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Question

Solve the following equations:
$$\sqrt {3x^{2} - 7x - 30} + \sqrt {2x^{2} - 7x - 5} = x + 5$$.


Solution

$$ \sqrt {3x^{2} - 7x - 30} - \sqrt {2x^{2} - 7x - 5} = x - 5 \qquad \qquad \ldots \ldots (1)$$

The identity, $$ (3x^{2} - 7x - 30) - (2x^{2} - 7x - 5) = x^2 - 25 \quad \text{is true for all }x \qquad \qquad \ldots \ldots (2)$$

Dividing (2) by (1), we get

$$ \sqrt {3x^{2} - 7x - 30} + \sqrt {2x^{2} - 7x - 5} = x + 5 \qquad \qquad \ldots \ldots (3)$$

Adding (1) and (3) we get

$$ 2\sqrt {3x^{2} - 7x - 30} = 2x $$

Squaring both sides, we get

$$3x^2-7x-30 = x^2 \implies 2x^2 - 7x - 30 = 0 \implies (2x+5)(x-6) = 0 \implies x=6, \dfrac{-5}{2}$$

Mathematics

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