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Question-120929

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Question Number 120929 by Canovas last updated on 04/Nov/20
Commented by Canovas last updated on 04/Nov/20
Please help me on number 11 and 15
$$\:\:\:{Please}\:{help}\:{me}\:{on}\:{number}\:\mathrm{11}\:{and}\:\mathrm{15} \\ $$$$ \\ $$
Commented by liberty last updated on 04/Nov/20
it is better if you write
Answered by Ar Brandon last updated on 04/Nov/20
15. ax^2 +bx+c=0 Let the roots be α and 2α SumOfRoots:3α=−(b/a)⇒α=−(b/(3a)) ProductOfRoots:2α^2 =(c/a) ⇒2((b/(3a)))^2 =(c/a) ⇒ 2b^2 =9ac
$$\mathrm{15}. \\ $$$$\mathrm{ax}^{\mathrm{2}} +\mathrm{bx}+\mathrm{c}=\mathrm{0} \\ $$$$\mathrm{Let}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{be}\:\alpha\:\mathrm{and}\:\mathrm{2}\alpha \\ $$$$\mathrm{SumOfRoots}:\mathrm{3}\alpha=−\frac{\mathrm{b}}{\mathrm{a}}\Rightarrow\alpha=−\frac{\mathrm{b}}{\mathrm{3a}} \\ $$$$\mathrm{ProductOfRoots}:\mathrm{2}\alpha^{\mathrm{2}} =\frac{\mathrm{c}}{\mathrm{a}} \\ $$$$\Rightarrow\mathrm{2}\left(\frac{\mathrm{b}}{\mathrm{3a}}\right)^{\mathrm{2}} =\frac{\mathrm{c}}{\mathrm{a}}\:\Rightarrow\:\mathrm{2b}^{\mathrm{2}} =\mathrm{9ac} \\ $$
Answered by Ar Brandon last updated on 04/Nov/20
11. x^2 +2px+q=0 Let roots be β and β+2 SumOfRoots: 2β+2=−2p⇒β=−(p+1) ProductOfRoots: β^2 +2β=q ⇒(p+1)^2 −2(p+1)=q ⇒(p+1)(p+1−2)=q ⇒(p+1)(p−1)=q ⇒p^2 −1=q, p^2 =1+q
$$\mathrm{11}. \\ $$$$\mathrm{x}^{\mathrm{2}} +\mathrm{2px}+\mathrm{q}=\mathrm{0} \\ $$$$\mathrm{Let}\:\mathrm{roots}\:\mathrm{be}\:\beta\:\mathrm{and}\:\beta+\mathrm{2} \\ $$$$\mathrm{SumOfRoots}:\:\mathrm{2}\beta+\mathrm{2}=−\mathrm{2p}\Rightarrow\beta=−\left(\mathrm{p}+\mathrm{1}\right) \\ $$$$\mathrm{ProductOfRoots}:\:\beta^{\mathrm{2}} +\mathrm{2}\beta=\mathrm{q} \\ $$$$\Rightarrow\left(\mathrm{p}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{2}\left(\mathrm{p}+\mathrm{1}\right)=\mathrm{q} \\ $$$$\Rightarrow\left(\mathrm{p}+\mathrm{1}\right)\left(\mathrm{p}+\mathrm{1}−\mathrm{2}\right)=\mathrm{q} \\ $$$$\Rightarrow\left(\mathrm{p}+\mathrm{1}\right)\left(\mathrm{p}−\mathrm{1}\right)=\mathrm{q} \\ $$$$\Rightarrow\mathrm{p}^{\mathrm{2}} −\mathrm{1}=\mathrm{q},\:\mathrm{p}^{\mathrm{2}} =\mathrm{1}+\mathrm{q} \\ $$
Answered by ebi last updated on 04/Nov/20
Q11 x^2 +2px+q=0 to show p^2 =1+q let α and β are the roots of the equation given that, α−β=2 →β=α−2 thus, x^2 −(α+β)x+αβ=0 x^2 −(α+α−2)x+(α−2)α=0 x^2 −(2α−2)x+(α^2 −2α)=0 2α−2=−2p → α=1−p.....(1) α^2 −2α=q......(2) substitute (1) into (2) (1−p)^2 −2(1−p)=q 1−2p+p^2 −2+2p=q p^2 −1=q p^2 =1+q (shown)
$${Q}\mathrm{11} \\ $$$${x}^{\mathrm{2}} +\mathrm{2}{px}+{q}=\mathrm{0} \\ $$$${to}\:{show}\:{p}^{\mathrm{2}} =\mathrm{1}+{q} \\ $$$$ \\ $$$${let}\:\alpha\:{and}\:\beta\:{are}\:{the}\:{roots}\:{of}\:{the} \\ $$$${equation} \\ $$$${given}\:{that}, \\ $$$$\alpha−\beta=\mathrm{2}\:\rightarrow\beta=\alpha−\mathrm{2} \\ $$$$ \\ $$$${thus}, \\ $$$${x}^{\mathrm{2}} −\left(\alpha+\beta\right){x}+\alpha\beta=\mathrm{0} \\ $$$${x}^{\mathrm{2}} −\left(\alpha+\alpha−\mathrm{2}\right){x}+\left(\alpha−\mathrm{2}\right)\alpha=\mathrm{0} \\ $$$${x}^{\mathrm{2}} −\left(\mathrm{2}\alpha−\mathrm{2}\right){x}+\left(\alpha^{\mathrm{2}} −\mathrm{2}\alpha\right)=\mathrm{0} \\ $$$$ \\ $$$$\mathrm{2}\alpha−\mathrm{2}=−\mathrm{2}{p}\:\rightarrow\:\alpha=\mathrm{1}−{p}…..\left(\mathrm{1}\right) \\ $$$$\alpha^{\mathrm{2}} −\mathrm{2}\alpha={q}……\left(\mathrm{2}\right) \\ $$$$ \\ $$$${substitute}\:\left(\mathrm{1}\right)\:{into}\:\left(\mathrm{2}\right) \\ $$$$\left(\mathrm{1}−{p}\right)^{\mathrm{2}} −\mathrm{2}\left(\mathrm{1}−{p}\right)={q} \\ $$$$\mathrm{1}−\mathrm{2}{p}+{p}^{\mathrm{2}} −\mathrm{2}+\mathrm{2}{p}={q} \\ $$$${p}^{\mathrm{2}} −\mathrm{1}={q} \\ $$$${p}^{\mathrm{2}} =\mathrm{1}+{q}\:\left({shown}\right) \\ $$
Answered by ebi last updated on 04/Nov/20
Q15 ax^2 +bx+c=0 x^2 +((b/a))x+((c/a))=0 to show: 2b^2 −9ac let α and β are the roots of the equation. given that α=2β thus, x−(α+β)x+αβ=0 x−(2β+β)x+(2β)β=0 x−(3β)x+(2β^2 )=0 3β=−(b/a) → β=−(b/(3a)).....(1) 2β^2 =(c/a).....(2) substitute (1) into (2) 2(−(b/(3a)))^2 =(c/a) 2((b^2 /(9a^2 )))=(c/a) 2b^2 =9ac → 2b^2 −9ac=0 (shown)
$${Q}\mathrm{15} \\ $$$${ax}^{\mathrm{2}} +{bx}+{c}=\mathrm{0} \\ $$$${x}^{\mathrm{2}} +\left(\frac{{b}}{{a}}\right){x}+\left(\frac{{c}}{{a}}\right)=\mathrm{0} \\ $$$${to}\:{show}:\:\mathrm{2}{b}^{\mathrm{2}} −\mathrm{9}{ac} \\ $$$$ \\ $$$${let}\:\alpha\:{and}\:\beta\:{are}\:{the}\:{roots}\:{of}\:{the} \\ $$$${equation}. \\ $$$${given}\:{that} \\ $$$$\alpha=\mathrm{2}\beta \\ $$$$ \\ $$$${thus}, \\ $$$${x}−\left(\alpha+\beta\right){x}+\alpha\beta=\mathrm{0} \\ $$$${x}−\left(\mathrm{2}\beta+\beta\right){x}+\left(\mathrm{2}\beta\right)\beta=\mathrm{0} \\ $$$${x}−\left(\mathrm{3}\beta\right){x}+\left(\mathrm{2}\beta^{\mathrm{2}} \right)=\mathrm{0} \\ $$$$\mathrm{3}\beta=−\frac{{b}}{{a}}\:\rightarrow\:\beta=−\frac{{b}}{\mathrm{3}{a}}…..\left(\mathrm{1}\right) \\ $$$$\mathrm{2}\beta^{\mathrm{2}} =\frac{{c}}{{a}}…..\left(\mathrm{2}\right) \\ $$$$ \\ $$$${substitute}\:\left(\mathrm{1}\right)\:{into}\:\left(\mathrm{2}\right) \\ $$$$\mathrm{2}\left(−\frac{{b}}{\mathrm{3}{a}}\right)^{\mathrm{2}} =\frac{{c}}{{a}} \\ $$$$\mathrm{2}\left(\frac{{b}^{\mathrm{2}} }{\mathrm{9}{a}^{\mathrm{2}} }\right)=\frac{{c}}{{a}} \\ $$$$\mathrm{2}{b}^{\mathrm{2}} =\mathrm{9}{ac}\:\rightarrow\:\mathrm{2}{b}^{\mathrm{2}} −\mathrm{9}{ac}=\mathrm{0}\:\left({shown}\right) \\ $$
Answered by mindispower last updated on 04/Nov/20
X^2 +2pX+q=0..E let a,b roots of b−a=2 we have b+a=−2p⇒p=−(((b+a)/2)) ab=q,p^2 =(1/4)(b^2 +a^2 +2ab)=(1/4)((b−a)^2 +4ab) =(1/4)(2^2 +4q)=1+q=p^2
$${X}^{\mathrm{2}} +\mathrm{2}{pX}+{q}=\mathrm{0}..{E} \\ $$$${let}\:{a},{b}\:{roots}\:{of}\:\:{b}−{a}=\mathrm{2} \\ $$$${we}\:{have}\:{b}+{a}=−\mathrm{2}{p}\Rightarrow{p}=−\left(\frac{{b}+{a}}{\mathrm{2}}\right) \\ $$$${ab}={q},{p}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{4}}\left({b}^{\mathrm{2}} +{a}^{\mathrm{2}} +\mathrm{2}{ab}\right)=\frac{\mathrm{1}}{\mathrm{4}}\left(\left({b}−{a}\right)^{\mathrm{2}} +\mathrm{4}{ab}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{2}^{\mathrm{2}} +\mathrm{4}{q}\right)=\mathrm{1}+{q}={p}^{\mathrm{2}} \\ $$

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