# (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. 27-Oct-2023)

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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.

 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

 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

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

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