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## Square Root Functions - Inequalities

Recall that in the case of quadratic inequalities ax2+bx+c>0 and ax2+bx+c<0 where a>0 (a is positive), a, b, c are real and we obtained the roots or solutions of ax2+bx+c=0. We also found that if ax2+bx+c>0, r1, r2 were the solutions, the solution set was of the form {x|x< r1} ∪{x|x>r2} where r2>r1 and for ax2+bx+c<0, the solution set was of the form {x | r1<x<r2}.

On the number line, we had

ax2+bx+c>0     ax2+bx+c<0      ax2+bx+c>0
+ve      r1        -ve         r2        +ve

In the cases we have just studied for square root functions, we are likely to obtain square root inequalities of the form

### Example 1

Squaring both sides

3x2-2<4x2-4x+1

0 < 4x2-4x+1-3x2+2

0 < x2-4x+3

0 < x2-3x-x+3

0 < x(x-3)-1(x-3)

0 < (x-1)(x-3)

This is of the form (x-1) (x-3)>0

r1 = 1,     r2 = 3

First, check whether 3r12-2 ≥ 0 and 2r1-1 ≥ 0

and 3r22-2 ≥ 0 and 2r2-1 ≥ 0

3x2-2 = 3*12-2 = 3-2 = 1>0

2x-1 = 2*1-1 = 2-1-1>0

3x2-2 = 3*32-2 = 27-2 = 25>0

2x-1 = 2*3-1 = 6-1 = 5>0

r1=1, r2 = 3

for (x-1) (x-3)>0

⇒      x-1>0 and x-3>0

⇒      x>1 and x>3

or x-1<0 and x-3<0

⇒      x<1 and x<3

∴ Solution set= {x|x<1} ∪ {x|x>3}

### Try this question

x2+8x+16<36(x-4)

x2+8x+16<36x-144

x2+8x+16-36x+144<0

x2-28x+160<0

x2-20x-8x+160<0

x(x-20)-8(x-20)<0

(x-8)(x–20)<0

∴ r1=8, r2=20

2x+9 = 2 *  8+9 = 16+9 = 25>0

x–4=8–4=4>0

and           2x+9=2*20+9=49>0

x–4=20–4=16>0

Consider     (x–8)(x–20)<0

⇒           x–8<0 and x–20>0

⇒          x<8 and x>20

or

⇒          x-8>0 and x-20<0

⇒          x>8 and x<20

Solution set ={x|8<x<20}

 Canceling 4 on both sides

9(6-x) = 64 - 16x + x2

64 - 16x + x2 - 54 + 9x = 0

x2 - 7x + 10 = 0

x2 - 5x - 2x + 10 = 0

x(x-5) -2 (x-5) = 0

(x-2) (x-5) = 0

⇒      x - 2 = 0 or x - 5 = 0

x = 2 or x = 5

If x = 2

3x + 10

=3*2+10

=6+10

=16 > 0

6 - x = 6 - 2

= 4 > 0
If x=5

3x + 10

= 3*5 + 10

= 15 + 10

= 25 > 0

6 - x = 6 - 5

=1>0

Solution set {2, 5}