Find roots of the quadratic equation 2x²+3x+1=0 without calculating the discriminant

Question

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 $x_2$?

Answer 1

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.

Answer 2

$$2x^2+3x+1=x^2+2x+1+x^2+x$$ $$\implies 2x^2+3x+1=(x+1)^2+x(x+1)$$$$\implies 2x^2+3x+1=(x+1)(x+1)+x(x+1)$$$$\implies 2x^2+3x+1=(x+1)((x+1)+x)$$$$\implies 2x^2+3x+1=(x+1)(2x+1)$$

Set to zero and you will get the roots!

Answer 3

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

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

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

Now you can solve for x.

Answer 4

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.

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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.

Answer 5

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.