x2+6x+5=0

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kingattaskus12's profile pic

kingattaskus12 | (Level 3) Adjunct Educator

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Factor the trinomial `x^2 + 6x + 5`

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


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

`x+5=0`
`x+1=0 `


Since `5` does not contain the variable to solve for, move it to the right-hand side of the equation by subtracting `5` from both sides.

`x=-5 `
`x+1=0 `


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

`x=-5 `
`x+1=0 `


Since 1 does not contain the variable to solve for, move it to the right-hand side of the equation by subtracting `1` from both sides.

`x=-5 `
`x=-1`


The complete solution is the set of the individual solutions.

`x=-5,-1`  

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baxthum8's profile pic

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

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Solve `x^2 + 6x + 5 = 0` .

Factor the quadratic.

`x^2 + 6x + 5 = 0`

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

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

`x+5 = 0` and `x+1 = 0`

`x = -5` and `x = -1`

The solutions are x = -5, -1.

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

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QUESTION:-

 `x^2+6x+5=0`

``

SOLUTION:-

METHOD # 1:-

Breaking down the quadratic equation;

`x^2+6x+5=0`

x^2 + 1x + 5x + 5 = 0

Take x  and 5 common; 

x(x+1) + 5(x+1) = 0

Therefore;

 x + 1= 0       ,        x+5 =0

 x = -1           ,        x = -5  

 Hence solved!

METHOD # 2:-

Solving for x by using quadratic formula;

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

x = {-(6)+- sqrt((6)^2 - 4(1)(5))} /(2*1)`

x = {-6+- sqrt (36-20)}/2

x = {-6 +- sqrt(16)} / 2

x = {-6 +- 4} /2

Take 2 common in numerator;

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

The 2 in numerator and the 2 in denominator are cancelled with each other;

x = -3 +-2

x = -3+2               ,                    x = -3-2

x = -1                   ,                    x = -5

By both methods the answers we are getting are as follows;

x = -1

x = -5

Hence Solved.

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malkaam | Student, Undergraduate | (Level 1) Valedictorian

Posted on

x2 + 6x + 5 = 0

Since this is a quadratic equation we can solve it with the help of quadratic formula,

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

Where 

a= 1

b= 6

c= 5

Input the values in the formula,

`x= [-6+-sqrt(6^2-4*1*5)] / (2*1)`

`x= [-6+-sqrt(36-20)] / 2`

`x= [-6+-sqrt(16)] / 2`

`x= [-6+-4] / 2`

`x= [-6+4] / 2 => -2/2 => -1 `

`x= [-6-4] / 2 => -10/2 => -5`

`Solution Set: {-1,-5}`

` `

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acompanioninthetardis | Student, Undergraduate | (Level 1) Valedictorian

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x^2+6x+5=0

what multiplies to get 5 and add to get 6

x^2+x+5x+5=0

(x^2+x)+(5x+5)=0

x(x+1)+5(x+1)=0

(X+5) (X+1)=0

x=-5, x=-1

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

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`x^2+6x+5=0`

`x^2+1x+5x+5=0`

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

`x(x+1)` `5(x+1)`

`(x+5) `  `(x+1)`

`x=-5`  `x=-1`

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