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if the equation $$(x-2)e^x+a(x-1)^2=0,x\in R$$ have two real roots.

show that $$a>0$$

Following is a solution

since $$-a=\dfrac{(x-2)e^x}{(x-1)^2}$$ Let $$g(x)=\dfrac{(x-2)e^x}{(x-1)^2}\Longrightarrow g'(x)=\dfrac{e^x(x^2-4x+5)}{(x-1)^3}$$ so we have $$x>1,g'(x)>0\Longrightarrow g(x)\in (-\infty,+\infty)$$ and $$x<1,g'(x)<0\Longrightarrow g(x)\in (-\infty,0)$$ then have following Fig

enter image description here

if $-a=g(x)$ have two real roots,then $-a<0$,so we have $a>0$

$\color{red} {Is ~there ~any ~other~ solution?and ~I~ want ~more ~solution}$

math110
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1 Answers1

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Let $$f(x)=(x-2)e^x+a(x-1)^2$$ Let's differentiate: $$f'(x)=(x-1)e^x+2a(x-1)=(x-1)(2a+e^x)$$

For $a=0$, clearly $f$ has a single zero, namely $x=2$.

Suppose now that $a< 0$. Then the first derivative has the zeros $x=1$ and $x=\ln(-2a)$. Let $I$ be the interval $[1,\ln(-2a)]$ or $[\ln(-2a),1]$; pick the one that makes sense, depending on the value of $a$.

Inside of $I$, $f$ is decreasing, and increasing outside. For brevity, let $b=\ln(-2a)$.

Now, $f(1)=-e<0$ and $f(b)=(b-2)(-2a)+a(b-1)^2=a(5-4b+b^2)<0$. This means that $f$ has at most one zero.

EDIT: Although the exercise is finished, perhaps it is interesting to show that if $a>0$ then $f$ has two zeros: that is, the implication is, actually, bidirectional.

If $a>0$ we see that $f'$ has only the zero $x=1$. Moreover, $f$ is decreasing in $(-\infty,1)$ and increasing in $(1,\infty)$. Since $f(-1)=-e<0$, and $\lim_{x\to\pm\infty}f(x)=\infty$, $f$ has two zeros.

ajotatxe
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