]/2*a

By substituting the quotients with ours:

x1=[-(-9)+sqrt(81+40)]/2*2

x1=(9+11)/4=20/4=5

x2=[-b-sqrt(b^2-4*a*c)]/2*a

x2=(9-11)/4=-2/4=-1/2

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