Solve the equations:
$x^{4} - 16 x^{3} + 86 x^{2} - 176 x + 105 = 0$ , two roots being $1$ and $7$ .

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Solution

$x^{4} - 16 x^{3} + 86 x^{2} - 176 x + 105 = 0$
Let the other two roots be $a, b$
$S_{1} = 1 + 7 + a + b = 16 \Rightarrow a + b = 8.... (i) S_{4} = 7 a b = 105 \Rightarrow a b = 15..... (i i)$
substituting $b$ from $(i)$
$a (8 - a) = 15 8 a - a^{2} = 15 a^{2} - 8 a + 15 = 0 a^{2} - 5 a - 3 a + 15 = 0 a (a - 5) - 3 (a - 5) = 0 (a - 3) (a - 5) = 0 \Rightarrow a = 3, 5$
substituting $a$ in $(i i)$
$\Rightarrow b = 5, 3$
So the other two roots are $3$ and $5$

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Similar questions

Solve the equation $x^{4} - 16 x^{3} + 86 x^{2} - 176 x + 105 = 0$ . If two roots being 1 and 7, Find the sum of the square of other two roots.

Q. The absolute difference of the maximum and the minimum value of

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x^{4} - 16 x^{3} + 86 x^{2} - 176 x + 105 = 0

Write bi-quadratic polynomial $x^{4} - 16 x^{3} + 86 x^{2} - 176 x + 105$ as a product of two quadratic polynomial If one quadratic polynomial has roots 1 and 7.

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x^{4} - 16 x^{3} + 86 x^{2} - 176 x + 105 = 0

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I) $\sum α$	a) $86$
II) $\sum α β$	b) $105$
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Solve the equations:x4−16x3+86x2−176x+105=0, two roots being 1 and 7.

x4−16x3+86x2−176x+105=0Let the other two roots be a,bS1=1+7+a+b=16⇒a+b=8....(i)S4=7ab=105⇒ab=15.....(ii)substituting b from (i)a(8−a)=158a−a2=15a2−8a+15=0a2−5a−3a+15=0a(a−5)−3(a−5)=0(a−3)(a−5)=0⇒a=3,5substituting a in (ii)⇒b=5,3So the other two roots are 3 and 5

Solve the equations:
$x^{4} - 16 x^{3} + 86 x^{2} - 176 x + 105 = 0$ , two roots being $1$ and $7$ .