Answer:
26. [tex]\frac{3}{4} \leq x <7[/tex]
27. [tex]x\leq 5[/tex]
28. [tex]0\leq b<4[/tex]
Step-by-step explanation:
26. [tex]\sqrt{4x-3} <5[/tex]:
Taking square on both the sides to get:
[tex](\sqrt{4x-3} )^2 < (5)^2[/tex]
[tex]4x-3<25[/tex]
[tex]4x<28[/tex]
[tex]x<\frac{28}{4}[/tex]
[tex]x<7[/tex]
For non-negative values for radical:
[tex]4x-3\geq 0[/tex]
[tex]x\geq \frac{3}{4}[/tex]
So solution for this: [tex]\frac{3}{4} \leq x <7[/tex]
27. [tex]2+\sqrt{4x-4}\leq 6[/tex]
Subtracting 2 from both the sides to get:
[tex]2+\sqrt{4x-4}-2\leq 6-2[/tex]
[tex]\sqrt{4x-4}\leq 4[/tex]
Taking square root on both sides:
[tex](\sqrt{4x-4})^2\leq (4)^2[/tex]
[tex]4x-4\leq 16[/tex]
[tex]x\leq \frac{20}{4}[/tex]
[tex]x\leq 5[/tex]
28. [tex]\sqrt{b+12} -\sqrt{b}>2[/tex]
Adding [tex]\sqrt{b}[/tex] to both the sides to get:
[tex]\sqrt{b+12} -\sqrt{b}+\sqrt{b}>2+\sqrt{b}[/tex]
[tex]\sqrt{b+12} >2+\sqrt{b}[/tex]
Taking square on both sides:
[tex](\sqrt{b+12})^2 >(2+\sqrt{b})^2[/tex]
[tex]b+12>(2+\sqrt{b})^2[/tex]
[tex]b+12>4+4\sqrt{b}+b[/tex]
[tex]4+4\sqrt{b} +b<b+12[/tex]
Subtracting [tex]b[/tex] from both sides to get:
[tex]4+4\sqrt{b} +b-b<b+12-b[/tex]
[tex]4+4\sqrt{b} <12[/tex]
Subtracting 4 from both sides:
[tex]4+4\sqrt{b}-4 <12-4[/tex]
[tex]4\sqrt{b} <8[/tex]
Square both sides again:
[tex](4\sqrt{b})^2 <(8)^2[/tex]
[tex]16b<8^2[/tex]
[tex]b<\frac{64}{16}[/tex]
[tex]b<4[/tex]
and for non-negative radical [tex]b\geq 0[/tex]
therefore, solution is [tex]0\leq b<4[/tex].
