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.
$I.$ $\sqrt{784} x + 1234 = 1486$
$II.$ $\sqrt{1089} y + 2081 = 2345$

If the product of the roots of the equation $2x^2 + 6x + \alpha^2 + 1 = 0$ is $-\alpha$, then the value of $\alpha$ will be

Construct a quadratic equation whose roots are $\sqrt{2}$ and $2 \sqrt{2}$.

The number of real roots of the equation $e^{4x} + e^{3x} - 4e^{2x} + e^x + 1 = 0$ is:

The equation $(1 + 2k){x^2} + (1 - 2k)x + (1 - 2k) = 0$ is a perfect square for how many values of $k$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module