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Question Number 203202 by Abdullahrussell last updated on 12/Jan/24

Answered by cortano12 last updated on 12/Jan/24

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Answered by mr W last updated on 12/Jan/24

$$abc+d(ab+bc+ca)=14$$$$\Rightarrow abc=14+4d$$$$abcd=(14+4d)d=30$$$$\Rightarrow 2d^2+7d-15=0$$$$\Rightarrow (2d-3)(d+5)=0$$$$\Rightarrow d=\frac{3}{2}or-5$$$$abc=\frac{30}{\frac{3}{2}}=20or\frac{30}{-5}=-6$$$$ab+(a+b)c=-4$$$$abc-3c^2=-4c$$$$3c^2-4c-abc=0$$$$\Rightarrow 3c^2-4c-20=0or3c^2-4c+6=0\left(rejected\right)$$$$\Rightarrow (3c-10)(c+2)=0$$$$\Rightarrow c=\frac{10}{3}or-2$$$$ab=-4+3c=6\left(rejected\right)or-10$$$$a+b=-3$$$$\Rightarrow (a,b)=(-5,2)or(2,-5)$$$$(a,b,c,d)=(-5,2,-2,\frac{3}{2})or(2,-5,-2,\frac{3}{2})$$$$a^2+b^2+c^2+d^2=25+4+4+\frac{9}{4}=\frac{141}{4}✓$$

Answered by ajfour last updated on 12/Jan/24

$$ab=m$$$$c=\frac{4+m}{3}$$$$d=\frac{30\times 3}{m(4+m)}$$$$\frac{m(4+m)}{3}+(m-3c)d=14$$$$\frac{m(4+m)}{3}-\frac{90\times 4}{m(4+m)}=14$$$$\Rightarrow m^2{\left(\frac{4+m}{3}\right)}^2=\frac{14m(4+m)}{3}+120$$$$t^2-14-120=0$$$${(t-7)}^2={13}^2$$$$t=20,-6$$$$m(4+m)=60,-18$$$${(m+2)}^2=64,-14$$$$letstakem=6,-10$$$$\mathrm{\&}$$$$9-2m+{\left(\frac{4+m}{3}\right)}^2+\frac{8100}{m^2{(4+m)}^2}=?=Q$$$$withm=6$$$$Q_1=-3+\frac{100}{9}+\frac{81}{36}=\frac{481-108}{36}$$$$Q_1=\frac{373}{36}$$$$\mathrm{\&}withm=-10$$$$Q_2=29+4+\frac{81}{36}=35+\frac{1}{4}=\frac{141}{4}$$

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