Multiple choice

If ${x}^{2}+px+q=0$ is the quadratic equation whose roots are $a-2$ and $b-2$, where a and b are the roots of ${x}^{2}-3x+1=0$

  1. $p=1q=5$
  2. $p=1q=-5$
  3. $p=1q=-1$
  4. $none of these$
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C Correct answer
Explanation

Roots of x^2-3x+1=0 are a, b. Sum a+b=3, product ab=1. New roots are a-2, b-2. New sum = (a+b)-4 = 3-4 = -1. New product = (a-2)(b-2) = ab - 2(a+b) + 4 = 1 - 2(3) + 4 = -1. Equation is x^2 - (-1)x + (-1) = 0, so x^2 + x - 1 = 0. Thus p=1, q=-1.

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