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Question Number 34184 by Rio Mike last updated on 02/May/18
if α and β are the roots of the equation 3x^2 + (x/2) − 4= 0 find p is α−β are the roots of x^2 −px + 7 =0
$$\:{if}\:\alpha\:{and}\:\beta\:{are}\:{the}\:{roots}\:{of}\:{the}\:{equation} \\ $$$$\mathrm{3}{x}^{\mathrm{2}} +\:\frac{{x}}{\mathrm{2}}\:−\:\mathrm{4}=\:\mathrm{0}\:{find}\:{p}\:{is}\:\alpha−\beta\:{are} \\ $$$${the}\:{roots}\:{of}\:\:{x}^{\mathrm{2}} −{px}\:+\:\mathrm{7}\:=\mathrm{0} \\ $$
Commented by candre last updated on 02/May/18
3x^2 +(x/2)−4=0 6x^2 +x−8=0 Δ=1^2 −4×6×−8=1+6×4×8=1+12×4×4 =1+24×2×4=1+48×2×2=1+96×2 =1+192=193 x=((−1±(√(193)))/6) x_1 =−((1+(√(193)))/6);x_2 =((−1+(√(193)))/6) x_1 +x_2 =−(1/3) ∣x_1 −x_2 ∣=((√(193))/3)
$$\mathrm{3}{x}^{\mathrm{2}} +\frac{{x}}{\mathrm{2}}−\mathrm{4}=\mathrm{0} \\ $$$$\mathrm{6}{x}^{\mathrm{2}} +{x}−\mathrm{8}=\mathrm{0} \\ $$$$\Delta=\mathrm{1}^{\mathrm{2}} −\mathrm{4}×\mathrm{6}×−\mathrm{8}=\mathrm{1}+\mathrm{6}×\mathrm{4}×\mathrm{8}=\mathrm{1}+\mathrm{12}×\mathrm{4}×\mathrm{4} \\ $$$$=\mathrm{1}+\mathrm{24}×\mathrm{2}×\mathrm{4}=\mathrm{1}+\mathrm{48}×\mathrm{2}×\mathrm{2}=\mathrm{1}+\mathrm{96}×\mathrm{2} \\ $$$$=\mathrm{1}+\mathrm{192}=\mathrm{193} \\ $$$${x}=\frac{−\mathrm{1}\pm\sqrt{\mathrm{193}}}{\mathrm{6}} \\ $$$${x}_{\mathrm{1}} =−\frac{\mathrm{1}+\sqrt{\mathrm{193}}}{\mathrm{6}};{x}_{\mathrm{2}} =\frac{−\mathrm{1}+\sqrt{\mathrm{193}}}{\mathrm{6}} \\ $$$${x}_{\mathrm{1}} +{x}_{\mathrm{2}} =−\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\mid{x}_{\mathrm{1}} −{x}_{\mathrm{2}} \mid=\frac{\sqrt{\mathrm{193}}}{\mathrm{3}} \\ $$
Answered by Joel578 last updated on 02/May/18
3x^2 + (1/2)x − 4 = 0 α − β = ∣((√D)/a)∣, where D is discriminant = ∣((√((193)/4))/3)∣ = ((√(193))/6) x = ((√(193))/6) is one of the root from equation x^2 − px + 7 = 0 Let the other root is a (x − ((√(193))/6))(x − a) = x^2 − px + 7 = 0 x^2 − (a + ((√(193))/6))x + ((a(√(193)))/6) = x^2 − px + 7 = 0 → ((a(√(193)))/6) = 7 → a = ((42)/( (√(193)))) → p = ((42)/( (√(193)))) + ((√(193))/6)
$$\mathrm{3}{x}^{\mathrm{2}} \:+\:\frac{\mathrm{1}}{\mathrm{2}}{x}\:−\:\mathrm{4}\:=\:\mathrm{0} \\ $$$$\alpha\:−\:\beta\:=\:\:\mid\frac{\sqrt{{D}}}{{a}}\mid,\:\:\:\:\mathrm{where}\:{D}\:\mathrm{is}\:\mathrm{discriminant} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\:\mid\frac{\sqrt{\frac{\mathrm{193}}{\mathrm{4}}}}{\mathrm{3}}\mid\:=\:\:\frac{\sqrt{\mathrm{193}}}{\mathrm{6}} \\ $$$$ \\ $$$${x}\:=\:\frac{\sqrt{\mathrm{193}}}{\mathrm{6}}\:\mathrm{is}\:\mathrm{one}\:\mathrm{of}\:\mathrm{the}\:\mathrm{root}\:\mathrm{from}\:\mathrm{equation}\:{x}^{\mathrm{2}} \:−\:{px}\:+\:\mathrm{7}\:=\:\mathrm{0} \\ $$$$\mathrm{Let}\:\mathrm{the}\:\mathrm{other}\:\mathrm{root}\:\mathrm{is}\:{a} \\ $$$$\left({x}\:−\:\frac{\sqrt{\mathrm{193}}}{\mathrm{6}}\right)\left({x}\:−\:{a}\right)\:=\:{x}^{\mathrm{2}} \:−\:{px}\:+\:\mathrm{7}\:=\:\mathrm{0} \\ $$$${x}^{\mathrm{2}} \:−\:\left({a}\:+\:\frac{\sqrt{\mathrm{193}}}{\mathrm{6}}\right){x}\:+\:\frac{{a}\sqrt{\mathrm{193}}}{\mathrm{6}}\:=\:{x}^{\mathrm{2}} \:−\:{px}\:+\:\mathrm{7}\:=\:\mathrm{0} \\ $$$$\rightarrow\:\frac{{a}\sqrt{\mathrm{193}}}{\mathrm{6}}\:=\:\mathrm{7}\:\rightarrow\:{a}\:=\:\frac{\mathrm{42}}{\:\sqrt{\mathrm{193}}} \\ $$$$\rightarrow\:{p}\:=\:\frac{\mathrm{42}}{\:\sqrt{\mathrm{193}}}\:+\:\frac{\sqrt{\mathrm{193}}}{\mathrm{6}} \\ $$

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