I don't know its roots. I only know the product of 2 of its roots

Algebra Level 4

x 4 18 x 3 + k x 2 + 200 x 1984 = 0 \large \color{#D61F06}{x^{4}}-\color{#EC7300}{18x^{3}}+\color{#20A900}{kx^{2}}+\color{#3D99F6}{200x}-\color{#69047E}{1984}=\color{#333333}{0}

For the equation above, the product of two of its roots (in x x ) is 32 -32 . Find k k .


The answer is 86.

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1 solution

Let the four roots be a a , b b , c c and d d . Using Vieta's formula , we have:

{ a + b + c + d = 18 . . . ( 1 ) a b + a c + a d + b c + b d + c d = k . . . ( 2 ) a b c + a b d + a c d + b c d = 200 . . . ( 3 ) a b c d = 1984 . . . ( 4 ) \begin{cases} a+b+c+d = 18 &...(1) \\ ab+ac+ad+bc+bd+cd = k &...(2) \\ abc+abd+acd+bcd = -200 &...(3) \\ abcd = -1984 &...(4) \end{cases}

Let a b = 32 ab=-32 , then from ( 4 ) : c d = 1984 a b = 1984 32 = 62 (4): cd = \dfrac {-1984}{ab} = \dfrac {-1984}{-32} = 62

From (3):

a b c + a b d + a c d + b c d = 200 a b ( c + d ) + c d ( a + b ) = 200 32 ( c + d ) + 62 ( a + b ) = 200 32 ( a + b + c + d ) + 94 ( a + b ) = 200 ( 1 ) : a + b + c + d = 18 32 ( 18 ) + 94 ( a + b ) = 200 \begin{aligned} abc+abd+acd+bcd & = -200 \\ ab(c+d)+cd(a+b) & = -200 \\ -32(c+d)+62(a+b) & = -200 \\ -32({\color{#3D99F6}a+b+c+d})+94 (a+b) & = -200 & \small \color{#3D99F6} (1): a+b+c+d = 18 \\ -32({\color{#3D99F6}18})+94 (a+b) & = -200 \end{aligned}

a + b = 32 × 18 200 94 = 4 \implies a+b = \dfrac {32\times 18-200} {94} = 4

From (1): c + d = 18 ( a + b ) = 18 4 = 14 c+d = 18-(a+b) = 18-4 = 14

From (2):

k = a b + a c + a d + b c + b d + c d = a b + ( a + b ) ( c + d ) + c d = 32 + ( 4 ) ( 14 ) + 62 = 86 \begin{aligned} k & = ab+ac+ad+bc+bd+cd \\ & = ab + (a+b)(c+d) + cd \\ & = -32 + (4)(14) + 62 \\ & = \boxed{86} \end{aligned}

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