# Solve `x^2 + 24 = 11x` by using the quadratic formula. Identify a,b, and c. Identify the solution set.

lemjay | Certified Educator

`x^2+24=11x`

To solve this using quadractic formula

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

the equation should be in the form `ax^2+bx+c=0` .

So, to have zero at the right side, subtract both sides by 11x.
`x^2 +24-11x = 11x -11x`

`x^2-11x + 24 = 0`

Base on that, the values of a, b and c are:

a=1, b=-11 and c=24

Now that their values are known, plug-in them to the quadratic formula.

`x=(-(-11)+-sqrt((-11)^2-4(1)(24)))/(2*1)`

`x=(11+-sqrt(121-96))/2`

`x=(11+-sqrt25)/2`
`x=(11+-5)/2`

`x_1=(11+5)/2=8`

`x_2=(11-5)/2=3`

Therefore, the solution set for `x^2+4x=11x`   is  `{3,8}` .

sid-sarfraz | Student

QUESTION:-

Solve `x^2+24=11x` by using the quadratic formula.

• Identify a,b, and c.
• Identify the solution set.

SOLUTION:-

The quadratic formula for a equation is;

`ax^2+bx+c=0`

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

The given equation;

`x^2+24=11x`

`x^2-11x+24=0`

Where;

a = 1

b = -11

c = 24

Insert the values of a , b and c in the quadratic equation;

`x={-(-11)+-sqrt[(-11)^2-4(1)(24)]}/(2*1)`

` `

Now simplify:

`x={11+-sqrt[121-96]}/2`

`x=[11+-sqrt(25)]/2`

`x=[11+-5]/2`

The two values of x are as follows:-

`x=[11+5]/2,x=[11-5]/2`

`x=16/2,x=6/2`

`x=8,x=3`

Hence the solution set is:{8,3}

Hence Solved!

malkaam | Student

In order to solve this through quadratic formula, we need to change it to this equation format ax^2+bx+c=0

x^2-11x+24=0

Where, a=1

b=-11

c=24

x=[-(-11)+sqrt{(-11)^2-4(1)(24)}]/2(1)

x=[-(-11)-sqrt{(-11)^2-4(1)(24)}]/2(1)   ` `

x=[11+sqrt{121-96}]/2

x=[11-sqrt{121-96}]/2

x=[11+sqrt{25}]/2

x=[11-sqrt{25}]/2

x=[11+5]/2

x=[11-5]/2

x=16/2

x=6/2

ayl0124 | Student

Look at the function:

`y = ax^2 + bx + c`

Change the form of your original equation.

`y = x^2 -11x + 24`

Your "a" would be 1, your "b" would be -11, and your "c" would be 24.

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

Plug in your a, b, and c values.

`x = (11 + sqrt(121+(4)(1)(24)))/((2)(1)) = 8`

`x = (11 - sqrt(121 + (4)(1)(24)))/((2)(1)) = 3`

You can also factor the equation.

`y = (x-8)(x-3)`

Jyotsana | Student

x^2-11x+24=0

x=(-(-11)+square root (-11)^2-4(1)(24))/2(1)

x=(-(-11)-square root(-11)^2-4(1)(24))/2(1)

x=8  or x=3