Get to the root!

Algebra Level 3

If b 1 b 2 = 2 ( c 1 + c 2 ) {b_1}{b_2} = 2({c_1} + {c_2}) then at least one of the equations x 2 + b 1 x + c 1 = 0 x^2 + {b_1}x + c_1 = 0 and x 2 + b 2 x + c 2 = 0 x^2 + {b_2}x + c_2 = 0 has __________ \text{\_\_\_\_\_\_\_\_\_\_} .

Real roots Imaginary roots Purely imaginary roots None

Rishi Raj by Rishi Raj , 23, India

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3 solutions

Arturo Presa
Apr 28, 2018

Let us prove it by contradiction. Assume that none of the equations has real solutions. This would imply that both discriminants have to be negative. So b 1 2 < 4 c 1 and b 1 2 < 4 c 2 b_1^2<4c_1\quad\text{and}\quad b_1^2<4c_2 From these two inequalities we can obtain that c 1 c_1 and c 2 c_2 have to be positive numbers, and then, using the condition given in the problem, the product b 1 b 2 b_1b_2 has to be positive too. Multiplying the inequalities that we mention above, we get that b 1 2 b 1 2 < 16 c 1 c 2 . ( ∗ ) b_1^2 b_1^2<16c_1c_2.\quad\quad(*) Besides that, we know that 4 c 1 c 2 ≤ ( c 1 + c 2 ) 2 , 4c_1c_2\leq(c_1+c_2)^2, and therefore 16 c 1 c 2 ≤ 4 ( c 1 + c 2 ) 2 ( ∗ ∗ ) 16c_1c_2\leq4(c_1+c_2)^2\quad\quad(**) Combining ( ∗ ) (*) and ( ∗ ∗ ) , (**), we get that b 1 2 b 1 2 < 4 ( c 1 + c 2 ) 2 . ( ∗ ∗ ∗ ) b_1^2 b_1^2<4(c_1+c_2)^2.\quad\quad(***) This inequality contradicts the condition b 1 b 2 = 2 ( c 1 + c 2 ) . {b_1}{b_2} = 2({c_1} + {c_2}) .

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Abhishek Sinha
May 31, 2018

Let D 1 D_1 and D 2 D_2 denote the discriminants of the first and second quadratic respectively. We have D 1 + D 2 = ( b 1 2 + b 2 2 ) − 4 ( c 1 + c 2 ) = ( b 1 2 + b 2 2 ) − 2 b 1 b 2 = ( b 1 − b 2 ) 2 ≥ 0. D_1+D_2= (b_1^2+b_2^2)-4(c_1+c_2)=(b_1^2+b_2^2)-2b_1b_2=(b_1-b_2)^2 \geq 0. Hence at least one of the the discriminant must be non-negative, i.e., at least one of the equations must have real roots.

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Tom Engelsman
Aug 13, 2016

If one adds these two quadratic equations together, then the result becomes:

2x^2 + (b1 + b2)*x + (c1 + c2) = 0;

or 2x^2 + (b1 + b2) x + (b1 b2)/2 = 0;

or x^2 + (b1 + b2) x/2 + (b1 b2)/4 = 0 (i).

The roots of (i) are found via the Quadratic Formula:

x = (1/2) * [ -(b1+b2)/2 (+/-) sqrt [((b1+b2)/2)^2 - 4(1)(b1*b2)/4]];

or x = (1/2) * [ -(b1+b2)/2 (+/-) sqrt [((b1^2 + 2 b1 b2 + b2^2) / 4 - (b1*b2)];

or x = (1/2) * [ -(b1+b2)/2 (+/-) sqrt [((b1^2 - 2 b1 b2 + b2^2) / 4];

or x = (1/2) * [ -(b1+b2)/2 (+/-) sqrt [(b1 - b2)^2) / 4];

or x = (1/2) * [ -(b1 + b2)/2 (+/-) (b1 - b2)/2];

or x = -b1/2 or -b2/2.

Both roots for (i) are real for all b1 & b2, hence at least one of the original quadratic equations must have real roots as well.

0 Helpful 0 Interesting 0 Brilliant 0 Confused

From Vieta's formulas, we know that if a quadratic equation has real coefficients, but complex roots, the roots must be conjugates. Both the sum and product of conjugates result in a real answer. The roots of the combined equation are potentially each the sum of complex conjugates. So, the conclusion that one of the original equations had to have real roots does not follow, although I believe the conclusion itself is correct.

A more convincing argument can be made by assuming that both original equations have complex conjugate roots. Call them (m + ni), (m - ni), (p + qi) and (p - qi). From Vieta, b1 = -[(m + ni) + (m - ni)] = -2 m. Similarly, b2 = -2 p. Note that both m and p are solutions to the combined equation above, but neither of the original equations has real roots. Continuing, c1 = (m + ni) (m - ni) = m^2 + n^2, and c2 = p^2 + q^2. Then b1 b2 = 4 m p, and 2 (c1 + c2) = 2 (m^2 + n^2 + p^2 + q^2). Equate these results and rearrange terms: m^2 - 2 m p + p^2 + n^2 + q^2 = (m - p)^2 + n^2 + q^2 = 0. This can only be true if all the squared terms are identically zero, contradicting the assumption that all 4 roots were complex conjugate pairs. Consequently, at least one of the original equations has real roots. QED.

Tom Capizzi - 4 years, 9 months ago

(Hello @Tom Engelsman , I tried here to write down your solutions in proper latex)

If one adds these two quadratic equations together, then the result becomes: 2 x 2 + ( b 1 + b 2 ) ∗ x + ( c 1 + c 2 ) = 0 ; 2x^2 + (b_1 + b_2)*x + (c_1 + c_2) = 0;

or 2 x 2 + ( b 1 + b 2 ) x + ( b 1 b 2 ) / 2 = 0 ; 2x^2 + (b_1 + b_2)x + (b_1b_2)/2 = 0;

or x 2 + ( b 1 + b 2 ) x / 2 + ( b 1 b 2 ) / 4 = 0 ( ∗ ) . x^2 + (b_1 + b_2)x/2 + (b_1b_2)/4 = 0 \quad(*).

The roots of (*) are found via the Quadratic Formula:

x = ( 1 / 2 ) ∗ [ − ( b 1 + b 2 ) / 2 ± ( ( b 1 + b 2 ) / 2 ) 2 − 4 ( 1 ) ( b 1 ∗ b 2 ) / 4 ] ; x = (1/2) * [ -(b_1+b_2)/2 \pm\sqrt {((b_1+b_2)/2)^2 - 4(1)(b_1*b_2)/4}];

or x = ( 1 / 2 ) ∗ [ − ( b 1 + b 2 ) / 2 ± ( ( b 1 2 + 2 b 1 b 2 + b 2 2 ) / 4 − ( b 1 ∗ b 2 ) ] ; x = (1/2) * [ -(b_1+b_2)/2 \pm\sqrt {((b_1^2 + 2b_1b_2 + b_2^2) / 4 - (b_1*b_2)}];

or x = ( 1 / 2 ) ∗ [ − ( b 1 + b 2 ) / 2 ± ( ( b 1 2 − 2 b 1 b 2 + b 2 2 ) / 4 ] ; x = (1/2) * [ -(b_1+b2)/2 \pm \sqrt {((b_1^2 - 2b_1b_2 + b_2^2) / 4}];

or x = ( 1 / 2 ) ∗ [ − ( b 1 + b 2 ) / 2 ± ( b 1 − b 2 ) 2 ) / 4 ] ; x = (1/2) * [ -(b_1+b_2)/2 \pm \sqrt {(b_1 - b_2)^2) / 4}];

or x = ( 1 / 2 ) ∗ [ − ( b 1 + b 2 ) / 2 ± ( b 1 − b 2 ) / 2 ] ; x = (1/2) * [ -(b_1 + b_2)/2 \pm (b_1 - b_2)/2];

or x = − b 1 / 2 or − b 2 / 2. x = -b_1/2 \quad\text{or}\quad -b_2/2. Both roots for (*) are real for all b 1 b_1 & b 2 , b_2, hence at least one of the original quadratic equations must have real roots as well.

Arturo Presa - 3 years, 1 month ago

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