To solve the expression [tex]\( b \sqrt{40 b^2} + 3 \sqrt{90 b^4} \)[/tex], let's break it down step-by-step:
1. Evaluate each term separately:
First term: [tex]\( b \sqrt{40 b^2} \)[/tex]
Second term: [tex]\( 3 \sqrt{90 b^4} \)[/tex]
2. Simplify each term:
For the first term [tex]\( b \sqrt{40 b^2} \)[/tex]:
- Recognize that [tex]\( \sqrt{40 b^2} \)[/tex] can be rewritten as [tex]\( \sqrt{40} \cdot \sqrt{b^2} \)[/tex].
- Since [tex]\( \sqrt{b^2} = |b| \)[/tex], we get [tex]\( b \sqrt{40} \cdot |b| \)[/tex].
Given that [tex]\( b \sqrt{40} \cdot |b| = b \cdot \sqrt{40} \cdot b = b^2 \sqrt{40} \)[/tex].
Since we assume [tex]\( b \geq 0 \)[/tex] (or ignore the absolute value as a common practice in such contexts), we have:
[tex]\( b \sqrt{40 b^2} = b^2 \sqrt{40} \)[/tex]
Now express [tex]\( \sqrt{40} \)[/tex] in its simpler form:
[tex]\( \sqrt{40} = \sqrt{4 \cdot 10} = \sqrt{4} \cdot \sqrt{10} = 2 \sqrt{10} \)[/tex].
Thus, the first term simplifies to:
[tex]\( b \sqrt{40 b^2} = b^2 \cdot 2\sqrt{10} = 2 \sqrt{10} b^2 \)[/tex].
For the second term [tex]\( 3 \sqrt{90 b^4} \)[/tex]:
- Recognize that [tex]\( \sqrt{90 b^4} \)[/tex] can be rewritten as [tex]\( \sqrt{90} \cdot \sqrt{b^4} \)[/tex].
- Since [tex]\( \sqrt{b^4} = b^2 \)[/tex], we get [tex]\( 3 \sqrt{90} \cdot b^2 \)[/tex].
Now express [tex]\( \sqrt{90} \)[/tex] in its simpler form:
[tex]\( \sqrt{90} = \sqrt{9 \cdot 10} = \sqrt{9} \cdot \sqrt{10} = 3 \sqrt{10} \)[/tex].
Thus, the second term simplifies to:
[tex]\( 3 \sqrt{90 b^4} = 3 \cdot 3 \sqrt{10} \cdot b^2 = 9 \sqrt{10} b^2 \)[/tex].
3. Add the simplified terms together:
Now combine the results of the first and second term:
[tex]\( 2 \sqrt{10} b^2 + 9 \sqrt{10} b^2 \)[/tex]
Factor out the common term [tex]\( \sqrt{10} b^2 \)[/tex]:
[tex]\( \sqrt{10} b^2 (2 + 9) = \sqrt{10} b^2 \cdot 11 = 11 \sqrt{10} b^2 \)[/tex].
Thus, the simplified form of the original expression is:
[tex]\[ b \sqrt{40 b^2} + 3 \sqrt{90 b^4} = 11 \sqrt{10} b^2 \][/tex]
