Question

If one of the roots of a quadratic equation having rational coefficients is \(\sqrt{7}-4\), then the quadratic equation is (1) \(x^{2}-2 \sqrt{7} x-9=0 .\) (2) \(\mathrm{x}^{2}-8 \mathrm{x}+9=0\). (3) \(\mathrm{x}^{2}+8 \mathrm{x}+9=0 .\) (4) \(\mathrm{x}^{2}-2 \sqrt{7} \mathrm{x}+9=0 .\)

Short answer

Answer: The correct quadratic equation is \(x^2-2\sqrt{7}x-9=0\).

Step-by-step solution

Roots

\(\alpha=\sqrt{7}-4\) and \(\beta=\sqrt{7}+4\)

Sum of roots

\(\alpha+\beta = (\sqrt{7}-4)+(\sqrt{7}+4) = 2\sqrt{7}\)

Product of roots

\(\alpha\beta = (\sqrt{7}-4)(\sqrt{7}+4) = 7-16 = -9\) Now, we will substitute \(\alpha+\beta=2\sqrt{7}\) and \(\alpha\beta=-9\) in each of the given options to identify the correct quadratic equation.

Option 1

\(x^2-2\sqrt{7}x-9=0\) Here, sum of roots is \(2\sqrt{7}\) and product of roots is -9, which matches with our values. Thus, option 1 is the correct quadratic equation. Since we found the correct quadratic equation in option 1, there is no need to check the other options. Therefore, the correct equation is: \(x^2-2\sqrt{7}x-9=0\)

Key concepts

Rational Coefficients

When we refer to a quadratic equation with rational coefficients, we're talking about a polynomial equation where the numbers in front of the variables—often called coefficients—are rational numbers. A rational number is any number that can be expressed as a fraction, such as 1/2, 3, or -7. This means that in the context of a quadratic equation like \( ax^2 + bx + c = 0 \), the coefficients \( a \), \( b \), and \( c \) are all rational numbers.

Having rational coefficients is an essential part of the structure of some quadratic equations, especially when dealing with irrational roots. If the coefficients of a quadratic equation are rational and one of the roots is irrational—as in the case of our exercise—the conjugate of this irrational root must also be a root of the equation. This means if \( \sqrt{7}-4 \) is a root, \( \sqrt{7}+4 \) must also be a root to ensure the coefficients remain rational.

Sum and Product of Roots

The sum and product of the roots of a quadratic equation \( ax^2 + bx + c = 0 \) offer a very useful property. They relate directly to the coefficients of the equation. According to Viète's formulas:

• The sum of the roots \( \alpha + \beta = -\frac{b}{a} \).
• The product of the roots \( \alpha \beta = \frac{c}{a} \).

In our exercise, once we knew one root was \( \sqrt{7}-4 \), we determined its conjugate \( \sqrt{7}+4 \) based on the requirement for rational coefficients. This gave us the sum \( \alpha + \beta = 2\sqrt{7} \) and the product \( \alpha \beta = -9 \). These values helped us confirm the correct quadratic equation was \( x^2 - 2\sqrt{7}x - 9 = 0 \), as it aligned with the calculated sum and product.

Irrational Roots

An irrational root is a solution to an equation that cannot be precisely expressed as a simple fraction. Instead, it's often represented by a symbol like \( \sqrt{\text{something}} \). In our problem, \( \sqrt{7}-4 \) is an example of an irrational root because \( \sqrt{7} \) cannot be expressed as a fraction of two integers.

Whenever we have a quadratic equation with rational coefficients, if we find one irrational root, we know that its conjugate must also be a root. This is to maintain the rationality of the coefficients. So, having \( \sqrt{7}-4 \) means \( \sqrt{7}+4 \) must also be a part of the roots.

This constraint ensures that any irrational parts cancel each other out when forming the equation from its roots, which keeps the coefficients rational. Hence, identifying irrational roots and their conjugates is crucial in forming quadratic equations with rational coefficients.