Square root and Approximations

Square root and Approximations

Scientific and Standard Notations

Linear inequalities

Linear inequalities with radicals

Linear inequalities with radicals

While dealing with inequalities, we also need to consider the domain of radical expressions. We take into consideration both the solutions, one which we get after solving inequality by squaring both sides and other by setting radicand ≥ 0. lets work on some examples to understand this process better.

Example1. Solve the following inequality and write the solution in interval form.

${\color{Red} \sqrt{x-1}+2\leq 5}$

Solution: $\sqrt{x-1}+2\leq 5$

$\sqrt{x-1}\leq 3$

Squaring both sides, $(\sqrt{x-1})^{2}\leq 3^{2}$

x – 1 ≤ 9

x ≤ 10 ——(i)

Since radicals are not defined for negative values so radicand must be positive. Therefore, x-1 ≥ 0

=> x ≥ 1 ——–(ii)

Combining the solutions (i) and (ii), we get final solution as,

1 ≤ x ≤ 10 => [1 ,10]

Example2. Solve the following inequality and write the solution in interval form.

${\color{Red} 4-\sqrt{x-3} \geq -2}$

Solution: $4-\sqrt{x-3}\geq -2$ Also, $x-3 \geq 0$

$-\sqrt{x-3}\geq -6$ x ≥ 3

$\sqrt{x-3}\leq 6$ (multiplied with -1 and inequality sign changed)

squaring both sides, $(\sqrt{x-3})^{2}\leq 6^{2}$

x -3 ≤ 36

x ≤ 39

combining two solutions we get, 3 ≤ x ≤ 39

Final answer in interval form [3,39]

Example3.Solve the following inequality and write the solution in interval form.

$\dpi{120} {\color{Red} \sqrt[2]{x}\leq \sqrt[3]{x}}$

Solution: We can rewrite the given inequality as,

$\dpi{120} x^{1/2} \leq x^{1/3}$

Since LCM of denominators 2 and 3 is 6 so we multiply exponents of both sides with 6 and get,

$\dpi{120} \left ( x^{1/2} \right )^{6}\leq \left ( x^{1/3} \right )^{6}$ => $\dpi{120} x^{3}\leq x^{2}$

$\dpi{120} x^{3}-x^{2}\leq 0$

$\dpi{120} x^{2}(x-1)\leq 0$

As $\dpi{120} x^{2}$ can never be negative so x-1 ≤ 0

x≤ 1 —— (i)

Also $\dpi{120} x^{1/2}$ , being an even radical is never defined for negative values which restrict the given solution to x ≥ 0 —— (ii)

So final solution, after combining (i) and (ii) is given as,

0≤x≤1 => [0,1]

Example4. Solve the inequality $\dpi{120} {\color{Red} \sqrt{x}\geq 2 \sqrt[4]{x}}$

Solution: Rewriting the given inequality as,

$\dpi{120} x^{1/2} \geq 2 x^{1/4}$

Since lcm of denominators 2 and 4 is 4, So raising exponents on both sides by 4 and we get,

$\dpi{120} \left ( x^{1/2} \right )^{4}\geq (2 \left ( x^{1/4} \right ))^{4}$

$\dpi{120} x^{2}\geq 2^{4}x$

$\dpi{120} x^{2}\geq 16x$

$\dpi{120} x^{2}-16x\geq 0$

x ( x-16) ≥ 0

As we know that even radicals are not defined for negative values which restrict domain to x ≥ 0,

Also, x-16≥0 => x≥ 16

So final solution is given as, [16,∞)

Practice problems:

Solve the following inequalities and write the answer in interval form.

$\dpi{120} \sqrt[4]{x}> \sqrt[3]{x}$
$3\sqrt{2x-1}+6\leq 15$
$\dpi{120} \sqrt{c+4}-3\geq 3$
$\dpi{120} -3\sqrt{11r+3}> -15$

Answers:

(0,1)
[1/2, 5]
[32,∞)
( -3/11 , 2)

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