In the real number system,the equation sqrt{x | vedclass

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

Question

(D) Given the equation: $\sqrt{x+3-4 \sqrt{x-1}}+\sqrt{x+8-6 \sqrt{x-1}}=1$ Let $u = \sqrt{x-1}$,where $u \ge 0$. Then $x-1 = u^2$,so $x = u^2+1$. Sub

Vedclass · Accepted Answer

infinitely many solutions

Vedclass · Answer

(D) Given the equation: $\sqrt{x+3-4 \sqrt{x-1}}+\sqrt{x+8-6 \sqrt{x-1}}=1$ Let $u = \sqrt{x-1}$,where $u \ge 0$. Then $x-1 = u^2$,so $x = u^2+1$. Substituting this into the equation: $\sqrt{u^2+1+3-4u} + \sqrt{u^2+1+8-6u} = 1$ $\sqrt{u^2-4u+4} + \sqrt{u^2-6u+9} = 1$ $\sqrt{(u-2)^2} + \sqrt{(u-3)^2} = 1$ $|u-2| + |u-3| = 1$ This equation holds if and only if $2 \le u \le 3$. Since $u = \sqrt{x-1}$,we have $2 \le \sqrt{x-1} \le 3$. Squaring the inequality: $4 \le x-1 \le 9$,which gives $5 \le x \le 10$. Thus,the equation has infinitely many solutions in the interval $[5, 10]$.

In the real number system,the equation $\sqrt{x+3-4 \sqrt{x-1}}+\sqrt{x+8-6 \sqrt{x-1}}=1$ has

Similar Questions

The equation $x^4-x^3-6x^2+4x+8=0$ has two equal roots. If $\alpha$ and $\beta$ are the other two roots of this equation,then $\alpha^2+\beta^2=$

The product and sum of the roots of the equation $|x|^2 - 5|x| - 24 = 0$ are respectively:

If one root of the equation $x(x + 2) = 3 - ax^2$ approaches infinity,then the value of $a$ approaches which of the following?

What are the real roots of the equation $x^2 + 5|x| + 4 = 0$?

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module