Find the real parameter so as the equation has no real solutions....

Question

My question is:

For which values of parameter $a\in \mathbb{R}$ the following equation $$25^x+(a-4) \,5^x-2a^2+a+3=0$$ has no real solutions?

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My idea is:

First of all we should transform the equation above in a quadratic equation but noting $5^x=t$ where $t\gt0$. So first of all I think that a quadratic equation has no real solutions when $D\lt 0$, but the second case , I think is when $D\gt0$, but the solutions are negative. But although it seems like logical, I still do not get the right, convenient solutions. I hope you help me solve this problem. Thank you very much!

Answer 1

For real solution the discriminant of $$(5^x)^2+5^x(a-4)+3+a-2a^2=0$$ must be $\ge0$

i.e., $$(a-4)^2-4\cdot1\cdot(3+a-2a^2)=9a^2-12a+4=(3a-2)^2$$ needs to be $$\ge0$$ which holds true for all real $a$

So, we have $$5^x=\frac{-(a-4)\pm(3a-2)}2$$

Now for real $\displaystyle x, 5^x>0$

Answer 2

\begin{align} 25^x+(a-4) \,5^x-2a^2+a+3&=5^{2x}+(a-4) \,5^x-2a^2+a+3\\ &=(5^x)^2+(a-4) \,5^x-2a^2+a+3 \end{align} Let $y=5^x$, then \begin{align} (5^x)^2+(a-4) \,5^x-2a^2+a+3=y^2+(a-4) \,y+3+a-2a^2=0 \end{align} The quadratic equation has no real roots iff its discriminant less than zero. Hence \begin{align} D&=0\\ (a-4)^2-4\cdot1\cdot(3+a-2a^2)&=0\\ a^2-8a+16+8a^2-4a-12&=0\\ 9a^2-12a+4&=0\\ (3a-2)^2&=0\\ a&=\frac{2}{3} \end{align} In this case, its discriminant will be negative if $a<\cfrac{2}{3}$.

Answer 3

Another answer got as far as \[5^x=\frac{-(a-4)\pm(3a-2)}2\] observed that $5^x$ must be positive, but didn't comment on what this implied about $a$.

Well, let's take the $\pm$ to be $+$. Then \begin{aligned} &\frac{-(a-4) + (3a-2)}2 > 0 \\ \implies&2a+2>0\\ \implies&a>-1,\end{aligned}

whereas, if that $\pm$ is $-$, then:

\begin{aligned} &\frac{-(a-4) - (3a-2)}2 > 0 \\ \implies&-4a+6>0\\ \implies&a<\textstyle\frac{2}{3}.\end{aligned}

Only one of these conditions need to be satisfied for a solution to exist, so there is always a solution for every value of $a$ (but, unusually for a quadratic, there is usually only one).