Recently I came across a sum which was of the type

$$25x+10\sqrt x+1 = 0.$$

Now I could solve this using factoring, but I'm wondering if this falls under the category of quadratic equation and if it does, is there any other way to solve this without factoring?

Also another sum of the same type :

$$4x^2+4\sqrt x+1.$$

  • $\begingroup$ Please use MathJax for future posts. $\endgroup$ Commented Jun 6, 2018 at 8:09
  • $\begingroup$ It's not a quadratic as you have you fractional power $\endgroup$ Commented Jun 6, 2018 at 8:09
  • $\begingroup$ How do you solve these sums? $\endgroup$ Commented Jun 6, 2018 at 8:10
  • 7
    $\begingroup$ For the first one, substitute $u=x^{1/2}$ $\endgroup$
    – Badr B
    Commented Jun 6, 2018 at 8:10
  • 3
    $\begingroup$ The second sum is always positive so it is never equal to $0$. $\endgroup$ Commented Jun 6, 2018 at 8:15

5 Answers 5


It's sort of a quadratic equation. Let $y=\sqrt x$. Then you want to solve the equation $25y^2+10y+1=0$. This equation has no non-negative roots. Therefore (since $y=\sqrt x$) the original equation has no solutions (in $\mathbb R$).


Technically speaking, this is not a quadratic equation, as the LHS is a "fractional" polynomial (of leading degree $1$).

With some care, you can indeed turn it to a quadratic problem

$$25t^2+10t+1=0$$ for which the resolution methods are well-known. After resolution for $t$, you still need to discuss the relevance of the solutions, knowing that $t=\sqrt x$.

Technically speaking again, finding the roots of a polynomial is always equivalent to factoring because if you have the roots you have the factors and conversely.

And actually you don't solve a quadratic equation by factoring, but by completing the square, in a way or another. A trigonometric solution is also possible, but... never used.

Your second equation can likewise be turned to a quartic one (fourth degree), the resolution of which is more tedious.

It can be achieved by factoring into two quadratic polynomials, which requires the resolution of a cubic equation (third degree). Interestingly, the latter is solved by turning it to a sextic equation (sixth degree, but in fact bi-cubic) that reduces to a quadratic one.

And for some values of the coefficients (when the cubic has three real roots), you can't avoid a recourse to trigonometry !



Let $t=\sqrt x$ than

$$25x+10\sqrt x+1 = 0\implies25t^2+10t+1=0$$

then solve for $t$ and finally determine $x$.

Note that the acceptable solutions are only the values $t\ge 0$.

In the other example we obtain

$$4x^2+4\sqrt x+1\implies 4t^4+4t+1=0$$

which can not be reduced to a quadratic.


Wrapping up the various comments and answers:

First equation: (see José Carlos Santos' answer)

Substitute $\sqrt{x}$ for $t$, then you need to solve $25t^2+10t+1=0$ which has no positive solutions. Hence, solving $t=\sqrt{x}$ for $x$ yields no solution and your initial equation has no solution.

Second equation:

Notice how $\forall x\in \mathbb{R}$, $x^2\geq 0$ and $\sqrt{x}\geq 0$ then, $4x^2 + 4\sqrt{x} +1\geq 1 > 0$ and $4x^2 + 4\sqrt{x} +1=0$ has no real solutions.


As $x\geq 0$, we have $$25x+10\sqrt x+1\geq1$$ Hence there is no real solution.


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