## Question 6, p.89 of Algebra guide (6th edition)

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MarcoR557
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### Question 6, p.89 of Algebra guide (6th edition)

Given (d/4)+(8/d)+3=0, what is d?

First step of the answer is to multiply the entire equation by 4d (denominator) to get d^2+32+12d=0. When I multiple the equation by 4d I get d^2+8d+3d. What am I doing wrong?

Thanks!
Marco
Sage Pearce-Higgins
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### Re: Question 6, p.89 of Algebra guide (6th edition)

When I multiple the equation by 4d I get d^2+8d+3d. What am I doing wrong?

(d/4)+(8/d)+3=0 If we multiply the equation by 4d, then we get:
(4d)(d/4) + (4d)(8/d) + (4d)(3) = 0
Take care to multiply the fractions correctly:
(4)(d^2)/4 + (32d)/d + 12d = 0
Then simplify the fractions by cancelling:
d^2 + 32 + 12d = 0

I don't see exactly what step you take to get your version; it looks like you're not multiplying each term by 4d.