Question

Resolve gy square + 33y + 10 by the method of splitting the middle term​

Answer 1

Answer :

y = -1/3 , -10/3

Solution :

• Given : 9y² + 33y + 10 = 0
• To find : Roots by middle term splitting method

We have ;

=> 9y² + 33y + 10 = 0

=> 9y² + 3y + 30y + 10 = 0

=> 3y(3y + 1) + 10(3y + 1) = 0

=> (3y + 1)(3y + 10) = 0

Here ,

Two cases arises . Either (3y + 1) = 0 or (3y + 10) = 0 .

Case 1 :

If 3y + 1 = 0 , then

=> 3y = -1

=> y = -1/3

Case 2 :

If 3y + 10 = 0 , then

=> 3y = -10

=> y = -10/3

Hence ,

y = -1/3 , -10/3

Answer 2

Correct Equation:-

${9y}^{2}+33y+10=0$

Required Answer:-

${9y}^{2}+33y+10=0$

• First we need to find two numbers by which If we multiply them we get (10×9)=90 ,and if we adds them we get 33
• The numbers are 30 and 3
• as 30×3=90 and 30+3=33

${:}\longmapsto$${9y}^{2}+3y+30y+10=0$

• Now take common polynomials

${:}\longmapsto$$3y (3y+1)+10 (3y+1)=0$

${:}\longmapsto$$(3y+1)(3y+10)=0$

${:}\longmapsto$$(3y+1)=0{\quad}or {\quad}(3y+10)=0$

• Now interchange sides in both equations and get possible values of y

${:}\longmapsto$$3y=-1{\quad}or {\quad}3y=10$

${:}\longmapsto$$y={\dfrac {-1}{3}}{\quad},y={\dfrac {-10}{3}}$

$\therefore$$\sf {Possible\;values\;of\;y={\boxed {{\dfrac{-1}{3}}{\quad},{\dfrac {-10}{3}}}}}$