quadratic functions: various questions

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Damo17

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Re: quadratic functions: various questions

« Reply #15 on: March 14, 2009, 11:20:00 am »

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$\sqrt{\frac{a^2}{16}+\frac{24}{16}}$

$= \sqrt{\frac{a^2}{16}} + \sqrt{\frac{24}{16}}$

$= \frac{\sqrt{a^2}}{4} + \frac{\sqrt{24}}{4}$

$= \frac{\sqrt{a^2+24}}{4}$

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TrueTears

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Re: quadratic functions: various questions

« Reply #16 on: March 14, 2009, 11:25:50 am »

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Quote from: Damo17 on March 14, 2009, 11:20:00 am

$\sqrt{\frac{a^2}{16}+\frac{24}{16}}$

$= \sqrt{\frac{a^2}{16}} + \sqrt{\frac{24}{16}}$

$= \frac{\sqrt{a^2}}{4} + \frac{\sqrt{24}}{4}$

$= \frac{\sqrt{a^2+24}}{4}$

2nd step is incorrect, you can't split up the surd. Although you did get the correct answer the working should be this:

$\sqrt{\frac{a^2}{16}+\frac{24}{16}}$

$= \frac{\sqrt{a^2+24}}{\sqrt{16}}$

$= \frac{\sqrt{a^2+24}}{4}$

« Last Edit: March 14, 2009, 11:30:02 am by TrueTears »

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Damo17

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Re: quadratic functions: various questions

« Reply #17 on: March 14, 2009, 11:49:56 am »

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Quote from: TrueTears on March 14, 2009, 11:25:50 am

Quote from: Damo17 on March 14, 2009, 11:20:00 am
$\sqrt{\frac{a^2}{16}+\frac{24}{16}}$

$= \sqrt{\frac{a^2}{16}} + \sqrt{\frac{24}{16}}$

$= \frac{\sqrt{a^2}}{4} + \frac{\sqrt{24}}{4}$

$= \frac{\sqrt{a^2+24}}{4}$

2nd step is incorrect, you can't split up the surd. Although you did get the correct answer the working should be this:

$\sqrt{\frac{a^2}{16}+\frac{24}{16}}$

$= \frac{\sqrt{a^2+24}}{\sqrt{16}}$

$= \frac{\sqrt{a^2+24}}{4}$

Yeah, always small errors. Thanks

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Hielly

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Re: quadratic functions: various questions

« Reply #18 on: March 14, 2009, 12:45:44 pm »

0

thnks guys for part the same question.
if the equation of the metal rod is modeled by the linear function y=11+bx, find the value of b

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