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Equation $3\sqrt{-2(x+3)}-1=|x+3|+a$ has exactly two real roots, then the maximum possible value of $|[a]| $ is? ( where [] denotes the greatest integer function )

My attempt is as follows:-

$$-(x+3)>=0, (x+3)<=0$$ $$3\sqrt{-2(x+3)}-1=|x+3|+a$$ $$3\sqrt{-2(x+3)}-1=-(x+3)+a$$ $$3\sqrt{-2(x+3)}=(a+1)-(x+3)$$

Squaring both sides $$9(-2(x+3))=(a+1)^2+(x+3)^2-2(a+1)(x+3)$$ $$-18(x+3)=x^2+9+6x+a^2+1+2a-2(ax+3a+x+3)$$ $$x^2+x(18+6-2a-2)+a^2+2a-6a+54+9+1-6=0$$ $$x^2+x(22-2a)+a^2-4a+58=0$$

As it is given that equation has exactly two given roots:-

$$D>0$$ $$(22-2a)^2-4(a^2-4a+58)>0$$ $$(11-a)^2-(a^2-4a+58)>0$$ $$121+a^2-22a-a^2+4a-58>0$$ $$-18a+63>0$$ $$a<\dfrac{63}{18}$$

Hence maximum value of $|[a]|=\left|\left[\left(\dfrac{63}{18}\right)^-\right]\right|$=$3$. But actual answer is $1$

prat
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1 Answers1

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Let $f(x)=x^2+(22-2a)x+a^2-4a+58$. $x$ is a real root of the original equation iff it's a real root of the quadratic equation $$f(x)=0 $$ and $-2(x+3)\geq 0\Leftrightarrow x\leq -3$ and $(a+1)-(x+3)\geq 0\Leftrightarrow x\leq a-2$. Notice that by squaring the original equation we ensure that $-2(x+3)\geq 0$, and we only have to worry about $x\leq a-2$. So, the quadratic equation needs to have distinct roots $x_1,x_2$ such that $x_1<x_2\leq a-2$. The condition for distinct real roots is $D>0\Leftrightarrow a<63/18=3.5$, and you've already done this part. The condition for $a-2$ being larger than the roots is $f(a-2)\geq 0$ and $a-2\geq \frac{x_1+x_2}{2}=a-11$. The second inequality is always true. I'll leave $f(a-2)\geq 0$ to you. The solutions for that are $a\geq -1$.

Combining all conditions for $a$, we get $a\in[-1,3.5)$. Thus, the maximum value of $\left|\lfloor a\rfloor\right|$ is $3$. You got the right answer but without a complete proof.

bjorn93
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