Solution of the equation sqrt{x + 3 - 4sqrt{x | vedclass

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

Question

(C) Let $u = x - 1$. Since $x - 1$ is under the square root,we must have $u \geq 0$.The equation becomes $\sqrt{u + 4 - 4\sqrt{u}} + \sqrt{u + 9 - 6\s

Vedclass · Accepted Answer

$x \in [5, 10]$

Vedclass · Answer

(C) Let $u = x - 1$. Since $x - 1$ is under the square root,we must have $u \geq 0$.The equation becomes $\sqrt{u + 4 - 4\sqrt{u}} + \sqrt{u + 9 - 6\sqrt{u}} = 1$.This simplifies to $\sqrt{(\sqrt{u} - 2)^2} + \sqrt{(\sqrt{u} - 3)^2} = 1$,which is $|\sqrt{u} - 2| + |\sqrt{u} - 3| = 1$.Let $y = \sqrt{u}$. Then $|y - 2| + |y - 3| = 1$.We know that $|y - a| + |y - b| = |b - a|$ if and only if $y$ lies between $a$ and $b$ (inclusive).Here,$|y - 2| + |y - 3| = |3 - 2| = 1$.Therefore,$2 \leq y \leq 3$.Substituting back $y = \sqrt{u}$,we get $2 \leq \sqrt{u} \leq 3$.Squaring the inequality,$4 \leq u \leq 9$.Since $u = x - 1$,we have $4 \leq x - 1 \leq 9$,which gives $5 \leq x \leq 10$.Thus,the solution is $x \in [5, 10]$.

Solution of the equation $\sqrt{x + 3 - 4\sqrt{x - 1}} + \sqrt{x + 8 - 6\sqrt{x - 1}} = 1$ is

Similar Questions

In a triangle $ABC$,the value of $\angle A$ is given by $5\cos A + 3 = 0$. Then,the equation whose roots are $\sin A$ and $\tan A$ is:

If the roots of $x^2 - bx + c = 0$ are two consecutive integers,then $b^2 - 4c$ is

If $ax^2 + bx + c = 0$,then $x =$

If $\alpha, \beta$ are the roots of $x^2 + px + 1 = 0$ and $\gamma, \delta$ are the roots of $x^2 + qx + 1 = 0$,then $q^2 - p^2$ =

Consider the equation $x^2+x-n = 0$,where $n \in N$ and $n \in [5, 100]$. The total number of different values of $n$ such that the given equation has integral roots is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module