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

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### 5 Answers

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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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`

**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:-

- Factorization Method
- 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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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}.**

`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`