Square Root Functions - Inequalities |
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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
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| ax2+bx+c>0 ax2+bx+c<0 ax2+bx+c>0 |
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+ve r1 -ve r2 +ve
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| In the cases we have just studied for square root functions, we are likely to obtain square root inequalities of the form |
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Example 1 |
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| Squaring both sides |
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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} |
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Try this question |
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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} |
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Canceling 4 on both sides
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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}
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