# (solved)Question 9 SSC-CGL 2020 March 4 Shift 3

If 2x² + y² + 8z² - 2√2xy + 4√2yz - 8zx = (Ax + y + Bz)², then the value of A² + B² - AB =?
(Rev. 04-Jun-2024)

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### Question 9SSC-CGL 2020 Mar 4 Shift 3

If 2x² + y² + 8z² - 2√2xy + 4√2yz - 8zx = (Ax + y + Bz)², then the value of A² + B² - AB =?

### Solution in Short

Coefficient of x² on LHS is 2, and orally we can see that it is A² on the RHS. So A² = 2, and therefore A = √2. Similarly comparing coeff. of z², we can get B² = 8, and therefore B = 2√2. Finally, A² + B² - AB = 2 + 8 - 2√2 x √2 = 6 ans!

### Solution in Detail

We shall use the well known algebraic principle that if two algebraic expressions are equal then the coefficients of like terms on both sides should be equal.

[1] on LHS coeff. of $\displaystyle x^2 = 2$

To obtain the coefficient on the RHS we do not have to actually open the square. We can orally observe that the term on RHS is Ax². Hence

[2] on RHS coeff. of $\displaystyle x^2 = A^2$

Equating, $\displaystyle A^2 = 2\text{. . . (1)}$

Similarly, equating for $\displaystyle z^2$

$\displaystyle B^2 = 8, \therefore B = 2\sqrt 2$

Both A and B are known. Substituting in A² + B² - AB, we can get

$\displaystyle 2 + 8 - 2\sqrt 2 \cdot \sqrt 2$

$\displaystyle = 6\:\underline{Ans}$

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