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Question

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


Solution

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


Mathematics

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