The value of x for which $| x - 3 | + | x - 4 | = 9$ is

8, 1

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8, -1

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-8, 1

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-8, -1

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Solution

The correct option is B
8, -1

Given: $| x - 3 | + | x - 4 | = 9$

$\Rightarrow$ $x - 3 + x - 4 = \pm 9$

Case 1

$x - 3 + x - 4 = 9$

$2 x - 7 = 9$

$2 x = 16$

$\Rightarrow$ $x = 8$

Case 2

$x - 3 + x - 4 = - 9$

$2 x - 7 = - 9$

$2 x = - 2$

$\Rightarrow$ $x = - 1$

Hence, $x$ can take the values 8,-1

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Similar questions

| x - 3 | + | x - 4 | = 9

\geq

x^{3} + 9 x^{2} - x + 4

x = - 5 + 2 \sqrt{- 4}

x

(\frac{3}{4})^{6} \times (\frac{16}{9})^{5} = (\frac{4}{3})^{x + 2}

Find the value of a for which (x + 1) is a factor of $(a x^{3} + x^{2} - 2 x + 4 a - 9)$ .

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