Solve :
$(x - 5)^{2} - (x + 2)^{2} = - 2$

\frac{23}{14}

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\frac{14}{23}

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\frac{5}{2}

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\frac{2}{5}

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Solution

The correct option is A $\frac{23}{14}$
Use the formula:
$(a + b)^{2} = a^{2} + b^{2} + 2 a b$
$(a - b)^{2} = a^{2} + b^{2} - 2 a b$

Now,
$(x - 5)^{2} - (x + 2)^{2} = - 2$
$(x^{2} + 25 - 2 \times 5 \times x) - (x^{2} + 4 + 2 \times 2 \times x) = - 2$

$(x^{2} + 25 - 10 x) - (x^{2} + 4 + 4 x) = - 2$

$25 - 10 x - 4 - 4 x = - 2$

$21 - 14 x = - 2$
$- 14 x = - 23$

$x = \frac{23}{14}$

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Similar questions

\frac{2}{3} (x - 5) - \frac{1}{4} (x - 2) = \frac{9}{2}

5 x^{2} - y^{2} = 5

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\frac{x + 4}{3} = \frac{y - 1}{5} = \frac{z + 3}{4} and \frac{x + 1}{1} = \frac{y - 4}{1} = \frac{z - 5}{2}

\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{- 3} and \frac{x + 3}{- 1} = \frac{y - 5}{8} = \frac{z - 1}{4}

\frac{5 - x}{- 2} = \frac{y + 3}{1} = \frac{1 - z}{3} and \frac{x}{3} = \frac{1 - y}{- 2} = \frac{z + 5}{- 1}

\frac{x - 2}{3} = \frac{y + 3}{- 2}, z = 5 and \frac{x + 1}{1} = \frac{2 y - 3}{3} = \frac{z - 5}{2}

\frac{x - 5}{1} = \frac{2 y + 6}{- 2} = \frac{z - 3}{1} and \frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 6}{5}

\frac{- x + 2}{- 2} = \frac{y - 1}{7} = \frac{z + 3}{- 3} and \frac{x + 2}{- 1} = \frac{2 y - 8}{4} = \frac{z - 5}{4}

x

\frac{2}{x + 1} + \frac{3}{2 (x - 2)} = \frac{23}{5}

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