The solution set of the in equation x2-6 x-7-x-4-0 is

The correct option is A ( 7, 4) ∪ ( 4,1) nbsp;(x + 7)(x 1) lt; 0 and x ≠ 4 So x ε nbsp; ( 7,1) ( 4) ...

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Roots of a Quadratic Equation

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The solution set of the in equation $\frac{x^{2} + 6 x - 7}{| x + 4 |} < 0$ is

( -7, 1)

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(-7, -4)

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$(- 7, - 4) \cup (- 4, 1)$

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$(- 7, - 4) \cup (4, 1)$

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The solution set of the in equation $\frac{x^{2} + 6 x - 7}{| x + 4 |} < 0$ is

\frac{x^{2} + 6 x - 7}{| x + 4 |} < 0

x \in R

4^{- x + 0.5} - 7 \times 2^{- x} - 4 < 0

{log}_{1 / 3} (x^{2} - 6 x + 18) - 2 {log}_{1 / 3} (x - 4) < 0

x^{2} + x - 2 = 0

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