Question

How do you use the quadratic formula to find the solutions of `x^2-21=4x` ?

Answer 1

`x = 7" "` or `" "x = -3`

Explanation:

Given:

`x^2-21=4x`

Subtract `4x` from both sides to get:

`x^2-4x-21 = 0`

This is now in standard form:

`ax^2+bx+c = 0`

with `a=1`, `b=-4` and `c=-21`.

The roots are given by the quadratic formula:

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

`color(white)(x) = (-(color(blue)(-4))+-sqrt((color(blue)(-4))^2-4(color(blue)(1))(color(blue)(-21))))/(2(color(blue)(1)))`

`color(white)(x) = (4+-sqrt(16+84))/2`

`color(white)(x) = (4+-sqrt(100))/2`

`color(white)(x) = (4+-10)/2`

`color(white)(x) = 2+-5`

That is:

`x = 7" "` or `" "x = -3`

Answer 2

`x=-3" or "x=7`

Explanation:

`"rearrange into standard form":ax^2+bx+c=0`

`rArrx^2-4x-21=0larrcolor(blue)"in standard form"`

`"with the equation in standard form we can solve using the"`
`color(blue)"quadratic formula"`

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

`"here "a=1,b=-4" and "c=-21`

`x=(4+-sqrt(16+84))/2`

`color(white)(x)=(4+-sqrt100)/2=(4+-10)/2=(-6)/2" or "14/2`

`rArrx=-3" or "x=7`

`"this can also be solved by factorising"`

`x^2-4x-21=0`

`"the factors of - 21 which sum to - 4 are - 7 and + 3"`

`rArr(x-7)(x+3)=0`

`"equate each factor to zero and solve for x"`

`x-7=0rArrx=7`

`x+3=0rArrx=-3`

Answer 3

See below.

Explanation:

First you need to make sure if the equation is in the form of `ax^2 + bx + c = 0` where `a != 0`.

`x^2 - 21 = 4x`
`rArr x^2 - 4x - 21 = 0`

Now it is in the desired form. Let's compare this equation with the general form.

After comparing, we get,

`a = 1`, `b = -4`, `c = -21`

Now Let's find out the Discriminant.

`D` = `b^2 - 4ac` = `(-4)^2 - 4 * 1 * (-21)` = `16 + 84` = `100 > 0`

That means, the equation has two real, and distinct roots.

Now Applying Quadratic Formula (Sridhar Acharya's Formula) :

`alpha = (- b + sqrt(D))/(2a) = (4 + sqrt(100))/(2) = 7`

`beta = (-b - sqrt(D))/(2a) = (4 - sqrt(100))/2 = -3`

These are the two roots, `7` and`-3`.

Hence Explained.