Question

Sa se determine x aparine lui R astfel incat tripetul de numere sa fie in progresie aritimetica:
a) 3x-10,5x-12,x^2-2
b) 2x^2-x-13, x^2+x-4,2x^2-3x+5
c) 5x-2, (2x-3)^2,7x+4

Answer 1

[tex]\displaystyle a).3x-10,~5x-12,~x^2-2 \\ 5x-12= \frac{3x-10+x^2-2}{2} \\ 2(5x-12)=3x-10+x^2-2 \\ 10x-24=3x-10+x^2-2 \\ -x^2-3x+10x=-10-2+24 \\ -x^2+7x=12 \\ -x^2+7x-12=0| \cdot (-1) \\ x^2-7x+12=0 \\ a=1,~b=-7,~c=12 \\ \Delta=b^2-4ac=(-7)^2-4 \cdot 1 \cdot 12=49-48=1\ \textgreater \ 0 \\ x_1= \frac{7- \sqrt{1} }{2 \cdot 1} = \frac{7-1}{2} = \frac{6}{2} =3 \\ x_2= \frac{7+ \sqrt{1} }{2 \cdot 1} = \frac{7+1}{2} = \frac{8}{2} =4[/tex]
[tex]\displaystyle b).2x^2-x-13,~x^2+x-4,~2x^2-3x+5 \\ x^2+x-4= \frac{2x^2-x-13+2x^2-3x+5}{2} \\ 2(x^2+x-4)=2x^2-x-13+2x^2-3x+5 \\ 2x^2+2x-8=2x^2-x-13+2x^2-3x+5 \\ 2x^2+2x-2x^2+x-2x^2+3x=-13+5+8 \\ -2x^2+6x=0 | \cdot (-1) \\ 2x^2-6x=0 \\ a=2,~b=-6,~c=0 \\ \Delta=b^2-4ac=(-6)^2-4 \cdot 2 \cdot 0=36-0=36\ \textgreater \ 0 \\ x_1= \frac{6- \sqrt{36} }{2 \cdot 2} = \frac{6-6}{4} = 0 \\ x_2= \frac{6+ \sqrt{36} }{2 \cdot 2} = \frac{6+6}{4} = \frac{12}{4} =3 [/tex]
[tex]\displaystyle c).5x-2,~(2x-3)^2,~7x+4 \\ (2x-3)^2= \frac{5x-2+7x+4}{2} \\ (2x)^2-2 \cdot 2x \cdot 3+3^2= \frac{12x+2}{2} \\ 4x^2-12x+9= \frac{12x+2}{2} \\ 2(4x^2-12x+9)=12x+2 \\ 8x^2-24x+18=12x+2 \\ 8x^2-24x-12x=2-18 \\ 8x^2-36x=-16 \\ 8x^2-36x+16=0|:2 \\ 4x^2-18x+8=0 \\ a=4,~b=-18,~c=8 \\ \Delta=b^2-4ac=(-18)^2-4 \cdot 4 \cdot 8=324-128=196\ \textgreater \ 0 [/tex]
[tex]\displaystyle x_1= \frac{18- \sqrt{196} }{2 \cdot 4} = \frac{18-14}{8} = \frac{4}{8} ^{(4} = \frac{1}{2} \\ x_2= \frac{18+ \sqrt{196} }{2 \cdot 4} = \frac{18+14}{8} = \frac{32}{8} =4[/tex]

Answer 2

Doneeee 
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