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3 years ago

F(x)=1/4x^2+1/2x-35/4 how do I factor fractions?

1 answer:

5 0

let's multiply both sides by the LCD of all fractions, in this case is 4, just to do away with the denominators for a few seconds

$\bf f(x) = \cfrac{1}{4}x^2+\cfrac{1}{2}x-\cfrac{35}{4}\implies \stackrel{\textit{multiplying both sides by }\stackrel{LCD}{4}}{4[f(x)]=4\left( \cfrac{1}{4}x^2+\cfrac{1}{2}x-\cfrac{35}{4} \right)} \\\\\\ 4f(x) = x^2+2x-35\implies 4f(x) = (x+7)(x-5) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill f(x) = \cfrac{1}{4}(x+7)(x-5)~\hfill$

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Answer:

1) $\sqrt{x^7}=x^{\frac{7}{2}$

2) $\sqrt[3]{y^5}=y^{\frac{5}{3}$

3) $\sqrt[5]{a^{12}}=a^{\frac{12}{5} }$

4) $\sqrt[4]{z^{9}}=z^\frac{9}{4}$

Step-by-step explanation:

1) $\sqrt{x^7}$

We know that $\sqrt{x}=x^{\frac{1}{2}$

So, $\sqrt{x^7}=x^{\frac{7}{2}$

2) $\sqrt[3]{y^5}$

We know that $\sqrt[3]{x}=x^{\frac{1}{3}$

So, $\sqrt[3]{y^5}=y^{\frac{5}{3}$

3) $\sqrt[5]{a^{12}}$

We know that $\sqrt[5]{x}=x^{\frac{1}{5}$

So, $\sqrt[5]{a^{12}}=a^{\frac{12}{5} }$

4) $\sqrt[4]{z^{9}}$

We know that $\sqrt[4]{x}=x^{\frac{1}{4}$

So, $\sqrt[4]{z^{9}}=z^\frac{9}{4}$

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2 years ago

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