Answer: Given: 2y²+3y-2=0 To find: The zeroes of the equation by middle term splitting Proof: = 2y²+3y-2=0 ⇒ 2y²+3y-2=0 ⇒ 2y²+4y-y-2=0 ⇒ 2y²+4y-y-2=0 ⇒ 2y(y+2) -1 (y +2) =0 ⇒ (2y-1) (y+2) =0 ⇒ (2y-1) = 0 or (y+2) =0 ⇒ 2y= 1 or (y+2) =0 ⇒ y= [tex]\frac{1}{2}[/tex] or y = -2 Hence, y= [tex]\frac{1}{2}[/tex] or y = -2 Hope you got that. Thank You. Reply
Answer: y=-2 or y=½ Step-by-step explanation: 2y²+3y-2=0 2y²+4y-1y-2=0 2y(y+2)-1(y+2)=0 (y+2) (2y-1)=0 y=-2 or y=½ this is the ans tried it yay Reply
Answer:
Given: 2y²+3y-2=0
To find: The zeroes of the equation by middle term splitting
Proof:
= 2y²+3y-2=0
⇒ 2y²+3y-2=0
⇒ 2y²+4y-y-2=0
⇒ 2y²+4y-y-2=0
⇒ 2y(y+2) -1 (y +2) =0
⇒ (2y-1) (y+2) =0
⇒ (2y-1) = 0 or (y+2) =0
⇒ 2y= 1 or (y+2) =0
⇒ y= [tex]\frac{1}{2}[/tex] or y = -2
Hence, y= [tex]\frac{1}{2}[/tex] or y = -2
Hope you got that.
Thank You.
Answer:
y=-2 or y=½
Step-by-step explanation:
2y²+3y-2=0
2y²+4y-1y-2=0
2y(y+2)-1(y+2)=0
(y+2) (2y-1)=0
y=-2 or y=½
this is the ans
tried it yay