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3x2 +10xy +8y2

Posted in Math, asked by erica, 6 years ago. 1837 hits.

1

given quadratic equation is 3x^2+10xy+8y^2

 we can solve this problem splitting the middle term, 

we cam multiply the first term and last term then ,24x^2y^2

we can rewrite the product equal to the middle term,(6xy)(4xy)=6xy+4xy

= 3x^2+4xy+6xy+8y^2

=x(3x+4y)+2y(3x+4y)

=(3x+4y)(x+2y),

if you want to continue the problem and has to find x,y values

    we equalize the two factors with zero then,3x+4y=0----(1),x+2y=0-----(2)

                                                                              x+2y=0

                                                                              x=-2y

now we can put the x value in equation in(1) then

                          3(-2y)+4y=0

                            -6y+4y=0=>-2y=0=> y=0 ,now x=0

 

 

 

 

 

 

Venkat Nagendra Thati
Venkat Nagendra Thati - 6 years ago
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3x2 +10xy +8y2

solution:-

Those types of question first we see to solve it by formula if that not solve with formula then we solve it by MIDDLE TERM BREAK 

BY MIDDLE TERM BREAK


for middle term break first we multiply first and last  term we get 3x²  x (+8y²) = 24x²y² 

now we break the middle term in such a manner that if we multiply we get +24x²y²  and add or subtract we return get 10xy 

3x2 +10xy +8y2

3x2 + 6xy + 4xy + 8y²

take  common in each pair (in common the term should b lowest power variable)

3x(x + 2y) + 4y(x + 2y)

again common

(x + 2y)(3x + 4y)

If you find the values of x and y then equalise it with zero.

Furqan Ali
Furqan Ali - 6 years ago
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