Quadratic equation: solving with the discriminant and Vieta's theorem

A quadratic equation is written in the form $ax^2+bx+c=0$, and $a\ne0$ must hold. On the set of real numbers, the discriminant determines the number of roots; when the coefficients are convenient, Vieta's theorem helps to find the roots quickly or to evaluate an expression involving them.

Below, the equation is first reduced to standard form, then a suitable method is chosen, and the answer is checked by substitution or by Vieta's identities.

The quadratic equation and its types

General form: $ax^2 + bx + c = 0$, where $a$ is the leading coefficient, $b$ is the second coefficient, and $c$ is the constant term. If $a = 0$ and $b\ne0$, the equation becomes linear, so the condition $a \neq 0$ is required.

If $b = 0$ or $c = 0$, the equation is called an incomplete quadratic equation, and it can also be solved without a formula:

• in the form $ax^2 + c = 0$, we obtain the root directly: from $3x^2 - 12 = 0$ we get $x^2 = 4$, so $x = \pm 2$;
• in the form $ax^2 + bx = 0$, we factor: from $x^2 + 5x = 0$ we get $x(x + 5) = 0$, so $x = 0$ or $x = -5$.

For a complete equation, the universal tool is the discriminant.

Solving using the discriminant

The discriminant is found by the following formula:

$D = b^2 - 4ac$

On the set of real numbers, its sign tells the number of roots in advance: if $D > 0$, there are two distinct real roots; if $D = 0$, there is one double root; if $D < 0$, there is no real root.

The formula for the roots: $x_{1,2} = \dfrac{-b \pm \sqrt{D}}{2a}$.

1-Example. Solve the equation $2x^2 + 3x - 5 = 0$.

Coefficients: $a = 2$, $b = 3$, $c = -5$. Discriminant: $D = 3^2 - 4 \cdot 2 \cdot (-5) = 9 + 40 = 49$. Since $\sqrt{49} = 7$:

$x_1 = \dfrac{-3 + 7}{4} = 1$, $\quad x_2 = \dfrac{-3 - 7}{4} = -\dfrac{5}{2}$.

We check: $2 \cdot 1^2 + 3 \cdot 1 - 5 = 0$. For the second root, Vieta's identities also hold: $1+(-2{,}5)=-1{,}5=-\frac{b}{a}$ and $1\cdot(-2{,}5)=-2{,}5=\frac{c}{a}$. Answer: $x_1 = 1$, $x_2 = -2{,}5$.

2-Example. Solve the equation $x^2 - 6x + 9 = 0$.

$D = (-6)^2 - 4 \cdot 1 \cdot 9 = 36 - 36 = 0$. Thus, there is one root: $x = \dfrac{6}{2} = 3$. This can also be seen in the form $(x - 3)^2 = 0$ — the left-hand side factors as a perfect square.

Vieta's theorem: the art of solving mentally

For a reduced quadratic equation, that is, of the form $x^2 + px + q = 0$ ($a = 1$), Vieta's theorem states: if real roots exist,

$x_1 + x_2 = -p \quad$ and $\quad x_1 \cdot x_2 = q$.

In other words, the sum of the roots equals the second coefficient with the opposite sign, and the product equals the constant term. When the coefficients are small integers, this method makes it possible to find the roots mentally.

3-Example. Solve the equation $x^2 + 7x + 12 = 0$.

We look for two numbers whose product is $12$ and whose sum is $-7$. The product is positive, so the roots have the same sign; the sum is negative, so both are negative. $(-3) \cdot (-4) = 12$ and $(-3) + (-4) = -7$. Answer: $x_1 = -3$, $x_2 = -4$.

To make sure, we check with the discriminant: $D = 49 - 48 = 1$, $x = \dfrac{-7 \pm 1}{2}$, that is, $-3$ and $-4$ — they match.

4-Example. Solve the equation $x^2 - 5x + 6 = 0$.

The numbers whose sum is $5$ and whose product is $6$ are $2$ and $3$. Answer: $x_1 = 2$, $x_2 = 3$. Checking by substitution also confirms this: $2^2 - 5 \cdot 2 + 6 = 0$.

If $a \neq 1$, Vieta's identities work directly in the form $x_1+x_2=-\dfrac{b}{a}$ and $x_1x_2=\dfrac{c}{a}$. If finding the roots mentally is inconvenient, the discriminant remains the universal choice.

Vieta's theorem also works in the reverse direction: it is very convenient for recovering an equation from given roots or for evaluating expressions such as $x_1^2 + x_2^2 = (x_1 + x_2)^2 - 2x_1 x_2$ without solving the equation.

Independent exercise

Solve the equation $3x^2-12x+9=0$ and check the roots with Vieta's identities. Short answer: If we divide the equation by 3, we obtain $x^2-4x+3=0$; the roots are $1$ and $3$, their sum is $4$, and their product is $3$.

Typical mistakes and ways to avoid them

In practice, the most common mistakes are the following:

• Losing the sign. In 1-Example, $a$ is positive and $c$ is negative, so the term $-4ac$ becomes positive. In 1-Example, replacing $9 + 40$ with $9 - 40$ spoils the entire solution. Let it become a habit to calculate by placing negative coefficients in parentheses.

• Forgetting $2a$ in the denominator. In the formula, the denominator is $2a$, not simply $2$: if $a = 2$, the denominator is $4$.

• A sign mix-up in Vieta's theorem. The sum equals $-p$, not $p$. In $x^2 + 7x + 12 = 0$, the sum of the roots is $-7$, and the product is $+12$.

• Taking $D = 0$ to mean “no root”. Over the real numbers there is one double root; the absence of a real root belongs to the case $D < 0$.

• Not reducing the equation to standard form. The coefficients cannot be taken directly from $x^2 = 5x - 6$; first it is reduced to the form $x^2 - 5x + 6 = 0$.

• Not checking. Substitute the roots found into the original equation, or compare their sum and product with Vieta's identities.

Conclusion and the next step

• First reduce the equation to the form $ax^2+bx+c=0$ and check $a\ne0$.
• The number of real roots is determined by the sign of $D=b^2-4ac$.
• For convenient integer coefficients, Vieta's theorem is a fast method, and the discriminant is the universal method.
• Check the answer independently by substitution or by Vieta's identities.
• Solve the tests on quadratic equations or take a new example in the daily problem.