Question Number 1039 by tera last updated on 22/May/15
![jumlah akar−akar persamaan [x]^2 −2[x]−3=0 sama dengan a.−10 b.−3 e.4 c.−1 d.0](Q1039.png)
$${jumlah}\:{akar}−{akar}\:{persamaan}\:\left[{x}\right]^{\mathrm{2}} −\mathrm{2}\left[{x}\right]−\mathrm{3}=\mathrm{0}\:{sama}\:{dengan} \\ $$$${a}.−\mathrm{10}\:\:\:\:\:\:\:\:\:\:{b}.−\mathrm{3}\:\:\:\:\:\:{e}.\mathrm{4} \\ $$$${c}.−\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:{d}.\mathrm{0} \\ $$
Commented by prakash jain last updated on 22/May/15
![[x] =⌊x⌋ =gint(x)?](Q1040.png)
$$\left[{x}\right]\:=\lfloor{x}\rfloor\:={gint}\left({x}\right)?\: \\ $$
Commented by 123456 last updated on 22/May/15
![y=[x] y^2 −2y−3=0 Δ=(−2)^2 −4(1)(−3)=4+12=16 y=((−(−2)±(√(16)))/(2(1)))=((2±4)/2)=1±2 y_1 =1−2=−1 y_2 =1+2=+3 [x_1 ]=−1⇔−2≤x_1 ≤−1 [x_2 ]=+3⇔3≤x_2 ≤4](Q1042.png)
$${y}=\left[{x}\right] \\ $$$${y}^{\mathrm{2}} −\mathrm{2}{y}−\mathrm{3}=\mathrm{0} \\ $$$$\Delta=\left(−\mathrm{2}\right)^{\mathrm{2}} −\mathrm{4}\left(\mathrm{1}\right)\left(−\mathrm{3}\right)=\mathrm{4}+\mathrm{12}=\mathrm{16} \\ $$$${y}=\frac{−\left(−\mathrm{2}\right)\pm\sqrt{\mathrm{16}}}{\mathrm{2}\left(\mathrm{1}\right)}=\frac{\mathrm{2}\pm\mathrm{4}}{\mathrm{2}}=\mathrm{1}\pm\mathrm{2} \\ $$$${y}_{\mathrm{1}} =\mathrm{1}−\mathrm{2}=−\mathrm{1} \\ $$$${y}_{\mathrm{2}} =\mathrm{1}+\mathrm{2}=+\mathrm{3} \\ $$$$\left[{x}_{\mathrm{1}} \right]=−\mathrm{1}\Leftrightarrow−\mathrm{2}\leqslant{x}_{\mathrm{1}} \leqslant−\mathrm{1} \\ $$$$\left[{x}_{\mathrm{2}} \right]=+\mathrm{3}\Leftrightarrow\mathrm{3}\leqslant{x}_{\mathrm{2}} \leqslant\mathrm{4} \\ $$
Commented by prakash jain last updated on 22/May/15
$$\mathrm{Please}\:\mathrm{check}. \\ $$$$\lfloor−\mathrm{2}\rfloor=−\mathrm{2} \\ $$$$\lfloor−\mathrm{1}.\mathrm{5}\rfloor=−\mathrm{2} \\ $$$$\lfloor{x}\rfloor=−\mathrm{1}\Rightarrow−\mathrm{1}\leqslant{x}<\mathrm{0} \\ $$
Commented by 123456 last updated on 23/May/15
$$\mathrm{what}\:\mathrm{if}\:\mathrm{its}\:\mathrm{is}\:\mathrm{integer}\:\mathrm{part}? \\ $$
Answered by prakash jain last updated on 22/May/15
![Taking [x] as floor function y=⌊x⌋ y^2 −2y−3=0 (y+1)(y−3)=0 y=−1 or y=3 x∈[−1,0)∪[3,4)](Q1045.png)
$$\mathrm{Taking}\:\left[{x}\right]\:{as}\:{floor}\:{function} \\ $$$${y}=\lfloor{x}\rfloor \\ $$$${y}^{\mathrm{2}} −\mathrm{2}{y}−\mathrm{3}=\mathrm{0} \\ $$$$\left({y}+\mathrm{1}\right)\left({y}−\mathrm{3}\right)=\mathrm{0} \\ $$$${y}=−\mathrm{1}\:\mathrm{or}\:{y}=\mathrm{3} \\ $$$${x}\in\left[−\mathrm{1},\mathrm{0}\right)\cup\left[\mathrm{3},\mathrm{4}\right) \\ $$