Quadratic Equations#1

Algebra Level 3

If equation x 2 − 2 a x − c = 0 x^2 - 2ax - c = 0 have exactly one root common with each of equation x 2 − 4 x + 1 = 0 x^2 - 4x + 1 = 0 and x 2 + 8 x + 13 = 0 x^2 + 8x + 13 = 0 . Which will always be correct:

  1. a a is a rational number
  2. a a is an irrational number
  3. c c is an irrational number
  4. c c is a rational number


The answer is 3.

Aakhyat Singh by Aakhyat Singh , 20, India

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1 solution

Tom Engelsman
Jan 21, 2018

Let r 1 , r 2 r_1, r_2 be the roots of x 2 − 4 x + 1 = 0 x^2 - 4x +1 = 0 and x 2 + 8 x + 13 = 0 x^2 + 8x + 13 = 0 respectively, which compute to:

r 1 = 2 ± 3 , r 2 = − 4 ± 3 r_1 = 2 \pm \sqrt{3}, r_2 = -4 \pm \sqrt{3}

If the quadratic equation x 2 − 2 a x − c = 0 x^2 - 2ax - c = 0 contains one root of each, then we have:

x 2 − 2 a x − c = ( x − r 1 ) ( x − r 2 ) = 0 ⇒ x 2 − ( r 1 + r 2 ) x + r 1 r 2 = 0 x^2 - 2ax - c = (x - r_1)(x - r_2) = 0 \Rightarrow x^2 - (r_1 + r_2)x + r_1 r_2 = 0 ; or 2 a = r 1 + r 2 , c = − r 1 r 2 \boxed{2a = r_1 + r_2, c = -r_1 r_2} .

Checking each possible root combination for 2 a , c 2a, c yields:

2 a = ( 2 + 3 ) + ( − 4 + 3 ) ⇒ a = − 1 + 3 2a = (2+\sqrt{3}) + (-4 + \sqrt{3}) \Rightarrow a = -1 + \sqrt{3} (IRRATIONAL);

2 a = ( 2 + 3 ) + ( − 4 − 3 ) ⇒ a = − 1 2a = (2+\sqrt{3}) + (-4 - \sqrt{3}) \Rightarrow a = -1 (RATIONAL);

2 a = ( 2 − 3 ) + ( − 4 + 3 ) ⇒ a = − 1 2a = (2-\sqrt{3}) + (-4 + \sqrt{3}) \Rightarrow a = -1 (RATIONAL);

2 a = ( 2 − 3 ) + ( − 4 − 3 ) ⇒ a = − 1 − 3 2a = (2-\sqrt{3}) + (-4 - \sqrt{3}) \Rightarrow a = -1 - \sqrt{3} (IRRATIONAL);

c = − ( 2 + 3 ) ( − 4 + 3 ) ⇒ c = 5 + 2 3 c = -(2+\sqrt{3})(-4+\sqrt{3}) \Rightarrow c = 5 + 2\sqrt{3} (IRRATIONAL);

c = − ( 2 + 3 ) ( − 4 − 3 ) ⇒ c = 11 + 6 3 c = -(2+\sqrt{3})(-4-\sqrt{3}) \Rightarrow c = 11 + 6\sqrt{3} (IRRATIONAL);

c = − ( 2 − 3 ) ( − 4 + 3 ) ⇒ c = 11 − 6 3 c = -(2-\sqrt{3})(-4+\sqrt{3}) \Rightarrow c = 11 - 6\sqrt{3} (IRRATIONAL);

c = − ( 2 − 3 ) ( − 4 − 3 ) ⇒ c = 5 − 2 3 c = -(2-\sqrt{3})(-4-\sqrt{3}) \Rightarrow c = 5 - 2\sqrt{3} (IRRATIONAL).

After this exhaustive exercise, only Choice (3) is the correct one.

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