The roots of the equation sqrt{3x + 1} + 1 = | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given equation is $\sqrt{3x + 1} + 1 = \sqrt{x}$.Rearranging the terms,we get $\sqrt{3x + 1} = \sqrt{x} - 1$.Squaring both sides:$(\sqrt{3x + 1})^

Vedclass · Accepted Answer

None

Vedclass · Answer

(D) Given equation is $\sqrt{3x + 1} + 1 = \sqrt{x}$.Rearranging the terms,we get $\sqrt{3x + 1} = \sqrt{x} - 1$.Squaring both sides:$(\sqrt{3x + 1})^2 = (\sqrt{x} - 1)^2$$3x + 1 = x + 1 - 2\sqrt{x}$.Simplifying the equation:$2x = -2\sqrt{x}$$x = -\sqrt{x}$.Squaring again:$x^2 = x$$x^2 - x = 0$$x(x - 1) = 0$,which gives $x = 0$ or $x = 1$.Checking $x = 0$: $\sqrt{3(0) + 1} + 1 = 1 + 1 = 2$,while $\sqrt{0} = 0$. Since $2 
eq 0$,$x = 0$ is not a solution.Checking $x = 1$: $\sqrt{3(1) + 1} + 1 = \sqrt{4} + 1 = 2 + 1 = 3$,while $\sqrt{1} = 1$. Since $3 
eq 1$,$x = 1$ is not a solution.Therefore,there are no real roots for the given equation.

The roots of the equation $\sqrt{3x + 1} + 1 = \sqrt{x}$ are

Similar Questions

Solve the given two equations and select the correct answer from the given options.
$I.$ $7x^2 + 16x - 15 = 0$
$II.$ $y^2 - 6y - 7 = 0$

If $\alpha$ and $\beta$ are roots of $x^2 - 3x + 1 = 0$,then the equation whose roots are $\frac{1}{\alpha - 2}$ and $\frac{1}{\beta - 2}$ is

The equation $x^{(3/4)(\log_2 x)^2 + (\log_2 x) - 5/4} = \sqrt{2}$ has

Solve the given two equations and select the correct answer from the given options.
$I.$ $x = \sqrt[3]{2197}$
$II.$ $2y^2 - 54y + 364 = 0$

If $\alpha, \beta, \gamma$ are the roots of the equation $x^3 + qx - r = 0$,then find the equation whose roots are $\left( \beta \gamma + \frac{1}{\alpha} \right), \left( \gamma \alpha + \frac{1}{\beta} \right), \left( \alpha \beta + \frac{1}{\gamma} \right)$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module