Question and Answers Forum

All Questions      Topic List

Others Questions

Previous in All Question      Next in All Question      

Previous in Others      Next in Others      

Question Number 69778 by Rio Michael last updated on 27/Sep/19

prove that the equation     (b^2 −4ac)x^2  + 4(a + c)x −4 = 0 is always real.

$${prove}\:{that}\:{the}\:{equation}\: \\ $$$$\:\:\left({b}^{\mathrm{2}} −\mathrm{4}{ac}\right){x}^{\mathrm{2}} \:+\:\mathrm{4}\left({a}\:+\:{c}\right){x}\:−\mathrm{4}\:=\:\mathrm{0}\:{is}\:{always}\:{real}. \\ $$

Commented by prakash jain last updated on 28/Sep/19

Hi Rasheed  So glad to see that you are still here.

$$\mathrm{Hi}\:\mathrm{Rasheed} \\ $$$$\mathrm{So}\:\mathrm{glad}\:\mathrm{to}\:\mathrm{see}\:\mathrm{that}\:\mathrm{you}\:\mathrm{are}\:\mathrm{still}\:\mathrm{here}. \\ $$

Commented by prakash jain last updated on 27/Sep/19

Δ=(4(a+c))^2 −4×(−4)(b^2 −4ac)  =16a^2 +16c^2 +32ac+16b^2 −64ac  =16a^2 +16c^2 −32ac+16b^2   =16(a−c)^2 +16b^2   (a−c)^2  is always ≥0  b^2  is always ≥0  hence  Δ≥0  ⇒roots are always real  ■

$$\Delta=\left(\mathrm{4}\left({a}+{c}\right)\right)^{\mathrm{2}} −\mathrm{4}×\left(−\mathrm{4}\right)\left({b}^{\mathrm{2}} −\mathrm{4}{ac}\right) \\ $$$$=\mathrm{16}{a}^{\mathrm{2}} +\mathrm{16}{c}^{\mathrm{2}} +\mathrm{32}{ac}+\mathrm{16}{b}^{\mathrm{2}} −\mathrm{64}{ac} \\ $$$$=\mathrm{16}{a}^{\mathrm{2}} +\mathrm{16}{c}^{\mathrm{2}} −\mathrm{32}{ac}+\mathrm{16}{b}^{\mathrm{2}} \\ $$$$=\mathrm{16}\left({a}−{c}\right)^{\mathrm{2}} +\mathrm{16}{b}^{\mathrm{2}} \\ $$$$\left({a}−{c}\right)^{\mathrm{2}} \:\mathrm{is}\:\mathrm{always}\:\geqslant\mathrm{0} \\ $$$${b}^{\mathrm{2}} \:\mathrm{is}\:\mathrm{always}\:\geqslant\mathrm{0} \\ $$$$\mathrm{hence} \\ $$$$\Delta\geqslant\mathrm{0} \\ $$$$\Rightarrow\mathrm{roots}\:\mathrm{are}\:\mathrm{always}\:\mathrm{real} \\ $$$$\blacksquare \\ $$

Commented by Abdo msup. last updated on 28/Sep/19

hello prakash you are absent for a long time...

$${hello}\:{prakash}\:{you}\:{are}\:{absent}\:{for}\:{a}\:{long}\:{time}... \\ $$

Commented by Rasheed.Sindhi last updated on 28/Sep/19

Welcome sir prakash.Happy to see  you in the forum! The forum needs you.

$${Welcome}\:\boldsymbol{{sir}}\:\boldsymbol{{prakash}}.{Happy}\:{to}\:{see} \\ $$$${you}\:{in}\:{the}\:{forum}!\:{The}\:{forum}\:{needs}\:{you}. \\ $$

Commented by Rio Michael last updated on 28/Sep/19

thanks sir

$${thanks}\:{sir} \\ $$

Answered by mind is power last updated on 27/Sep/19

a=b=c=0 false

$${a}={b}={c}=\mathrm{0}\:{false} \\ $$$$ \\ $$

Commented by Rio Michael last updated on 27/Sep/19

what do you mean sir?

$${what}\:{do}\:{you}\:{mean}\:{sir}? \\ $$

Commented by mind is power last updated on 27/Sep/19

your quation mean ax^2 +bx+c=0 ..Elet y root of E  ⇒(b^2 −4ac)y^2 +4(a+c)y−4=0?

$${your}\:{quation}\:{mean}\:{ax}^{\mathrm{2}} +{bx}+{c}=\mathrm{0}\:..{Elet}\:{y}\:{root}\:{of}\:{E} \\ $$$$\Rightarrow\left({b}^{\mathrm{2}} −\mathrm{4}{ac}\right){y}^{\mathrm{2}} +\mathrm{4}\left({a}+{c}\right){y}−\mathrm{4}=\mathrm{0}? \\ $$

Commented by JDamian last updated on 28/Sep/19

In that case, there is no equation at all.

$${In}\:{that}\:{case},\:{there}\:{is}\:{no}\:{equation}\:{at}\:{all}. \\ $$

Answered by MJS last updated on 27/Sep/19

usually when dealing with  ax^2 +bx+c=0  it is clearly implied that a≠0  ⇒  x^2 +((4(a+c))/(b^2 −4ac))x−(4/(b^2 −4ac))=0; a, b, c ∈R∧b^2 −4ac≠0  x^2 +px+q=0 ⇔ x=−(p/2)±(√((p^2 /4)−q))  this is real for D=(p^2 /4)−q≥0  ⇒ D=4((a^2 −2ac+b^2 +c^2 )/((b^2 −4ac)^2 ))=4(((a−c)^2 +b^2 )/((b^2 −4ac)^2 ))>0 ⇒  ⇒ x∈R

$$\mathrm{usually}\:\mathrm{when}\:\mathrm{dealing}\:\mathrm{with} \\ $$$${ax}^{\mathrm{2}} +{bx}+{c}=\mathrm{0} \\ $$$$\mathrm{it}\:\mathrm{is}\:\mathrm{clearly}\:\mathrm{implied}\:\mathrm{that}\:{a}\neq\mathrm{0} \\ $$$$\Rightarrow \\ $$$${x}^{\mathrm{2}} +\frac{\mathrm{4}\left({a}+{c}\right)}{{b}^{\mathrm{2}} −\mathrm{4}{ac}}{x}−\frac{\mathrm{4}}{{b}^{\mathrm{2}} −\mathrm{4}{ac}}=\mathrm{0};\:{a},\:{b},\:{c}\:\in\mathbb{R}\wedge{b}^{\mathrm{2}} −\mathrm{4}{ac}\neq\mathrm{0} \\ $$$${x}^{\mathrm{2}} +{px}+{q}=\mathrm{0}\:\Leftrightarrow\:{x}=−\frac{{p}}{\mathrm{2}}\pm\sqrt{\frac{{p}^{\mathrm{2}} }{\mathrm{4}}−{q}} \\ $$$$\mathrm{this}\:\mathrm{is}\:\mathrm{real}\:\mathrm{for}\:{D}=\frac{{p}^{\mathrm{2}} }{\mathrm{4}}−{q}\geqslant\mathrm{0} \\ $$$$\Rightarrow\:{D}=\mathrm{4}\frac{{a}^{\mathrm{2}} −\mathrm{2}{ac}+{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }{\left({b}^{\mathrm{2}} −\mathrm{4}{ac}\right)^{\mathrm{2}} }=\mathrm{4}\frac{\left({a}−{c}\right)^{\mathrm{2}} +{b}^{\mathrm{2}} }{\left({b}^{\mathrm{2}} −\mathrm{4}{ac}\right)^{\mathrm{2}} }>\mathrm{0}\:\Rightarrow \\ $$$$\Rightarrow\:{x}\in\mathbb{R} \\ $$

Commented by Rio Michael last updated on 27/Sep/19

thank you sir

$${thank}\:{you}\:{sir} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com