Question

Middle term split
=5y²-7y+1​

Answer 1

Step-by-step explanation:

Given:-

Expression 5y^2-7y+1

To find:-

Factorise the given expression?

Solution:-

Given expression is 5y^2-7y+1

On comparing with the standard quadratic pilynomial ax^2+bx+c then we have

a = 5

b=-7

c= 1

To get zeroes we can write it as 5y^2-7y+1 = 0

=> On dividing by 5 both sides

=>(5y^2/5)-(7y/5)+(1/5)=0

=>y^2 -(7y/5) +(1/5) = 0

=>y^2 -(7y/5) = -1/5

=> y^2 -(2/2)(7y/5) = -1/5

=>y^2 -2(y)(7/10) = -1/5

=>On adding both sides (7/10)^2 then

=>y^2 -2(y)(7/10) + (7/10)^2 = -1/5 +(7/10)^2

=>[y-(7/10)]^2 = (-1/5)+(49/100)

=>[y-(7/10)]^2 = (-20+49)/100

=>[y-(7/10)]^2 = 29/100

=>[y-(7/10)] = ±√(29/100)

=>y = ±√(29/100)+(7/10)

=>y = (7/10)±√29/10

=>y = (7±√29)/10

=>y = (7+√29 )/10 and (7-√29)/10

=>y - {(7+√29 )/10} = 0 and y - { (7-√29)/10} = 0

=>[y-{(7+√29)/10}][y-{(7-√29)/10}]

=>[(10y-7-√29)/10]([10y-7+√29)/10]

=>[(10y-7-√29)][10y-7+√29]/100

Answer:-

5y^2-7y+1 = [y-{(7+√29)/10}][y-{(7-√29)/10}]