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# Algebra

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What is the largest value of k such that the equation 6x - x^2 = k has at least one real solution?

Feb 27, 2022

#1
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Rearrange the equation, x- 6x + k = 0 , then solve using the quadratic formula.
You will realise that the only possibility for x to be real is when the term under the square root (b- 4ac) must be greater than or equal to 0, so you do not get a negative result under the square root (which makes it imaginary).

Feb 27, 2022
#2
+361
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ok, so

6x - x^2 = k

everytime x is increased by 1, 6x is increased by 6

everytime x is increased by 1, x^2 is increased by 2x+1, since,

$$(x+1)\times(x+1)$$

$$=x(x+1)+1(x+1)$$

$$=x^2+x+x+1$$

I'll let you figure this out

p.s also trial and error will work just as fine

Feb 27, 2022