Solve each equation:  2x^2+9x+9=0  the x^2 is x raised to the second power! 

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baxthum8's profile pic

baxthum8 | High School Teacher | (Level 3) Associate Educator

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Solve `2x^2 + 9x + 9 = 0` .

Factor:  `(2x + 3)(x+3) = 0` .

Using the zero product property, set each factor equal to zero and solve.

`2x+3 = 0`   and `x + 3 = 0`

`2x = -3`      and `x = -3`

`x = -3/2` and` x=-3`

The solutions are `x = - 3` and `x = -3/2.`

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kingattaskus12 | (Level 3) Adjunct Educator

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We need to solve the equation for `x`

Remove the fraction by multiplying the first term of the factor by the denominator of the second term.

`(x+3)(2x+3)=0 `


Set each of the factors of the left-hand side of the equation equal to `0` .

`x+3=0`
`2x+3=0 `


Since `3` does not contain the variable to solve for, move it to the right-hand side of the equation by subtracting `3` from both sides.

`x=-3 `
`2x+3=0 `


Set each of the factors of the left-hand side of the equation equal to `0` .

`x=-3 `
`2x+3=0`


Since `3` does not contain the variable to solve for, move it to the right-hand side of the equation by subtracting `3` from both sides.

`x=-3 `
`2x=-3 `


Divide each term in the equation by `2` .

`x=-3 `

`x = -3/2`



The complete solution is the set of the individual solutions.

`x = -3, -3/2`

sid-sarfraz's profile pic

sid-sarfraz | Student, Graduate | (Level 2) Salutatorian

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QUESTION:-

Solve each equation: 

2x^2+9x+9=0 

the x^2 is x raised to the second power! 

SOLUTION:-

We can solve this problem by two methods, which are as follows:-

  1. Factorization Method
  2. Quadratic Equation Formula Method

1.  FACTORIZATION METHOD

`2x^2+9x+9=0`

`2x^2+6x+3x+9=0`

`2x(x+3)+3(x+3)=0`

`Hence,`

`2x+3=0,x+3=0`

`2x=-3,x=-3`

`x=-3/2,x=-3`

`Solution Set:[-3/2,-3]`

2.   QUADRATIC EQUATION FORMULA METHOD

Quadratic Equation is;

`x={-b+-sqrt(b^2-4ac)}/(2a)`

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Where,

a = 2 

b = 9

c = 9

Now insert the values of a,b and c in the formula;

`x={-9+-sqrt((9)^2-4(2)(9))}/(2*2)` 

Now Simplify,

`x={-9+-sqrt((81)-(72))}/4`

`x={-9+-sqrt(9)}/4`

`x=(-9+-3)/4`

`Hence,`

`x=(-9-3)/4,x=(-9+3)/4`

`x=-12/4,x=-6/4`

`x=-3,x=-3/2`

`SolutionSet:{-3,-3/2}`

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malkaam's profile pic

malkaam | Student, Undergraduate | (Level 1) Valedictorian

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2x^2 + 9x + 9 = 0

This can solved with the help of quadratic formula,

`x = [-b+-sqrt(b^2-4ac)] / (2a)`

Where

a= 2

b= 9

c= 9

Input the values in the formula:

`x = [-9+-sqrt(9^2-4*2*9)]/ (2*2)`

`x = [-9+-sqrt(81-72)] / 4`

`x = [-9+-sqrt(9)] / 4`

`x = [-9+-3] / 4`

`x = [-9+3] / 4 => -6/4 => -3/2`

`x= [-9-3] / 4 => -12/4 => -3`

Solution Set: {-3/2,-3}.

atyourservice's profile pic

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

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`2x^2+9x+9=0 `

a=2  b=9  c=9 

multiply a by b
`2xx9=18`

find factors of 18 that add up to 9, which would be 6 and 3, plug those in as b

`2x^2+6x+3x+9 `   group the numbers

`(2x^2+6x)(3x+9)` factor out factors the numbers in the parenthesis have in common.

`2x(x+3) 3(x+3)`   now put the numbers outside in parentheses

`(2x+3) (x+3)`

set them equal to 0 and solve

x+3=0

  -3   -3

x=-3

2x+3=0

    -3  -3

`2x=-3`

`(2x)/2 = (-3)/2`

`x=(-3)/2`

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