One root of the $4 x^{3} - 5 x^{2} - 6 x + 7 = 0$ is $x = 1$

True

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False

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Solution

The correct option is A True
Given equation is,
$4 x^{3} - 5 x^{2} - 6 x + 7 = 0$

$⟹ 4 x^{3} - 4 x^{2} - x^{2} + x - 7 x + 7 = 0$

$⟹ 4 x^{2} (x - 1) - x (x - 1) - 7 (x - 1) = 0$

$⟹ (x - 1) (4 x^{2} - x - 7) = 0$

$⟹ x = 1$

Hence $x = 1$ is the root of given equation.

Suggest Corrections

Similar questions

x^{4} + 4 x^{3} - 5 x^{2} - 6 x + 7

x - 1

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\sqrt{5 x^{2} - 6 x + 8} - \sqrt{5 x^{2} - 6 x - 7} = 1

\sqrt{5 x^{2} - 6 x + 8} - \sqrt{5 x^{2} - 6 x - 7} = 1

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