Solve the equation x^2 - 6x + 5 = 0 use the quadratic formula and factorization.

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aespinoza71's profile pic

aespinoza71 | (Level 1) Adjunct Educator

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In determining the factorization, it is a good trick to know that when the coefficient of the x^2 is indeed one (as it is in this case), solving for the two factors is much easier.

Basically, what two numbers, when multiplied, equal positive 5, but when added together, equal negative 6?

The only possibility is -1 and -5.  Therefore (x-1), (x-5)

To check your answer, you can use the FOIL method of multiplying your two factors (first, outside, inside, last), to arrive back to the original equation:

First = x times x equals x^2.

Outside = x times -5 equals -5x.

Inside = -1 times x equals -x.

Last = -1 times -5 equals 5.

Combined then you have x^2 -5x -x +5, simplified to x^2-6x+5.

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justaguide | College Teacher | (Level 2) Distinguished Educator

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The solution of a quadratic equation ax^2 + bx + c = 0 is given by `(-b+-sqrt(b^2-4ac))/(2*a)`

In the problem, the solution of the equation x^2 - 6x + 5 = 0 has to be determined. Using the formula given earlier, a = 1, b = -6 and c = 5.

The roots of the equation are `(6 +-sqrt(36-20))/(2*1) = (6+-4)/2`

The roots are 5 and 1.

It is also possible to factorize the expression x^2 - 6x + 5 to determine the roots of the equation.

x^2 - 6x + 5 = 0

=> x^2 - 5x - x + 5 = 0

=> x(x - 5) - 1(x - 5) = 0

=> (x - 1)(x - 5) = 0

=> x = 1 and x = 5

Again, the same roots are obtained.

The roots of the equation x^2 - 6x + 5 = 0 are 1 and 5

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

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Question:-

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

SOLUTION:-

Factorization Method:-

Now according to this method first we gotta separate the middle number in such a way that it can be further simplified;

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

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

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Now divide the two solution sets like this:-

x - 5 = 0                   ,             x - 1 = 0

x = 5                        ,             x = 1

Hence the Solution set is (5,1)

Quadratic Equation Formula Method:-

In this method we will be using the formula;

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

Where;

a = 1

b = -6

c = 5

Insert values in the equation:-

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

Since (- * - = +) i.e, minus sign when multiplied to minus sign, becomes an addition sign;

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

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

Since the sqrt of 16 is 4, therefore;

x = (6 +-4)/2

Separating the two solutions;

x = (6 + 4)/2                   ,             x = (6 - 4)/2

x = 10/2                         ,              x = 2/2

x = 5                              ,              x = 1

Hence the solution set is (5,1)

Hence Solved

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malkaam's profile pic

malkaam | Student, Undergraduate | (Level 1) Valedictorian

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x^2 - 6x + 5 = 0

Applying quadratic formula:

Where a=1

          b=-6

          c=5

Input the values in the equation

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

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

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

  =(6+4)/2                                                 =(6-4)/2

  =10/2                                                      =2/2

  =5                                                           =1

Solution Set= (5,1)

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malkaam | Student, Undergraduate | (Level 1) Valedictorian

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x^2 - 6x + 5 = 0

The above equation is to be solved through the quadratic formula.

Quadratic Equation:     ax^2+bx+c=0

 Where a=1

           b=6

           c=5

We are then supposed to input the values into the quadratic equation which is:

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

  =[-6+sqrt{36-20}]/2                          =[-6-sqrt{36-20}]/2 

  =[-6+16]/2                                        =[-6-16]/2 

  =10/2                                                =-22/2

  =5                                                     =-11

Solution set: (5,-11)

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malkaam's profile pic

malkaam | Student, Undergraduate | (Level 1) Valedictorian

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This one is incorrect I've posted the correct one below.

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