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Physics Guru is an online education and competitive examinations preparation portal founded by Manish Verma, an IIT Madras alumnus, providing learning resources for competitive examinations.

The emphasis is on conceptual understanding, analytical thinking and problem solving rather than simply memorising formulas and standard methods.

The objective is to help students understand what they are learning and develop the ability to solve problems they have not seen before.

Solve for x & y

$\sqrt x + y = 7$

$x + \sqrt y = 11$

Putting $y = {(11 - x)^2}$ obtained from the 2nd equation into the 1st equation,

We have, $\sqrt x  + {(11 - x)^2} = 7$

$ \Rightarrow x = {[7 - {(11 - x)^2}]^2}$

$ = 49 + {(11 - x)^4} - 14{(11 - x)^2}$

Let, $11 - x = t$

So, $11 - t = 49 + {t^4} - 14{t^2}$

$ \Rightarrow {t^4} - 14{t^2} + t + 38 = 0$

t=2 satisfies the above equation. So, using factor theorem

${t^3}(t - 2) + 2{t^2}(t - 2) - 10t(t - 2) - 19(t - 2) = 0$

$ \Rightarrow (t - 2)({t^3} + 2{t^2} - 10t - 19) = 0$ ........... (A)

Let, $f(t) = {t^3} + 2{t^2} - 10t - 19$

$t = 11 - x = \sqrt y  \ge 0$

$y = 7 - \sqrt x , \Rightarrow y \le 7$

So, $t = \sqrt y  \le \sqrt 7 $

Thus, $0 \le t \le \sqrt 7 $

Or, $0 \le t < 3$

Let us investigate the behaviour of f in the interval [0, 3].

f(0) = -19

f(1) = -26

f(2) = -23

f(3) = -4

The function decreases initially, then increases but is unable to cut the t-axis in the interval [0, 3).

t = 2 is the only solution as obtained earlier as per (A).

Now, $11 - x = 2$ or $x = 9$

$y = {t^2} = 4$