How do you solve #4x^2 - 5x=0# using the quadratic formula?

3 Answers
Apr 11, 2018

Answer:

#x=0 or x=5/4#

Explanation:

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#

Apr 11, 2018

Answer:

#x=5/4 or x=0#

Explanation:

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#

Apr 11, 2018

Answer:

#x=0,5/4#

Explanation:

#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#