Inequalities under square roots

To solve these questions, we have to make sure that the expression within the square root must always be positive.  All the rules related to inequalities apply to these questions. Let us have a look at some solved examples.

1. (x−3)(x2)−2x−15−−−−−−−−−−−−√≥0$(x-3)\sqrt{(x^2)-2x-15}\ge 0$
1.x≥5$x\ge 5$
2. x≤5$x\le 5$
3.x≥−8$x\ge -8$
4.x≤3$x\le 3$
5.x≥−5$x\ge -5$
Ans:
(x−3)(x2)−2x−15−−−−−−−−−−−−√≥0$(x-3)\sqrt{(x^2)-2x-15}\ge 0$
(x−3)(x2)−5x+3x−15−−−−−−−−−−−−−−−−√≥0$(x-3)\sqrt{(x^2)-5x+3x-15}\ge 0$
(x−3)(x−5)(x+3)−−−−−−−−−−−√≥0$(x-3)\sqrt{(x-5)(x+3)}\ge 0$
The above inequality is valid for
x−5≥0$x-5\ge 0$
That is, x≥5$x\ge 5$
Hence option 1

2.  (x−8)x2−3x+40−−−−−−−−−−√≥0$(x-8)\sqrt{x^2-3x+40}\ge 0$
1.x≥5$x\ge 5$
2.x≥5$x\ge 5$
3.x≥−8$x\ge -8$
4.x≥13$x\ge 13$
5.x≥−5$x\ge -5$
Ans:
(x−8)(x2)+8x−5x−40−−−−−−−−−−−−−−−−√≥0$(x-8)\sqrt{(x^2)+8x-5x-40}\ge 0$
(x−8)(x+8)(x−5)−−−−−−−−−−−√≥0$(x-8)\sqrt{(x+8)(x-5)}\ge 0$
The above inequality is vallid for
That is, x≥5$x\ge 5$
Hence option 2

3. 27x−13−x−−−−−−√>1$2\sqrt{{\displaystyle \frac{7x-1}{3-x}}}>1$
(1) 0<x<3
(2) 0.5<x<6
(3) 0.5<x<3
(4) 0<x<6
(5) 1<x<3
Ans:
27x−13−x−−−−−−√>1$2\sqrt{{\displaystyle \frac{7x-1}{3-x}}}>1$
7x−13−x−−−−−−√>1$\sqrt{{\displaystyle \frac{7x-1}{3-x}}}>1$
7x−13−x>1${\displaystyle \frac{7x-1}{3-x}>1}$
7x−13−x−1>0${\displaystyle \frac{7x-1}{3-x}-1>0}$
8x−43−x>0${\displaystyle \frac{8x-4}{3-x}>0}$
8x−4x−3<0${\displaystyle \frac{8x-4}{x-3}<0}$
2x−1x−3<0${\displaystyle \frac{2x-1}{x-3}<0}$
The above inequality is negative, when numerator is positive and denominator is negative or numerator is negative and denominator is positive, So 2x - 1 > 0 and x - 3 < 0, ⇒$\Rightarrow$ x > 1/2 and x < 3 ⇒$\Rightarrow$ for 0.5 <x<3 the above inequality is satisfied.
Hence option 2

4.  6x−85−2x−−−−−−√>2$\sqrt{{\displaystyle \frac{6x-8}{5-2x}}}>2$
(1)0.75<x<25
(2) 0<x<5
(3) 2<x<5
(4) 5<X<11
Ans:
6x−85−2x>22${\displaystyle \frac{6x-8}{5-2x}>2^2}$

6x−85−2x>4${\displaystyle \frac{6x-8}{5-2x}>4}$

6x−85−2x−4>0${\displaystyle \frac{6x-8}{5-2x}-4>0}$

14x−285−2x>0${\displaystyle \frac{14x-28}{5-2x}>0}$

x−22x−5<0${\displaystyle \frac{x-2}{2x-5}<0}$
So, we have,
2<x<2.5
Hence option 5

5. 1−(2x−1x2)−−−−−−−−−−−−√<12$\sqrt{1-\left({\displaystyle \frac{2x-1}{x^2}}\right)}<{\displaystyle \frac{1}{2}}$
1.23<x<2${\displaystyle \frac{2}{3}<x<2}$
2.x>23$x>{\displaystyle \frac{2}{3}}$
3.x>2$x>2$
4.x<2$x<2$
5.None of these
Ans:
1−(2x−1x2)−−−−−−−−−−−−√<12$\sqrt{1-\left({\displaystyle \frac{2x-1}{x^2}}\right)}<{\displaystyle \frac{1}{2}}$
1−(2x−1)x2<14$1-{\displaystyle \frac{(2x-1)}{x^2}<{\displaystyle \frac{1}{4}}}$
x2−2x+1x2<14${\displaystyle \frac{x^2-2x+1}{x^2}<{\displaystyle \frac{1}{4}}}$
As x2$x^2$ is positive, we can cross multiply.
4x2−8x+4<x2$4x^2-8x+4<x^2$
3x2−8x+4<0$3x^2-8x+4<0$
(3x−2)(x−2)<0$(3x-2)(x-2)<0$
x<2$\mathrm{x}<2$ and x>23$x>{\displaystyle \frac{2}{3}}$
Hence option 4

6. 1−(x+3x2)−−−−−−−−−−−√<34$\sqrt{1-\left({\displaystyle \frac{x+3}{x^2}}\right)}<{\displaystyle \frac{3}{4}}$
1. −127<x≤1−3√2$-{\displaystyle \frac{12}{7}<x\le {\displaystyle \frac{1-\sqrt{3}}{2}}}$and 1−13−−√2≤x<4${\displaystyle \frac{1-\sqrt{13}}{2}\le x<4}$
2. x≥1−13−−√2$x\ge {\displaystyle \frac{1-\sqrt{13}}{2}}$
3. x≤1−13−−√2$x\le {\displaystyle \frac{1-\sqrt{13}}{2}}$
4. −127≤x≤4$-{\displaystyle \frac{12}{7}\le x\le 4}$
5. None of these
Ans:
1−(x+3x2)−−−−−−−−−−−√<34$\sqrt{1-\left({\displaystyle \frac{x+3}{x^2}}\right)}<{\displaystyle \frac{3}{4}}$
1−x+3x2<916$1-{\displaystyle \frac{x+3}{x^2}<{\displaystyle \frac{9}{16}}}$
x2−x−3x2<916${\displaystyle \frac{x^2-x-3}{x^2}<{\displaystyle \frac{9}{16}}}$
16x2−16x−48<9x2$16x^2-16x-48<9x^2$
7x2−16x−48<0$7x^2-16x-48<0$
7x2−28x+12x−48<0$7x^2-28x+12x-48<0$
(7x+12)(x−4)<0$(7x+12)(x-4)<0$
7x+12>0andx−4<0$7x+12>0\mathrm{a}\mathrm{n}\mathrm{d}x-4<0$
x>−127andx<4$x>-{\displaystyle \frac{12}{7}andx<4}$ -------(1)
Also we check the numerator is greater than 0 or not.
x2−x−3≥0$x^2-x-3\ge 0$
−b±b2−4ac−−−−−−−√2a=1±1−4.1.−3−−−−−−−−−√2=1±1+13−−−−−√2${\displaystyle \frac{-b\pm \sqrt{b^2-4ac}}{2a}={\displaystyle \frac{1\pm \sqrt{1-\mathrm{4.1.}-3}}{2}={\displaystyle \frac{1\pm \sqrt{1+13}}{2}}}}$
x<1−1+13−−−−−√2$x<{\displaystyle \frac{1-\sqrt{1+13}}{2}}$ and x>1+1+13−−−−−√2$x>{\displaystyle \frac{1+\sqrt{1+13}}{2}}$ ----- (2)
Combining the above two results we get −127<x≤1−3√2$-{\displaystyle \frac{12}{7}<x\le {\displaystyle \frac{1-\sqrt{3}}{2}}}$and 1−13−−√2≤x<4${\displaystyle \frac{1-\sqrt{13}}{2}\le x<4}$
Hence option 1