Given 3−2(x+3)−1=|x+3|+a has exactly two real roots, then the maximum possible value of |[a]| is? ( where [] denotes the greatest integer function )

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Korbin Rivera · Accepted Answer

Let f(x)=x2+(22−2a)x+a2−4a+58. x is a real root of the original equation iff it&#039;s a real root of the quadratic equationf(x)=0and −2(x+3)≥0⇔x≤−3 and (a+1)−(x+3)≥0⇔x≤a−2. Notice that by squaring the original equation we ensure that −2(x+3)≥0, and we only have to worry about x≤a−2. So, the quadratic equation needs to have distinct roots x1,x2 such that x1&amp;lt;x2≤a−2. The condition for distinct real roots is D&amp;gt;0⇔a&amp;lt;6318=3.5, and you&#039;ve already done this part. The condition for a−2 being larger than the roots is f(a−2)≥0 and a−2≥x1+x22=a−11. The second inequality is always true. I&#039;ll leave f(a−2)≥0 to you. The solutions for that are a≥−1.Combining all conditions for a, we get a∈[−1,3.5). Thus, the maximum value of |⌊a⌋| is 3. You got the right answer but without a complete proof.

Given \(\displaystyle{3}\sqrt{{-{2}{\left({x}+{3}\right)}}}-{1}={\left|{x}+{3}\right|}+{a}\) has exactly two real roots,

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