Question

Please solve 4th one as soon as possible............

Answer 1

Answer :-

$\implies \sqrt{3x + 1} + 1 = \sqrt{x} \\ \\ \\</p><p><em><u>B</u></em><em><u>y</u></em><em><u> </u></em><em><u>\</u></em><em><u>:</u></em><em><u>Squaring</u></em><em><u> </u></em><em><u>\</u></em><em><u>:</u></em><em><u>both</u></em><em><u> </u></em>\:sides, \\ \\</p><p></p><p>\implies (3x + 1) + 1 + 2( \sqrt{ 3x + 1}) \times 1 = x \\ \\\implies 2x + 2 = - 2 \sqrt{3x + 1} \\ \\\implies - 2x - 2 = - 2 \sqrt{3x + 1} \\ \\\implies - 2(x + 1) = - 2 \sqrt{3x + 1} \\ \\\implies x + 1 = \sqrt{3x + 1} \\ \\ <em><u>Taking</u></em><em><u> </u></em><em><u>\</u></em><em><u>:</u></em><em><u> </u></em><em><u>square</u></em><em><u> </u></em><em><u>\</u></em><em><u>:</u></em><em><u> </u></em><em><u>root</u></em><em><u> </u></em><em><u>\</u></em><em><u>:</u></em><em><u> </u></em><em><u>left</u></em><em><u> </u></em><em><u>\</u></em><em><u>:</u></em><em><u> </u></em><em><u>side</u></em><em><u> </u></em><em><u>,</u></em><em><u> </u></em><em><u>\</u></em><em><u>\</u></em> \\ \implies ( {x + 1})^{2} = 3x + 1 \\ \\ \implies {x}^{2} + 1 + 2x = 3x + 1 \\ \\ {x}^{2} - x = 0 \\ \\ x(x - 1) = 0$

By equating factors to zero we get ,

$\implies$X = 0

(X-1) =0

$\implies$ x=1

So, Correct answer is (c) 0,1

_______________________________________

IDENTITIES USED:-

(a + b)² = a² + b² + 2ab

Hope it helps

Answer 2

$\huge\mathcal\underline{Question}$

The roots of the given eqn .

$\sqrt{3x + 1} + 1 = \sqrt{x}$

are :-

☆ 0

☆ 1

☆ 0,1

☆ None

$\huge\mathcal\underline{Solution}$

$\sqrt{3x + 1} + 1 = \sqrt{x}$

By squaring both sides .....

$( \sqrt{3x + 1} + 1) ^{2} = (\sqrt{x} ) ^{2}$

By using the identity ...

$(a + b) ^{2} = {a}^{2} + {b}^{2} + 2ab$

$= > ( \sqrt{3x + 1} ) ^{2} + (1) ^{2} + 2 \times ( \sqrt{3x + 1} )(1)= (\sqrt{x} ) ^{2}$

$= > (3x + 1) + 1 + 2( \sqrt{3x + 1} ) \times 1 = x$

$= > 3x - x + 2 = - 2 \sqrt{3x + 1}$

$= > 2x + 2 = - 2 \sqrt{3x + 1}$

$= > - 2x - 2 = - 2 \sqrt{3x + 1}$

$= > - 2(x + 1) = - 2 \sqrt{3x + 1}$

$= > (x + 1) = \sqrt{3x + 1}$

Now by squaring both sides ....

$= > {(x + 1)}^{2} = 3x + 1$

Now , by using

$(a + b) ^{2} = {a}^{2} + {b}^{2} + 2ab$

$= > {x}^{2} + 1 + 2 \times x \times 1 = 3x + 1$

$= > {x}^{2} + 1 + 2x = 3x + 1$

$= > {x}^{2} = 3x - 2x + 1 - 1$

$= > {x}^{2} = 1x + 0$

$= > {x}^{2} - x = 0$

$= > x(x - 1) = 0$

Now ,

$= > x = 0$

and

$= > x - 1 = 0 = x = 1$

The correct ans is option (c)

The roots of the given equation are 0 and 1 .