Question

find the roots of 4x square +3x+5=0
by completing the square method

Answer 1

Here, 4x square +3x+5=0
=> (2x) square +2 ✖ 2x ✖ 3/4x+(3) square -
(3) square +5 = 0
=> (2x+3) square - (3) square + 5 = 0
=> (2x+3) square = (3)square - 5
=> (2x+3) square = 9 - 5
=> (2x+3) square = 4
=> (2x+3) = +2,(-2)
=> 2x = (2 - 3) or (-2 -3)
=> x = (-1)/2 or (-5)/2

Thus, either :- x = (-1)/2. or :- x = (-5)/2

Answer 2

Heya friend !!!!

we have ,
4x^2+3x+5=0.

=>4x^2+3x+5=0

=>(2x)^2+3x=-5

◆for finding in the form of complete square we will be thinking that what can I do it become in the form of complete square..then as you know ,
(a+b)^2=a^2+b^2+2a*b then we set up this equation in form of (a+b)^2 so we can add, subtract multiply and divide by some no..but my equation must be as equal to as ago


so,we saw ,(2x)^2+2*2x*3/4+(3/4)^2=-5+(3/4)^2
adding (3/4)^2on both side..

(2x+3/4)^2=-5+9/16

=>-80+9/16

=>-71/16<0
but here -71/16 is smaller than 0.it means (2x+3/4)^2 is smaller than 0 also.

so ,(2x+3/4)^2 cannot be negative for any real value if x.

so there is no real value of x satisfying the equation .


Hence ,the given equation has no real root.......



hope it help you...
.
@rajukumar ☺☺☺☺