Multiple choice

If $a$ and $b$ are rational and $b$ is not a perfect square, then the quadratic equation with rational coefficients whose one root is $\displaystyle \frac{1}{a+\sqrt{b}}$ is

  1. $\displaystyle x^{2}-2ax+(a^{2}-b)=0$
  2. $\displaystyle (a^{2}-b)x^{2}-2ax+1=0$
  3. $\displaystyle (a^{2}-b)x^{2}-2bx+1=0$

  4. none of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Root x = 1 / (a + sqrt(b)). Then 1/x = a + sqrt(b), so 1/x - a = sqrt(b). Squaring both sides: (1/x - a)^2 = b => 1/x^2 - 2a/x + a^2 = b. Multiply by x^2: 1 - 2ax + a^2x^2 = bx^2. Rearranging: (a^2 - b)x^2 - 2ax + 1 = 0.

AI explanation

Because the quadratic equation has rational coefficients, the irrational conjugate 1/(a - sqrt(b)) must also be a root. The sum of the roots is 1/(a + sqrt(b)) + 1/(a - sqrt(b)) = 2a / (a^2 - b), and the product of the roots is 1 / (a^2 - b). Using the standard form of a quadratic equation x^2 - (sum of roots)x + (product of roots) = 0, we substitute the values to get x^2 - (2a / (a^2 - b))x + 1 / (a^2 - b) = 0. Multiplying the entire equation by the non-zero quantity (a^2 - b) yields (a^2 - b)x^2 - 2ax + 1 = 0.

Keep practicing — related questions