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Here is the equation I am stuck at:

$$2x^2+3x+1=0$$

I need to find the roots without calculating the discriminant.

So this is where I stopped:

$$2x^2+3x+1=0 \Rightarrow x^2 + (3/2)x + 1/2 = 0. $$

Everything is divided by $2$ so I can get this form

$$x^2-Sx+P=0;\ S=x_1+x_2 \text{ and }P=x_1x_2$$ $$\Rightarrow S=x_1+x_2= -3/2 ,\ \ P=x_1.x_2= 1/2.$$

How can I get the value of $x_1$ and $x_2$?

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  • $\begingroup$ Do you know about completing the square? $\endgroup$ – amd Sep 27 '15 at 6:56
  • $\begingroup$ completing the square means by finding the discriminant, right? if yes i want to find the roots without finding the descriminant $\endgroup$ – evexoio Sep 27 '15 at 6:57
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    $\begingroup$ That might help: Call $y_1 = 2x_1$, $y_2 = 2x_2$ Then $y_1+ y_2 =-3$, $y_1y_2 = 2$. Is it easier to guess $y_1$, $y_2$ now? $\endgroup$ – user99914 Sep 27 '15 at 7:12
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    $\begingroup$ Actually, even more fundamental, it's factorable in integers (by grouping). -- 2012books.lardbucket.org/books/beginning-algebra/… $\endgroup$ – Daniel R. Collins Sep 27 '15 at 7:16
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    $\begingroup$ @DanielR.Collins OMG! that worked and is so easy! thanks $\endgroup$ – evexoio Sep 27 '15 at 7:25
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Just to turn the comment into an answer: it's solvable by factoring in integers. To factor $ax^2 + bx + c$, specifically $2x^2+3x+1$, note that $a \cdot c = 2$, and we can factor this as $2 \cdot 1$, whose factors conveniently sum to $b = 3$. Thus by grouping we have: $2x^2+3x+1 = 2x^2 + 2x + 1x + 1 = 2x(x+1) + 1 (x+1) = (x+1)(2x+1)$.

So the original equation is equivalent to $(x+1)(2x+1) = 0$; by the zero product property either $x+1=0$ or $2x+1=0$, and from these follow the solutions $x = \{-1, -1/2\}$.

Note that this process of solving by factoring is the fundamental tool to solve higher-degree equations, and points in the direction of the Fundamental Theorem of Algebra: each factor of a polynomial generates one root.

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Vieta's formulas might be useful here. They are more general, but let us have a look at the quadratic case.

If $x_1,x_2$ are roots of $x^2+ax+b$, then we have $$(x-x_1)(x-x_2)=x^2+ax+b.$$ After bit of algebraic manipulation we can see that this means that $$ \begin{align*} x_1+x_2&=-a\\ x_1x_2&=b \end{align*} $$ If $a$ and $b$ are some simple numbers, then we can guess the numbers $x_1$, $x_2$ simply by trial and error.

In this case: Can you think of numbers $x_{1,2}$ such that their sum is $-\frac32$ and their product is $\frac12$?

It is not very difficult to see that $-1$ and $-\frac12$ fulfill these conditions.

If you know rational root theorem, this can make things a bit easier for you.

Notice that the original equation $2x^2+3x+1$ has integer coefficients, so this theorem is applicable for this polynomial.

Rational root theorem tells us in this case that if a rational number $\frac pq$ is a root of this polynomial, then $p$ divides $1$ and $q$ is a divisor of $2$. So the only possible candidates for rational roots are $\pm1$ and $\pm\frac12$.

We can simply plug these numbers into the original equation, to see if some of them is root. Or we can use Vieta's formulas instead.

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  • $\begingroup$ You meant the possible candidates for rational roots are $\pm 1$ and $\pm \color{red}{\frac{1}{2}}$. It looks like you forgot to use the \frac command. $\endgroup$ – N. F. Taussig Sep 27 '15 at 9:58
  • $\begingroup$ Thanks @N.F.Taussig, I have corrected it. (I should probably read more carefully what I post here.) $\endgroup$ – Martin Sleziak Sep 27 '15 at 10:00
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$$2x^2+3x+1=x^2+2x+1+x^2+x$$ $$\implies 2x^2+3x+1=(x+1)^2+x(x+1)$$$$\implies 2x^2+3x+1=(x+1)(x+1)+x(x+1)$$$$\implies 2x^2+3x+1=(x+1)((x+1)+x)$$$$\implies 2x^2+3x+1=(x+1)(2x+1)$$

Set to zero and you will get the roots!

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  • $\begingroup$ nice, but how did you convert (x+1)^2 + x(x+1) into (x+1)(x+1+x), please simplify more $\endgroup$ – evexoio Sep 27 '15 at 7:31
  • $\begingroup$ @AbdallahSamman take a look now, do you get it. $x+1$ is the common factor!! I used the property $ab+ac=a(b+c)$ In your case $a=x+1$,$b=x+1$,$c=x$ $\endgroup$ – Wanderer Sep 27 '15 at 7:39
  • $\begingroup$ you are welcome! $\endgroup$ – Wanderer Sep 27 '15 at 7:46
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Your polynomial has integer coefficients, so you can use the following theorem very easy to prove:

Theorem. The integer roots of a polynomial with integer coefficients are divisors of its constant coefficient.

So the possible integer roots of $2x^2 + 3x + 1$ are divisors of 1, namely -1 and 1. Test them, and find the second root using one of the relations between coefficients and roots.

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$x^2+2(\frac{3}{4})x+\frac{9}{16}-\frac{9}{16}+\frac{1}{2}=0$

$(x+\frac{3}{2})^2-\frac{1}{16}=0$

$(x+\frac{3}{2})=\pm{\frac{1}{4}}$

Now you can solve for x.

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  • $\begingroup$ from where did all of this come, my equation is 2x^2 + 3x + 1 = 0 $\endgroup$ – evexoio Sep 27 '15 at 7:33
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    $\begingroup$ This is “completing the square,” although I’ve usually seen it done with the constant term moved to the r.h.s. Once the equation is in the form $x^2-2Bx=-C$, you add $B^2$ to both sides so that you can factor the l.h.s. as $(x-B)^2$. This is, incidentally, one way to derive the quadratic formula. $\endgroup$ – amd Sep 27 '15 at 17:39
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    $\begingroup$ This answer has the clearly desired method as the question posed to the poster carefully avoided demanding factoring. (And if one is queasy about ±(square root), one can take the positive square root and use Vieta for x = –1.) $\endgroup$ – Roy Aug 3 '16 at 4:08
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Different method altogether, though I stand by the idea that Display name's method is the desired solution.

I call this the over/under method. For instance: x^2 + 3x + 2 = 0. Add to it (both sides) to get the square using one more "x"... more clearly stated, get x^2 + 4x + 4 = x + 2 ("Over"). Solving, the left side is (x + 2)^2 and divide each side by (x + 2) leaving x + 2 = 1, then solve to get x = –1.

Second step is the other direction, going down an "x": x^2 + 2x + 1 = –x – 1 with the left side being (x + 1)^2 ("Under"). Divide each side by (x + 1) to get x + 1 = –1 and solve to get x = –2.

So x = –1, –2. No discriminants, no factoring of the nasty, ugly kind. I.e.: the hard kind.

Over/under method. Nothing complicated, just add an x, subtract an x, complete each square, divide to get the squared term equals 1 or –1, solve each.

(For a ≠ 1, as in this problem, divide through by "a" first for the easy solution (c'mon, fractions aren't hard to work with), or just do the slightly harder completing of the square, if you like "harder.")

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