How do you solve #h²=32-4h#?

Question

Avery Alexander · Accepted Answer

Convert to standard form then either factor or use the quadratic formula to get
#color(white)("XXX")h=-8 or h=4# Given #color(white)("XXX")h^2=32-4h# Re-write into standard form: #color(white)("XXX")color(red)((1)h^2color(blue)(+4)hcolor(green)(-32) = 0# Option 1 Recognize the factoring: #color(white)("XXX")(h+8)(h-4)=0# #rarrcolor(white)("XXX")h=-8 or h=4# Option 2 Apply the quadratic formula for roots: #color(white)("XXX")h=(-color(blue)(b)+-sqrt(color(blue)(b)^2-4color(red)(a)color(green)(c)))/(2color(red)(a))# in this specific case: #color(white)("XXX")h=(-color(blue)(4)+-sqrt((color(blue)(4))^2-4(color(red)(1))(color(green)(-32))))/(2(color(red)(1)))# #color(white)("XXX")=(-4+-sqrt(16+128))/2# #color(white)("XXX")=(-4+-sqrt(144))/2# #color(white)("XXX")=-2+-6# #rarrcolor(white)("XXX")h=+4 or h=-8#

Kennedy Adams · Answer

#h=4,-8#

#h^2=32-4h#

Gather all terms to one side of the equation and arrange the equation in standard form.

#h^2+4h-32=0#

Determine two numbers that when added equal #4# and when multiplied equal #-32#. The numbers #8# and #-4#.

Rewrite the equation in factored form.

#color(red)((h-4))color(blue)((h+8))=0#

Set each binomial equal to zero and solve for #h#.

#color(red)(h-4)=0#

#color(red)(h=4)#

#color(blue)(h+8)=0#

#color(blue)(h=-8)#

#h=color(red)4, color(blue)(-8)#

Genesis Allen · Answer

undefined To solve the equation h² = 32 - 4h, follow these steps:

1. Move all terms to one side of the equation to set it equal to zero:
   h² + 4h - 32 = 0

2. Factor the quadratic equation:
   (h - 4)(h + 8) = 0

3. Set each factor equal to zero and solve for h:
   h - 4 = 0  -->  h = 4
   h + 8 = 0  -->  h = -8

Therefore, the solutions to the equation are h = 4 and h = -8.