Question

How do you solve `h^2+48=16h`?

Answer 1

See a solution process below:

Explanation:

First, subtract `color(red)(16h)` from each side of the equation to put the equation in standard quadratic form:

`h^2 - color(red)(16h) + 48 = 16h - color(red)(16h)`

`h^2 - 16h + 48 = 0`

Now, we can use the quadratic equation to solve this problem:

The quadratic formula states:

For `color(red)(a)x^2 + color(blue)(b)x + color(green)(c) = 0`, the values of `x` which are the solutions to the equation are given by:

`x = (-color(blue)(b) +- sqrt(color(blue)(b)^2 - (4color(red)(a)color(green)(c))))/(2 * color(red)(a))`

Substituting:

`color(red)(1)` for `color(red)(a)`

`color(blue)(-16)` for `color(blue)(b)`

`color(green)(48)` for `color(green)(c)` gives:

`h = (-color(blue)(-16) +- sqrt(color(blue)(-16)^2 - (4 * color(red)(1) * color(green)(48))))/(2 * color(red)(1))`

`h = (16 +- sqrt(256 - 192))/2`

`h = (16 +- sqrt(64))/2`

`h = (16 - 8)/2` and `h = (16 + 8)/2`

`h = 8/2` and `h = 24/2`

`h = 4` and `h = 12`

The Solution Set Is: `h = {4, 12}`

Another method is to factor the quadratic as:

`(h - 12)(h - 4) = 0`

The solve each term on the left for `0`:

Solution 1:

`h - 12 = 0`

`h - 12 + color(red)(12) = 0 + color(red)(12)`

`h - 0 = 12`

`h = 12`

Solution 2:

`h - 4 = 0`

`h - 4 + color(red)(4) = 0 + color(red)(4)`

`h - 0 = 4`

`h = 4`

The Solution Set Is: `h = {4, 12}`

Answer 2

`h=4" or "h=12`

Explanation:

`"rearrange equation into "color(blue)"standard form"`

`rArrh^2-16h+48=0larrcolor(blue)"in standard form"`

`"the factors of + 48 which sum to - 16 are - 12 and - 4"`

`rArr(h-12)(h-4)=0`

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

`h-12=0rArrh=12`

`h-4=0rArrh=4`

`rArrh=12" or "h=4" are the solutions"`