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NDA 2024 Maths · Mathematics · Quadratic Equations · easy

The roots α,β of x^2-6x+4=0 are used to form the sum of reciprocals of their squares, (1)/(α^2+(1)/(β^2. Its value is

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The roots α,β\alpha,\beta of x2−6x+4=0x^2-6x+4=0 are used to form the sum of reciprocals of their squares, 1α2+1β2\dfrac1{\alpha^2}+\dfrac1{\beta^2}. Its value is
  1. A. 32\dfrac32
  2. B. 94\dfrac{9}{4}
  3. C. 72\dfrac{7}{2}
  4. D. 74\dfrac74
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Answer: (D) 74\dfrac74

1α2+1β2=α2+β2(αβ)2=36−816=74\frac1{\alpha^2}+\frac1{\beta^2}=\frac{\alpha^2+\beta^2}{(\alpha\beta)^2}=\frac{36-8}{16}=\frac74. Trap: using αβ\alpha\beta rather than (αβ)2(\alpha\beta)^2 in the denominator gives 72\frac72.
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