Question

Suppose one of the roots of the equation \(a x^2 - b x + c = 0 \) is \( 2 + \sqrt{3}\), where a, b and c are rational numbers and \(a ≠ 0\). If \(b = c^3\) then |a| equals

• A. 2
• B. 3
• C. 4
• D. 1

Answer 1

The Correct Option is A

In a quadratic equation of the form \(ax^2 + bx + c = 0\), the roots can be calculated using the formula: \(x = \frac{(-b ± \sqrt{(b^2 - 4ac)}) }{ (2a)}.\)

Since all the coefficients a, b, and c are rational, and one of the roots, 2 + √3, is irrational, it implies that the other root must also be irrational and be the conjugate of \(2 + \sqrt{3}\), which is \(2 - \sqrt{3}\)

The sum of the roots can be found as: \(\frac{(-b) }{ a} = (2 + \sqrt3) + (2 - \sqrt3) = 4.\)

The product of the roots can be calculated as: \(\frac{c }{ a} = (2 + \sqrt3)(2 - \sqrt3) = 4 - 3 = 1.\)

However, it's important to note that the given quadratic equation is actually of the form \(ax^2 - bx + c = 0.\)

Hence, we can conclude that: \(\frac{b }{ a} = \frac{4 c }{ a} = 1\)

And it's also given that \(b = c^3.\)

Substituting \(b = c^3\) into the equation \(\frac{b }{ a} = 4\), we get: \(\frac{(c^3) }{ a} = 4 c^3 = 4a\)

Now, combining \(\frac{c }{ a} = 1\) with \(c^3 = 4a\), we have: \(c^3 = 4 c = ∛4 c = ∓2\) (taking cube root on both sides)

So, we've found that \(c = ∓2.\)

Finally, calculating the absolute value of a, we get: |a| = 2.

Answer 2

Given that \(a, b, c\) are rational numbers, which can be written in the form of \(\frac pq\).
So, if one root is \(2+\sqrt 3\) and let the other root to be \(x\).
The sum and product of the two roots must also be rational numbers.
Then, the other root must be the conjugate of \(2+\sqrt 3\) = \(2-\sqrt 3\)

Sum of the roots \(= 2+\sqrt 3 + 2-\sqrt 3\)
\(⇒\frac ba = 4\)
\(⇒b=4a\)

The product of the roots = \(=( 2+\sqrt 3) \times( 2-\sqrt 3)\)
\(⇒ \frac ca=1\)
\(⇒ c=a\)

Given that,
\(b=c^3\)
\(⇒4a=a^3\)
\(⇒a^2=4\)
\(⇒a=\sqrt 4\)
\(⇒a=+2, \ -2\)
\(⇒|a|=2\)

So, the correct option is (A): \(2\)