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Question Number 10933 by Joel576 last updated on 02/Mar/17

If p and q are the roots for the x^2 − (a + 1)x + (−a − (5/2)) = 0 The minimum value of p^(2 ) + q^2 is ...

$$\mathrm{If}\:\:{p}\:\:\mathrm{and}\:\:{q}\:\:\mathrm{are}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{for}\:\mathrm{the} \\ $$$${x}^{\mathrm{2}} \:−\:\left({a}\:+\:\mathrm{1}\right){x}\:+\:\left(−{a}\:−\:\frac{\mathrm{5}}{\mathrm{2}}\right)\:=\:\mathrm{0} \\ $$$$\mathrm{The}\:\mathrm{minimum}\:\mathrm{value}\:\mathrm{of}\:\:{p}^{\mathrm{2}\:} +\:{q}^{\mathrm{2}} \:\:\mathrm{is}\:... \\ $$

Answered by ridwan balatif last updated on 02/Mar/17

there 2 solution to solve this question x^2 −(a+1)x+(−a−(5/2))=0 p+q=a+1 & p×q=−(a+(5/2)) let p^2 +q^2 =k,then k=(p+q)^2 −2p×q k=(a+1)^2 −2×{−(a+(5/2))} k=a^2 +2a+1+(2a+5) k=a^2 +4a+6 ∗first solution k will minimum if k′=0 k′=0 2a+4=0→a=−2, so k=(−2)^2 +4×(−2)+6=2 ∗∗second solution k=a^2 +4a+6 k=(a+2)^2 +2 if we want k is minimum, (a+2)^2 must be zero so k=0+2=2

$$\mathrm{there}\:\mathrm{2}\:\mathrm{solution}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{this}\:\mathrm{question} \\ $$$$\mathrm{x}^{\mathrm{2}} −\left(\mathrm{a}+\mathrm{1}\right)\mathrm{x}+\left(−\mathrm{a}−\frac{\mathrm{5}}{\mathrm{2}}\right)=\mathrm{0} \\ $$$$\mathrm{p}+\mathrm{q}=\mathrm{a}+\mathrm{1}\:\&\:\mathrm{p}×\mathrm{q}=−\left(\mathrm{a}+\frac{\mathrm{5}}{\mathrm{2}}\right) \\ $$$$\mathrm{let}\:\mathrm{p}^{\mathrm{2}} +\mathrm{q}^{\mathrm{2}} =\mathrm{k},\mathrm{then} \\ $$$$\mathrm{k}=\left(\mathrm{p}+\mathrm{q}\right)^{\mathrm{2}} −\mathrm{2p}×\mathrm{q} \\ $$$$\mathrm{k}=\left(\mathrm{a}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{2}×\left\{−\left(\mathrm{a}+\frac{\mathrm{5}}{\mathrm{2}}\right)\right\} \\ $$$$\mathrm{k}=\mathrm{a}^{\mathrm{2}} +\mathrm{2a}+\mathrm{1}+\left(\mathrm{2a}+\mathrm{5}\right) \\ $$$$\mathrm{k}=\mathrm{a}^{\mathrm{2}} +\mathrm{4a}+\mathrm{6} \\ $$$$\ast\mathrm{first}\:\mathrm{solution} \\ $$$$\mathrm{k}\:\mathrm{will}\:\mathrm{minimum}\:\mathrm{if}\:\mathrm{k}'=\mathrm{0} \\ $$$$\mathrm{k}'=\mathrm{0} \\ $$$$\mathrm{2a}+\mathrm{4}=\mathrm{0}\rightarrow\mathrm{a}=−\mathrm{2},\:\mathrm{so} \\ $$$$\mathrm{k}=\left(−\mathrm{2}\right)^{\mathrm{2}} +\mathrm{4}×\left(−\mathrm{2}\right)+\mathrm{6}=\mathrm{2} \\ $$$$\ast\ast\mathrm{second}\:\mathrm{solution} \\ $$$$\mathrm{k}=\mathrm{a}^{\mathrm{2}} +\mathrm{4a}+\mathrm{6} \\ $$$$\mathrm{k}=\left(\mathrm{a}+\mathrm{2}\right)^{\mathrm{2}} +\mathrm{2} \\ $$$$\mathrm{if}\:\mathrm{we}\:\mathrm{want}\:\mathrm{k}\:\mathrm{is}\:\mathrm{minimum},\:\left(\mathrm{a}+\mathrm{2}\right)^{\mathrm{2}} \:\mathrm{must}\:\mathrm{be}\:\mathrm{zero} \\ $$$$\mathrm{so}\:\mathrm{k}=\mathrm{0}+\mathrm{2}=\mathrm{2} \\ $$

Commented by Joel576 last updated on 02/Mar/17

thank you very much

$${thank}\:{you}\:{very}\:{much} \\ $$

Commented by Joel576 last updated on 02/Mar/17

thank you very much

$${thank}\:{you}\:{very}\:{much} \\ $$

Answered by bahmanfeshki last updated on 02/Mar/17

p^2 +q^2 =S^2 −2P=(a+1)^2 +2a+5=a^2 +4a+6 minimum value=((4×1×6−4^2 )/(4×1))=2

$${p}^{\mathrm{2}} +{q}^{\mathrm{2}} ={S}^{\mathrm{2}} −\mathrm{2}{P}=\left({a}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{2}{a}+\mathrm{5}={a}^{\mathrm{2}} +\mathrm{4}{a}+\mathrm{6} \\ $$$${minimum}\:{value}=\frac{\mathrm{4}×\mathrm{1}×\mathrm{6}−\mathrm{4}^{\mathrm{2}} }{\mathrm{4}×\mathrm{1}}=\mathrm{2} \\ $$

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