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Question-56818




Question Number 56818 by Gulay last updated on 24/Mar/19
Commented by Gulay last updated on 24/Mar/19
sir plz help me
$$\mathrm{sir}\:\mathrm{plz}\:\mathrm{help}\:\mathrm{me} \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 24/Mar/19
for eqn ax^2 +bx+c=0  x=((−b±(√(b^2 −4ac)))/(2a))  now eqn  x^2 −2004x+50=0    comparing with ax^2 +bx+c=0  a=1   b=−2004   c=50  x=((2004±(√((2004)^2 −4×1×50)))/2)  =((2004±(√((2004)^2 −200)))/2)  =((2004±2003.95)/2)  =((4007.95)/2) and ((0.05)/2)  =2003.975 and 0.025  in this way solve others two...  2)a=1   b=2006   c=−3  x=((−2006±(√((2006)^2 −4×1×(−3))))/2)  =((−2006±(√((2006)^2 +12)))/2)  now pls use calculator..
$${for}\:{eqn}\:{ax}^{\mathrm{2}} +{bx}+{c}=\mathrm{0} \\ $$$${x}=\frac{−{b}\pm\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}}{\mathrm{2}{a}} \\ $$$${now}\:{eqn}\:\:{x}^{\mathrm{2}} −\mathrm{2004}{x}+\mathrm{50}=\mathrm{0}\:\: \\ $$$${comparing}\:{with}\:{ax}^{\mathrm{2}} +{bx}+{c}=\mathrm{0} \\ $$$${a}=\mathrm{1}\:\:\:{b}=−\mathrm{2004}\:\:\:{c}=\mathrm{50} \\ $$$${x}=\frac{\mathrm{2004}\pm\sqrt{\left(\mathrm{2004}\right)^{\mathrm{2}} −\mathrm{4}×\mathrm{1}×\mathrm{50}}}{\mathrm{2}} \\ $$$$=\frac{\mathrm{2004}\pm\sqrt{\left(\mathrm{2004}\right)^{\mathrm{2}} −\mathrm{200}}}{\mathrm{2}} \\ $$$$=\frac{\mathrm{2004}\pm\mathrm{2003}.\mathrm{95}}{\mathrm{2}} \\ $$$$=\frac{\mathrm{4007}.\mathrm{95}}{\mathrm{2}}\:{and}\:\frac{\mathrm{0}.\mathrm{05}}{\mathrm{2}} \\ $$$$=\mathrm{2003}.\mathrm{975}\:{and}\:\mathrm{0}.\mathrm{025} \\ $$$${in}\:{this}\:{way}\:{solve}\:{others}\:{two}… \\ $$$$\left.\mathrm{2}\right){a}=\mathrm{1}\:\:\:{b}=\mathrm{2006}\:\:\:{c}=−\mathrm{3} \\ $$$${x}=\frac{−\mathrm{2006}\pm\sqrt{\left(\mathrm{2006}\right)^{\mathrm{2}} −\mathrm{4}×\mathrm{1}×\left(−\mathrm{3}\right)}}{\mathrm{2}} \\ $$$$=\frac{−\mathrm{2006}\pm\sqrt{\left(\mathrm{2006}\right)^{\mathrm{2}} +\mathrm{12}}}{\mathrm{2}} \\ $$$${now}\:{pls}\:{use}\:{calculator}.. \\ $$

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