Pergunta
Express in simplest radical form. -xsqrt (28)+10sqrt (63x^2) Answer Attempt 1 out of 3 square
Solução
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JoaoProfissional · Tutor por 6 anos
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To simplify the expression -x\sqrt{28} + 10\sqrt{63x^2}, we need to break down the radicals into their simplest forms.
First, let's simplify \sqrt{28}:
\sqrt{28} = \sqrt{4 \times 7} = \sqrt{4} \times \sqrt{7} = 2\sqrt{7}
So, -x\sqrt{28} becomes:
-x\sqrt{28} = -x \cdot 2\sqrt{7} = -2x\sqrt{7}
Next, let's simplify \sqrt{63x^2}:
\sqrt{63x^2} = \sqrt{9 \times 7 \times x^2} = \sqrt{9} \times \sqrt{7} \times \sqrt{x^2} = 3\sqrt{7} \times x = 3x\sqrt{7}
So, 10\sqrt{63x^2} becomes:
10\sqrt{63x^2} = 10 \cdot 3x\sqrt{7} = 30x\sqrt{7}
Now, combine the simplified terms:
-x\sqrt{28} + 10\sqrt{63x^2} = -2x\sqrt{7} + 30x\sqrt{7}
Combine the like terms:
-2x\sqrt{7} + 30x\sqrt{7} = (30x - 2x)\sqrt{7} = 28x\sqrt{7}
Thus, the simplified form of the expression is:
\boxed{28x\sqrt{7}}
First, let's simplify \sqrt{28}:
\sqrt{28} = \sqrt{4 \times 7} = \sqrt{4} \times \sqrt{7} = 2\sqrt{7}
So, -x\sqrt{28} becomes:
-x\sqrt{28} = -x \cdot 2\sqrt{7} = -2x\sqrt{7}
Next, let's simplify \sqrt{63x^2}:
\sqrt{63x^2} = \sqrt{9 \times 7 \times x^2} = \sqrt{9} \times \sqrt{7} \times \sqrt{x^2} = 3\sqrt{7} \times x = 3x\sqrt{7}
So, 10\sqrt{63x^2} becomes:
10\sqrt{63x^2} = 10 \cdot 3x\sqrt{7} = 30x\sqrt{7}
Now, combine the simplified terms:
-x\sqrt{28} + 10\sqrt{63x^2} = -2x\sqrt{7} + 30x\sqrt{7}
Combine the like terms:
-2x\sqrt{7} + 30x\sqrt{7} = (30x - 2x)\sqrt{7} = 28x\sqrt{7}
Thus, the simplified form of the expression is:
\boxed{28x\sqrt{7}}
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