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What should be taken away from 3x 2 – 4y 2 + 5xy + 20 to obtain – x 2 – y 2 + 6xy + 20?

What should be taken away from 3x 2 – 4y 2 + 5xy + 20 to obtain – x 2 – y 2 + 6xy + 20?

Grade:12

1 Answers

Harshit Singh
askIITians Faculty 5964 Points
one year ago
Dear Student

Let us assume a be the required term
Then,
3x^2–4y^2+ 5xy + 20–a = -x^2–y^2+ 6xy + 20
a = 3x^2–4y^+ 5xy + 20–(-x^2–y^2+ 6xy + 20)
a = 3x^2–4y^2+ 5xy + 20 + x^2+ y^2–6xy–20
a = 3x^2+ x^2–4y^2+ y^2+ 5xy–6xy + 20–20
a = 4x^2–3y^2–xy

​Thanks

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