Relationship between Zeros and Coefficients of Quadratic Polynomials

The relationship between the zeros (roots) of a quadratic equation and its coefficients provides a quick and elegant way to find the sum and product of the roots without actually solving the equation. This is particularly useful when dealing with complex or messy quadratic equations.

Formulae

Consider a general quadratic equation in the form $ax^2 + bx + c = 0$, where $a$, $b$, and $c$ are coefficients, and $a \neq 0$. Let $\alpha$ and $\beta$ be the zeros (roots) of the quadratic equation.

• Sum of the roots: $\alpha + \beta = -\frac{b}{a}$
• Product of the roots: $\alpha \cdot \beta = \frac{c}{a}$

Examples

Example-1:

Find the sum and product of the roots of the quadratic equation $2x^2 + 5x – 3 = 0$.

Here, $a = 2$, $b = 5$, and $c = -3$.

• Sum of the roots: $-\frac{b}{a} = -\frac{5}{2}$
• Product of the roots: $\frac{c}{a} = \frac{-3}{2} = -\frac{3}{2}$

Example-2:

If the sum of the roots of a quadratic equation is 4 and the product of the roots is -12, and the coefficient of $x^2$ is 1, then find the quadratic equation.

We know that the sum of roots is $-\frac{b}{a} = 4$ and product of roots is $\frac{c}{a} = -12$. Since $a=1$ then $b = -4$ and $c=-12$.

The equation is $x^2 – 4x -12 = 0$

Theorem with Proof

Theorem: For a quadratic equation $ax^2 + bx + c = 0$, the sum of the roots is $-\frac{b}{a}$ and the product of the roots is $\frac{c}{a}$.

Proof:

We can use the quadratic formula to find the roots of the equation:

$x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}$

Let the two roots be $\alpha$ and $\beta$. Then:

$\alpha = \frac{-b + \sqrt{b^2 – 4ac}}{2a}$

$\beta = \frac{-b – \sqrt{b^2 – 4ac}}{2a}$

Sum of the roots:

$\alpha + \beta = \frac{-b + \sqrt{b^2 – 4ac}}{2a} + \frac{-b – \sqrt{b^2 – 4ac}}{2a} = \frac{-2b}{2a} = -\frac{b}{a}$

Product of the roots:

$\alpha \cdot \beta = \left(\frac{-b + \sqrt{b^2 – 4ac}}{2a}\right) \cdot \left(\frac{-b – \sqrt{b^2 – 4ac}}{2a}\right) = \frac{(-b)^2 – (\sqrt{b^2 – 4ac})^2}{(2a)^2} = \frac{b^2 – (b^2 – 4ac)}{4a^2} = \frac{4ac}{4a^2} = \frac{c}{a}$

Common mistakes by students

• Forgetting the negative sign: A common mistake is forgetting the negative sign in the formula for the sum of the roots, leading to calculating $\frac{b}{a}$ instead of $-\frac{b}{a}$.
• Misidentifying coefficients: Students sometimes misidentify the coefficients $a$, $b$, and $c$ from the quadratic equation, especially when the equation is not in standard form or when terms are missing.
• Confusing sum and product: Students sometimes confuse the formula for the sum with the formula for the product or vice versa.

Real Life Application

The relationship between the zeros and coefficients can be used in various real-life applications, such as:

• Engineering: Analyzing the stability of systems, modeling the behavior of circuits, and designing structures.
• Physics: Describing projectile motion, analyzing oscillations, and solving problems related to energy.
• Financial modeling: Analyzing investment returns, predicting market trends, and calculating portfolio risk.

Fun Fact

The relationship between the roots and coefficients of a polynomial is not limited to quadratic equations. Similar relationships exist for higher-degree polynomials, and they are known as Vieta’s formulas. For example, for a cubic equation $ax^3 + bx^2 + cx + d = 0$, the sum of the roots is $-\frac{b}{a}$, and the sum of the products of the roots taken two at a time is $\frac{c}{a}$, and the product of the roots is $-\frac{d}{a}$.

Recommended YouTube Videos for Deeper Understanding

Q.1 If the sum of the zeros of the quadratic polynomial $kx^2 + 2x + 3k$ is equal to their product, then the value of k is:

Check Solution

Ans: D

Sum of zeros $= -\frac{2}{k}$, Product of zeros $= \frac{3k}{k} = 3$. Given: $-\frac{2}{k} = 3$, hence $k = -\frac{2}{3}$

Q.2 The quadratic polynomial whose zeros are $2 + \sqrt{3}$ and $2 – \sqrt{3}$ is:

Check Solution

Ans: A

Sum of zeros $= (2 + \sqrt{3}) + (2 – \sqrt{3}) = 4$. Product of zeros $= (2 + \sqrt{3})(2 – \sqrt{3}) = 4 – 3 = 1$. Required polynomial: $x^2 – (\text{sum})x + \text{product} = x^2 – 4x + 1$.

Q.3 If $\alpha$ and $\beta$ are the zeros of the polynomial $f(x) = x^2 – p(x + 1) – c$, then $(\alpha + 1)(\beta + 1)$ is equal to:

Check Solution

Ans: A

$f(x) = x^2 – px – p – c$. Sum of zeros: $\alpha + \beta = p$. Product of zeros: $\alpha\beta = -(p + c)$. $(\alpha + 1)(\beta + 1) = \alpha\beta + \alpha + \beta + 1 = -(p + c) + p + 1 = 1 – c$.

Q.4 If one zero of the quadratic polynomial $x^2 – 4x + k$ is 6, then the value of k is:

Check Solution

Ans: D

Let the zeros be 6 and $\alpha$. Sum of zeros: $6 + \alpha = 4$, so $\alpha = -2$. Product of zeros: $6 \times -2 = k$, therefore $k = -12$.

Q.5 If the zeros of the quadratic polynomial $ax^2 + bx + c$ are reciprocal of each other, then:

Check Solution

Ans: A

Let the zeros be $\alpha$ and $\frac{1}{\alpha}$. Product of zeros $= \alpha \times \frac{1}{\alpha} = 1$. From the formula, product of zeros $= \frac{c}{a}$. Hence, $\frac{c}{a} = 1$, so $a = c$.

Next Topic: Forming a Quadratic Polynomial from Zeros