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 $\sqrt{3x + 1} + 1 = \sqrt{x}$.Subtracting $1$ from both sides,we get $\sqrt{3x + 1} = \sqrt{x} - 1$.Squaring both sides,we get

Vedclass · Accepted Answer

None

Vedclass · Answer

(D) Given equation is $\sqrt{3x + 1} + 1 = \sqrt{x}$.Subtracting $1$ from both sides,we get $\sqrt{3x + 1} = \sqrt{x} - 1$.Squaring both sides,we get $3x + 1 = x + 1 - 2\sqrt{x}$.Simplifying,we get $2x = -2\sqrt{x}$,which implies $x = -\sqrt{x}$.Squaring again,$x^2 = x$,so $x^2 - x = 0$,which gives $x(x - 1) = 0$,so $x = 0$ 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 $x = 1$: $\sqrt{3(1) + 1} + 1 = 2 + 1 = 3$,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

If two roots of the equation $x^3 - 3x + 2 = 0$ are the same,then the roots are:

Let $x = (\sqrt{50} + 7)^{1/3} - (\sqrt{50} - 7)^{1/3}$. Then,

If $f(x) \in \mathbb{Q}[x]$ is a non-zero polynomial such that all its roots are irrational,then the degree of $f(x)$ is

If a polynomial $f(x)$ is divided by $(x + 1)$,$(x - 2)$,and $(x + 2)$,the remainders are $6$,$3$,and $15$ respectively. Find the remainder when $f(x)$ is divided by $(x + 1)(x - 2)(x + 2)$.

Let $\alpha, \beta$ be the roots of the quadratic equation $12x^{2}-20x+3\lambda=0$, where $\lambda \in \mathbb{Z}$. If $\frac{1}{2} \le |\beta-\alpha| \le \frac{3}{2}$, then the sum of all possible values of $\lambda$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module