Question

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SOLVE THESE:-

$(1)32x {}^{10} - 31 {x}^{5} - 1 = 0$
$(2)10x + \sqrt{x} - 2 = 0$
$(3)3 \sqrt{x} +( 5 \div \sqrt{x} ) = 16$
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Answer 1

Answer:

Step-by-step explanation:

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Answer 2

Answer :-

(1)

$32x^{10} -31x^5 - 1 = 0$

Let $x^5 = t$

◾Then equation :-

$\implies 32t^2 -31t - 1 = 0$

$\implies 32t^2 -32t + t -1 = 0$

$\implies 32t(t-1) +1(t-1) = 0$

$\implies (32t + 1)(t-1) = 0$

$\implies t = \dfrac{-1}{32} ,\: 1$

$\implies x^5 = \dfrac{-1}{32} ,\: 1$

$\bold{ \implies x = 1 ,\: \dfrac{-1}{2}}$

(2).

$10x + \sqrt{x} -2 = 0$

$\implies 10x + 5\sqrt{x} -4\sqrt{x} -2 = 0$

$\implies 5\sqrt{x}(2\sqrt{x} + 1) -2(2\sqrt{x} + 1) = 0$

$\implies (2\sqrt{x} + 1)(5\sqrt{x} - 2)= 0$

$\implies \sqrt{x} = \dfrac{-1}{2} ,\: \dfrac{2}{5}$

◾ But as square root of any Real Number cannot be negative .

◾Hence

$\sqrt{x} = \dfrac{2}{5} \\\\ Or \:\bold{ x = \dfrac{4}{25} }$

(3).

$3\sqrt{x} + \dfrac{5}{\sqrt{x}} = 16$

◾On multiplication of both sides by √x

$\implies 3x +5 = 16\sqrt{x}$

$\implies 3x - 16\sqrt{x} + 5 = 0$

$\implies 3x - 15\sqrt{x} - \sqrt{x} + 5 = 0$

$\implies 3\sqrt{x}(\sqrt{x} - 5) -1(\sqrt{x} - 5) = 0$

$\implies (\sqrt{x} - 5)(3\sqrt{x} -1) = 0$

$\implies \sqrt{x} = \dfrac{1}{3} ,\: 5$

$\implies \bold{x = \dfrac{1}{9} ,\: 25}$