Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 113684 by ZiYangLee last updated on 14/Sep/20

A nice question <3 If a quadratic equation (1−q+(p^2 /2))x^2 +p(1+q)x+q(q−1)+(p^2 /2)=0 has equal roots, prove that p^2 =4q

$$\mathrm{A}\:\mathrm{nice}\:\mathrm{question}\:<\mathrm{3} \\ $$ $$ \\ $$ $$\mathrm{If}\:\mathrm{a}\:\mathrm{quadratic}\:\mathrm{equation}\: \\ $$ $$\left(\mathrm{1}−{q}+\frac{{p}^{\mathrm{2}} }{\mathrm{2}}\right){x}^{\mathrm{2}} +{p}\left(\mathrm{1}+{q}\right){x}+{q}\left({q}−\mathrm{1}\right)+\frac{{p}^{\mathrm{2}} }{\mathrm{2}}=\mathrm{0} \\ $$ $$\mathrm{has}\:\mathrm{equal}\:\mathrm{roots},\:\mathrm{prove}\:\mathrm{that}\:{p}^{\mathrm{2}} =\mathrm{4}{q} \\ $$

Commented byDwaipayan Shikari last updated on 14/Sep/20

p^2 =4q (1−q+((4q)/2))x^2 +p(1+q)+q^2 −q+2q=0 (1+q)x^2 +p(1+q)+q(1+q)=0 It has roots α+β=−p((1+q)/(1+q))=−p αβ=q=(p^2 /4) (α−β)=(√(p^2 −4.(p^2 /4))) =0 α=β (Which is true)

$${p}^{\mathrm{2}} =\mathrm{4}{q} \\ $$ $$\left(\mathrm{1}−{q}+\frac{\mathrm{4}{q}}{\mathrm{2}}\right){x}^{\mathrm{2}} +{p}\left(\mathrm{1}+{q}\right)+{q}^{\mathrm{2}} −{q}+\mathrm{2}{q}=\mathrm{0} \\ $$ $$\left(\mathrm{1}+{q}\right){x}^{\mathrm{2}} +{p}\left(\mathrm{1}+{q}\right)+{q}\left(\mathrm{1}+{q}\right)=\mathrm{0} \\ $$ $${It}\:{has}\:{roots} \\ $$ $$\alpha+\beta=−{p}\frac{\mathrm{1}+{q}}{\mathrm{1}+{q}}=−{p} \\ $$ $$\alpha\beta={q}=\frac{{p}^{\mathrm{2}} }{\mathrm{4}} \\ $$ $$\left(\alpha−\beta\right)=\sqrt{{p}^{\mathrm{2}} −\mathrm{4}.\frac{{p}^{\mathrm{2}} }{\mathrm{4}}}\:=\mathrm{0} \\ $$ $$\alpha=\beta\:\:\:\left({Which}\:{is}\:{true}\right) \\ $$ $$ \\ $$

Answered by MJS_new last updated on 14/Sep/20

(1−q+(p^2 /2))x^2 +p(1+q)x+q(q−1)+(p^2 /2)=0 x^2 +((2p(q+1))/(p^2 −2(q−1)))x+((p^2 +2q(q−1))/(p^2 −2(q−1)))=0 x=t−((p(q+1))/(p^2 −2(q−1))) t^2 −(((p^2 −4q)(p^2 +q^2 −2q+1))/((p^2 −2(q−1))^2 ))=0 x_1 =x_2 ⇔ t=0 ⇒ p^2 −4q=0∨p^2 +q^2 −2q+1=0 ⇒ p^2 =4q (p^2 +q^2 −2q+1=0 ⇒ p^2 =−(q−1)^2 ⇒ p∉R)

$$\left(\mathrm{1}−{q}+\frac{{p}^{\mathrm{2}} }{\mathrm{2}}\right){x}^{\mathrm{2}} +{p}\left(\mathrm{1}+{q}\right){x}+{q}\left({q}−\mathrm{1}\right)+\frac{{p}^{\mathrm{2}} }{\mathrm{2}}=\mathrm{0} \\ $$ $${x}^{\mathrm{2}} +\frac{\mathrm{2}{p}\left({q}+\mathrm{1}\right)}{{p}^{\mathrm{2}} −\mathrm{2}\left({q}−\mathrm{1}\right)}{x}+\frac{{p}^{\mathrm{2}} +\mathrm{2}{q}\left({q}−\mathrm{1}\right)}{{p}^{\mathrm{2}} −\mathrm{2}\left({q}−\mathrm{1}\right)}=\mathrm{0} \\ $$ $${x}={t}−\frac{{p}\left({q}+\mathrm{1}\right)}{{p}^{\mathrm{2}} −\mathrm{2}\left({q}−\mathrm{1}\right)} \\ $$ $${t}^{\mathrm{2}} −\frac{\left({p}^{\mathrm{2}} −\mathrm{4}{q}\right)\left({p}^{\mathrm{2}} +{q}^{\mathrm{2}} −\mathrm{2}{q}+\mathrm{1}\right)}{\left({p}^{\mathrm{2}} −\mathrm{2}\left({q}−\mathrm{1}\right)\right)^{\mathrm{2}} }=\mathrm{0} \\ $$ $${x}_{\mathrm{1}} ={x}_{\mathrm{2}} \:\Leftrightarrow\:{t}=\mathrm{0} \\ $$ $$\Rightarrow\:{p}^{\mathrm{2}} −\mathrm{4}{q}=\mathrm{0}\vee{p}^{\mathrm{2}} +{q}^{\mathrm{2}} −\mathrm{2}{q}+\mathrm{1}=\mathrm{0} \\ $$ $$\Rightarrow\:{p}^{\mathrm{2}} =\mathrm{4}{q} \\ $$ $$\left({p}^{\mathrm{2}} +{q}^{\mathrm{2}} −\mathrm{2}{q}+\mathrm{1}=\mathrm{0}\:\Rightarrow\:{p}^{\mathrm{2}} =−\left({q}−\mathrm{1}\right)^{\mathrm{2}} \:\Rightarrow\:{p}\notin\mathbb{R}\right) \\ $$

Commented byZiYangLee last updated on 15/Sep/20

Cool!

$$\mathrm{Cool}! \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com