Question

How do you solve `9r^2+49=42r`?

Answer 1

Please see the explanation for steps leading to the solution `r = 7/3`

Explanation:

Subtract 42r from both sides:

`9r^2 - 42r + 49 = 0`

Please notice that this is a quadratic of the form `ar^2 + br + c = 0`, therefore, the values of r can be determined, using the quadratic formula:

`r = {-b + sqrt(b^2 - 4(a)(c))}/(2a)` and `r = {-b - sqrt(b^2 - 4(a)(c))}/(2a)`

But, before we do that, let's check the value under the `sqrt`, called the determinant:

`b^2 - 4(a)(c) = (-42)^2 - 4(9)(49) = 0`

This means that the quadratic factors into a perfect square:

`(3r - 7)^2 = 0`

This is called a repeated root and it yields only one value for r:

`3r - 7 = 0`

`3r = 7`

`r = 7/3`

Answer 2

`r = 7/3`

Explanation:

The `r^2` is the indication that this is a quadratic equation. so make it equal to 0.

`9r^2 -42r +49 = 0`

Find factors of `9 and 49` which ADD to give 42.

Try different combinations of the factors and cross-multiply until the sum of the products gives 42.

`color(white)(xxxx)9 and 49`

`color(white)(xxxx)3" "7" "rarr 3 xx 7 =21`
`color(white)(xxxx)3" "7" "rarr 3 xx 7 =ul21`
`color(white)(xxxxxxxxxxxxxxxxxxxxx)42`

The signs will be the same, they are both negative.

`(3r-7)(3r-7) = 0`

There are two equal solutions..

`3r - 7 =0`

`3r = 7`

`r = 7/3`