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

2 Answers
Mar 17, 2018

Answer:

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}#

Mar 17, 2018

Answer:

#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"#