Question

Can someone please help with these 3 I am so confused on what to do thank you!!!!

Answer 1

Answer:

26. $\frac{3}{4} \leq x$

27. $x\leq 5$

28. $0\leq b$

Step-by-step explanation:

26. $\sqrt{4x-3}$:

Taking square on both the sides to get:

$(\sqrt{4x-3} )^2 < (5)^2$

$4x-3$

$4x$

$x$

$x$

For non-negative values for radical:

$4x-3\geq 0$

$x\geq \frac{3}{4}$

So solution for this: $\frac{3}{4} \leq x$

27. $2+\sqrt{4x-4}\leq 6$

Subtracting 2 from both the sides to get:

$2+\sqrt{4x-4}-2\leq 6-2$

$\sqrt{4x-4}\leq 4$

Taking square root on both sides:

$(\sqrt{4x-4})^2\leq (4)^2$

$4x-4\leq 16$

$x\leq \frac{20}{4}$

$x\leq 5$

28. $\sqrt{b+12} -\sqrt{b}>2$

Adding $\sqrt{b}$ to both the sides to get:

$\sqrt{b+12} -\sqrt{b}+\sqrt{b}>2+\sqrt{b}$

$\sqrt{b+12} >2+\sqrt{b}$

Taking square on both sides:

$(\sqrt{b+12})^2 >(2+\sqrt{b})^2$

$b+12>(2+\sqrt{b})^2$

$b+12>4+4\sqrt{b}+b$

$4+4\sqrt{b} +b$

Subtracting $b$ from both sides to get:

$4+4\sqrt{b} +b-b$

$4+4\sqrt{b}$

Subtracting 4 from both sides:

$4+4\sqrt{b}-4$

$4\sqrt{b}$

Square both sides again:

$(4\sqrt{b})^2$

$16b$

$b$

$b$

and for non-negative radical $b\geq 0$

therefore, solution is $0\leq b$.