Answer:
1) [tex]\mathbf{\sqrt{50}= 5\sqrt{2} }[/tex]
2) [tex]\mathbf{\sqrt{28}= 2\sqrt{7} }[/tex]
3) [tex]\mathbf{\sqrt{252}= 6\sqrt{7} }[/tex]
Step-by-step explanation:
We need to simplify the radicals:
1) [tex]\sqrt{50}[/tex]
First we find prime factors of 50
50 = 2 x 5 x 5
Now,
[tex]\sqrt{50}\\=\sqrt{2\times 5\times 5}\\=\sqrt{2\times 5^2}\\ =\sqrt{2} \sqrt{5^2}\\=5\sqrt{2}[/tex]
So, [tex]\mathbf{\sqrt{50}= 5\sqrt{2} }[/tex]
2) [tex]\sqrt{28}[/tex]
First we find prime factors of 28
28 = 2 x 2 x 7
Now,
[tex]\sqrt{28}\\=\sqrt{2\times 2\times 7}\\=\sqrt{2^2\times 7}\\ =\sqrt{2^2} \sqrt{7}\\=2\sqrt{7}[/tex]
So, [tex]\mathbf{\sqrt{28}= 2\sqrt{7} }[/tex]
3) [tex]\sqrt{252}[/tex]
First we find prime factors of 252
50 = 2 x 2 x 3 x 3 x 7
Now,
[tex]\sqrt{252}\\=\sqrt{2\times 2\times 3\times 3 \times7}\\=\sqrt{2^2\times 3^2 \times7}\\ =\sqrt{2^2} \sqrt{3^2}\sqrt{7} \\=2\times 3\sqrt{7}\\=6\sqrt{7}[/tex]
So, [tex]\mathbf{\sqrt{252}= 6\sqrt{7} }[/tex]
