Question

solve the following quadratic equation by factorization method 10y2-y-2=0​

Answer 1

Answer:

${10y}^{2} - y - 2 = 0 \\ = > {10y}^{2} - 5y + 4y - 2 = 0 \\ = > 5y(2y - 1) + 2(2y - 1) \\ = >(2y - 1)(5y + 2) \\ \\ y = \frac{1}{2} \\ y = \frac{ - 2}{5}$

Tge above question is solved using splitting the middle term .

Here , a=10 and c=1

acc. to splitting the middle term the middle term term, here. it is 2y should be divided into to such numbers which when multiplied results in ac and SUBTRACTED or ADDED results the middle term.

ac ,= 10 × 2 =20

middle term is 1

two such numbers are 5 and 4.

So , you divide the middle term as -5+4

that is 1.

select common from the two pairs of equation.

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Answer 2

Answer:

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A) 10y^2-y-2

=10y^2-(5y-4y)-2

=10y^2-5y+4y-2                       (breaking the middle term)

=5y(2y-1)+2(2y-1)                    (taking common)

=(2y-1)(5y+2)

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