How do you solve #4x^2 - 5x=0# using the quadratic formula?
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The quadratic formula for #ax^2+bx+c=0# is given by #x=(-b+-sqrt(b^2-4ac))/(2a)#
#a=4, b=-5, c=0#
#therefore x=(-(-5)+-sqrt((-5)^2-4(4)(0)))/(2(4))#
#x=(5+-sqrt(25))/8#
#x=(5+-5)/8=>x=0 or x=10/8=5/4#
The equation #y=4x^2-5x=0# is written in the form #y=ax^2+bx+c#,
so
#a=4#, #b=-5#, #c=0#
The quadratic formula is #x=(-b+-sqrt(b^2-4ac))/(2a)#
Substitute the values of a, b and c into the formula
#x=(5+-sqrt(25))/(8)#
#x=(5+sqrt(25))/(8)# or #x=(5-sqrt(25))/(8)#
#x=(10)/(8)# or #x=0/(8)#
#x=5/4 or x=0#
#4x^2-5x=0# is a quadratic equation in standard form:
#ax^2+bx+c=0#,
where:
#a=4#, #b=-5#, #c=0#
Quadratic Formula
#x=(-b+-sqrt(b^2-4ac))/(2a)#
Plug in the known values and solve.
#x=(-(-5)+-sqrt((-5)^2-4*4*0))/(2*4)#
Simplify.
#x=(5+-sqrt25)/8#
#x=(5+-5)/8#
#x=(5+5)/8=10/8=5/4#
#x=(5-5)/8=0/8=0#
#x=0,5/4#
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#4x^2-5x=0# can also be solved by factoring.
Factor out the common #x#.
#x(4x-5)=0#
#x=0#
#4x-5=0#
#4x=5#
#x=5/4#
#x=0,5/4#
