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Marely Raymond
Mathematics
30 June, 23:01
Rewrite the equation in ax^2 + bx+c form: x^2=5
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Gideon Mcbride
30 June, 23:15
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Answer = x² - 5
All you really have to do is move 5 to the other side, which makes it negative.
There is no B in this equation ... although technically it is 0.
If you prefer to write x² + 0x - 5, that's fine.
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Brynlee Black
30 June, 23:17
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=x2 (1-d) + 5x - (1+d)
Step-by-step explanation:
Most of us would call this "combining like terms", which is in fact equivalent to factoring common monomial factors.
Hopefully you are aware of the distribution property of multiplying over addition: for all a, b, c, a (b+c) = ab+ac. This can be used in reverse to combine like terms: ca+ba = (b+c) a, or perhaps more familiarly with renamed variables, ax+bx = (a+b) x=x (a+b).
In the given equation, one grouping is made with x2-dx2=1x2-dx2=x2 (1-d). A second grouping is made with - 1-d = - (1+d). You can check these via distribution if you like. Therefore: x2+5x-1-dx2-d=x2-dx2+5x-1-d=x2 (1-d) + 5x - (1+d).
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