If $x^{2} - 6 x + 5 < 0$ , then set of solution for that
equation is

1 < x < 5

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2 < x < 7

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- 1 < x <5

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1 < x < 3

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Solution

The correct option is A 1 < x < 5
$x^{2} - 6 x + 5 < 0$
$⟹ x^{2} - x - 5 x + 5 < 0$
$⟹ x (x - 1) - 5 (x - 1) < 0$
$⟹ (x - 1) (x - 5) < 0$
Hence $1 < x < 5$

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Similar questions

5 \frac{1}{3}

5 \frac{2}{3}

\div

\frac{1}{3}

\div

x + 3

\frac{1}{5}

\div

1 \frac{1}{5}

= 7

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

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

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

\frac{2 x - 5}{(x - 1) (x - 7)} \leq 0

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