Symmetrical Quadratics!

Algebra Level 4

Let a a and b b be the roots of the quadratic equation x 2 − 100 c x − 101 d = 0 x^2-100cx-101d=0 .

Let c c and d d be the roots of the quadratic equation x 2 − 100 a x − 101 b = 0 x^2-100ax-101b=0 .

If a , b , c a,b,c and d d are distinct, non zero real numbers, then find the value of:

a + b + c + d \huge{\color{#69047E}{{a+b+c+d}}}


The answer is 1020100.

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Yash Singhal by Yash Singhal , 22, India

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1 solution

Chew-Seong Cheong
Mar 21, 2015

Using Vieta's formulas, we have:

{ a + b = 100 c . . . ( 1 ) a b = − 101 d . . . ( 2 ) c + d = 100 a . . . ( 3 ) c d = − 101 b . . . ( 4 ) ⇒ { E q . 1 : b = 100 c − a . . . ( 1 a ) E q . 3 : d = 100 a − c . . . ( 3 a ) E q . 2 : a ( 100 c − a ) = − 101 ( 100 a − c ) . . . ( 2 a ) E q . 4 : c ( 100 a − c ) = − 101 ( 100 c − a ) . . . ( 4 a ) ⇒ { E q . 2 a : 100 a c − a 2 = − 10100 a + 101 c . . . ( 2 b ) E q . 4 a : 100 a c − c 2 = − 10100 c + 101 a . . . ( 4 b ) E q . 4 b − E q . 2 b : { a 2 − c 2 = 10100 ( a − c ) + 101 ( a − c ) ( a + c ) ( a − c ) = 10201 ( a − c ) a + c = 10201 . . . ( 5 ) ⇒ { a + b + c + d = a + 100 c − a + c + 100 a − c = 100 ( a + c ) = 100 × 10201 = 1020100 \begin{cases} a+b = 100c & ...(1) \\ ab = -101d & ...(2) \\ c+d = 100a & ...(3) \\ cd = -101b & ...(4) \end{cases} \\ \Rightarrow \begin{cases} Eq.1: & b = 100c - a & ...(1a) \\ Eq.3: & d = 100a - c & ...(3a) \\ Eq.2: & a(100c-a) = -101(100a-c) & ...(2a) \\ Eq.4: & c(100a-c) = -101(100c-a) & ...(4a) \end{cases} \\ \Rightarrow \begin{cases} Eq.2a: & 100ac-a^2 = -10100a+101c & ...(2b) \\ Eq.4a: & 100ac-c^2 = -10100c+101a & ...(4b) \end{cases} \\ Eq.4b-Eq.2b: \begin{cases} a^2-c^2 & = 10100(a-c)+101(a-c) \\ (a+c)(a-c) & = 10201(a-c) \\ a+c & = 10201 \quad ...(5) \end{cases} \\ \Rightarrow \begin{cases} a+b+c+d & = a+100c-a+c+100a-c \\ & = 100(a+c) \\ & = 100 \times 10201 \\ &= \boxed{1020100} \end{cases}

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Sir @Chew-Seong Cheong ,I'm getting 1020200.On multiplying eqn 2 and 4 a*c=10201.Since a,c are distinct hence a=10201 and c=1 or vice versa.

Utkarsh Bansal - 6 years, 2 months ago

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a a and c c are distinct reals but not necessary integers.

From a + c = 10201 a+c=10201 and a c = 10201 ac=10201 , we have a 2 − 10201 a + 10201 = 0 ⇒ a , c = 10201 ± 104019597 2 a^2-10201a+10201=0\quad \Rightarrow a,c = \dfrac {10201 \pm \sqrt{104019597}}{2} .

⇒ a , c = { 1.000098049 10199.9999 ⇒ b , d = { − 10099.9901 1019998.99 \Rightarrow a,c = \begin{cases} 1.000098049 \\ 10199.9999 \end{cases} \Rightarrow b,d = \begin{cases} -10099.9901 \\ 1019998.99 \end{cases}

⇒ a + b + c + d = 1020100 \Rightarrow a+b+c+d = \boxed{1020100}

Chew-Seong Cheong - 6 years, 2 months ago

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Ok I got it sir I mistaken a and c for being intergers.

Utkarsh Bansal - 6 years, 2 months ago

sir the solutions isnt exact,you must have some decimals

Omar El Mokhtar - 5 years, 11 months ago

not a problem of 400 points

Omar El Mokhtar - 5 years, 11 months ago

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It's rather LEVEL 4 now

Aakash Khandelwal - 4 years, 12 months ago

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