Question

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Answer 1

Given that,

$x - 4 = 3 \sqrt{x}$

Squaring both sides,

$\longrightarrow (x - 4) {}^{2} = (3 \sqrt{x}) {}^{2}$

Expanding L.H.S. using (a - b)² = a² + b² - 2ab, with a = x and b = 4,

$\longrightarrow {x}^{2} + {4}^{2} - 2(x)(4) = 9x$

$\longrightarrow {x}^{2} + 16 - 8x = 9x$

$\longrightarrow {x}^{2} + 16 - 8x - 9x = 0$

$\longrightarrow {x}^{2} - 17x + 16 = 0$

Splitting the middle term,

$\longrightarrow {x}^{2} - 16x - x + 16 = 0$

$\longrightarrow x(x - 16) - 1(x - 16) = 0$

$\longrightarrow (x - 16)(x - 1) = 0$

So,

either $(x-16) = 0$

or $(x - 1) = 0$

If (x - 16) = 0:

$x - 16 = 0$

$\longrightarrow x_{1} = 16$

If (x - 1) = 0:

$x - 1 = 0$

$\longrightarrow x_{2} = 1$

Verification:

1st solution:

$x - 4 = 3 \sqrt{x}$

Putting $x = 16,$

$\longrightarrow 16 - 4 = 3 \sqrt{16}$

$\longrightarrow 12 = 3 \times 4$

$\longrightarrow 12 = 12$

As L.H.S. = R.H.S., it is verified that $x=16$ is a solution of this equation.

2nd solution:

$x - 4 = 3 \sqrt{x}$

Putting $x=1,$

$\longrightarrow 1 - 4 = 3 \sqrt{1}$

$\longrightarrow - 3 = 3$

As L.H.S. ≠ R.H.S., $x=1$ is not a solution of this equation.

Hence this equation has only one solution,

$\longrightarrow \underline{\underline{x=16}}$