Quadratic formula — derivation by completing the square

2.1.17 · Maths › Algebra — Introduction & Intermediate

Standard quadratic equation

Yeh letters kyun? Convention yeh hai: $a$ squared term ko multiply karta hai, $b$ linear term ko, aur $c$ constant hai. $a\ne 0$ ki restriction ensure karti hai ki hamare paas actually ek $x^2$ term ho.

Derivation — first principles se

Step 1: Leading coefficient ko normalize karo

Shuru karo: $a{x}^2+bx+c=0$

Yeh step kyun? Completing the square sabse aasaan hoti hai jab $x^2$ ka coefficient 1 ho. Hum sab kuch $a$ se divide kar dete hain: $x^2+\frac{b}{a}x+\frac{c}{a}=0$

Humne kya kiya? Dono sides ko $a$ se divide kiya (yeh valid hai kyunki $a\ne 0$). Ab $x^2$ term ka coefficient 1 hai.

Step 2: $x$ wale terms ko isolate karo

Constant ko right side par le jaao: $x^2+\frac{b}{a}x=-\frac{c}{a}$

Kyun? Left side par "completing the square" term add karne ke liye jagah chahiye.

Step 3: Square complete karo

Yeh magic number kyun? Kyunki $(x+h{)}^2=x^2+2hx+h^2$. Agar hum chahte hain ki middle term $kx$ ho, toh $2h=k$ chahiye, matlab $h=\frac{k}{2}$. Phir constant term $h^2={\left(\frac{k}{2}\right)}^2$ hona chahiye.

Hamare case mein, $k=\frac{b}{a}$ hai, toh hum dono sides mein ${\left(\frac{b}{2a}\right)}^2$ add karte hain: $x^2+\frac{b}{a}x+{\left(\frac{b}{2a}\right)}^2=-\frac{c}{a}+{\left(\frac{b}{2a}\right)}^2$

Kya ho raha hai? Left side ab ek perfect square hai: ${\left(x+\frac{b}{2a}\right)}^2$.

Step 4: Perfect square ko pehchaano

${\left(x+\frac{b}{2a}\right)}^2=-\frac{c}{a}+\frac{b^2}{4a^2}$

Hum kaise jaante hain? ${\left(x+\frac{b}{2a}\right)}^2=x^2+2\cdot x\cdot \frac{b}{2a}+{\left(\frac{b}{2a}\right)}^2=x^2+\frac{b}{a}x+\frac{b^2}{4a^2}$ expand karo. ✓

Step 5: Right side simplify karo

Common denominator lo: $-\frac{c}{a}+\frac{b^2}{4a^2}=-\frac{4ac}{4a^2}+\frac{b^2}{4a^2}=\frac{b^2-4ac}{4a^2}$

Combine kyun karte hain? Taaki square root cleanly le sakein.

Ab hamare paas hai: ${\left(x+\frac{b}{2a}\right)}^2=\frac{b^2-4ac}{4a^2}$

Step 6: Square roots lo

$x+\frac{b}{2a}=\pm \sqrt{\frac{b^2-4ac}{4a^2}}$

$\pm$ kyun? Kyunki agar $y^2=k$, toh $y=\pm \sqrt{k}$ (do solutions).

Square root simplify karo: $\sqrt{\frac{b^2-4ac}{4a^2}}=\frac{\sqrt{b^2-4ac}}{\sqrt{4a^2}}=\frac{\sqrt{b^2-4ac}}{2|a|}$

Kyunki hum formula mein generally $2a$ likhte hain (na ki $2|a|$), aur $\pm$ dono signs cover kar leta hai: $x+\frac{b}{2a}=\pm \frac{\sqrt{b^2-4ac}}{2a}$

Step 7: $x$ ke liye solve karo

$x=-\frac{b}{2a}\pm \frac{\sqrt{b^2-4ac}}{2a}$

Fractions combine karo:

Iska kya matlab hai? Kisi bhi quadratic $a{x}^2+bx+c=0$ ke liye, $a,b,c$ plug in karo, aur tumhe do solutions mil jaate hain (ya ek repeated solution agar discriminant zero ho).

Discriminant

Kyun? Discriminant square root ke andar hota hai. Agar yeh negative ho, toh $\sqrt{\Delta }$ real nahi hota.

Worked examples

Coefficients identify karo: $a=1$, $b=-5$, $c=6$.

Formula apply karo: $x=\frac{-(-5)\pm \sqrt{(-5{)}^2-4(1)(6)}}{2(1)}$

Yeh step kyun? Seedha quadratic formula mein substitute karo.

$x=\frac{5\pm \sqrt{25-24}}{2}=\frac{5\pm \sqrt{1}}{2}=\frac{5\pm 1}{2}$

Dono solutions evaluate karo: $x=\frac{5+1}{2}=3\text{or}x=\frac{5-1}{2}=2$

Check karo: $(3{)}^2-5(3)+6=9-15+6=0$ ✓, aur $(2{)}^2-5(2)+6=4-10+6=0$ ✓

Coefficients: $a=1$, $b=-6$, $c=9$.

Discriminant: $\Delta =(-6{)}^2-4(1)(9)=36-36=0$.

Discriminant check kyun karte hain? $\Delta =0$ batata hai ki exactly ek solution hai (ek repeated root).

Formula apply karo: $x=\frac{-(-6)\pm \sqrt{0}}{2(1)}=\frac{6\pm 0}{2}=3$

Dhyaan do: Yeh $(x-3{)}^2=0$ hai, jo perfectly factor hota hai.

Coefficients: $a=2$, $b=7$, $c=-4$.

Formula apply karo: $x=\frac{-7\pm \sqrt{7^2-4(2)(-4)}}{2(2)}$

Yeh step kyun? Negative $c$ aur denominator mein $2a$ ko dhyan se handle karo.

$x=\frac{-7\pm \sqrt{49+32}}{4}=\frac{-7\pm \sqrt{81}}{4}=\frac{-7\pm 9}{4}$

Do solutions: $x=\frac{-7+9}{4}=\frac{2}{4}=\frac{1}{2}\text{or}x=\frac{-7-9}{4}=\frac{-16}{4}=-4$

Check karo: $2{\left(\frac{1}{2}\right)}^2+7\left(\frac{1}{2}\right)-4=\frac{1}{2}+\frac{7}{2}-4=4-4=0$ ✓

Coefficients: $a=1$, $b=2$, $c=5$.

Discriminant: $\Delta =2^2-4(1)(5)=4-20=-16<0$.

Iska kya matlab hai? Koi real solutions nahi. Complex numbers mein: $x=\frac{-2\pm \sqrt{-16}}{2}=\frac{-2\pm 4i}{2}=-1\pm 2i$

Common mistakes

Sahi lagta kyun hai: $-5$ dikhta hai aur woh seedha copy kar lete hain.

Fix yeh hai: Formula mein $-b$ hai, na ki $b$. Yahan $b=-5$ hai, toh $-b=-(-5)=+5$. Pehle $b$ identify karo, phir use negate karo.

Sahi lagta kyun hai: Numerator par saara dhyan jaata hai; denominator asaani se ignore ho jaata hai.

Fix yeh hai: Pehle poora formula structure likho: $x=\frac{\text{numerator}}{2a}$. POORA numerator ($\pm$ wala part bhi) $2a$ se divide hota hai, na ki sirf kuch hissa.

Sahi lagta kyun hai: "Real numbers mein negative number ka square root nahi ho sakta."

Fix yeh hai: Specify karo "koi real solutions nahi." Complex numbers mein solutions HOTE hain: $\sqrt{-k}=i\sqrt{k}$. Quadratic ke hamesha do roots hote hain (multiplicity count karte hue) $\mathbb{C}$ mein.

Sahi lagta kyun hai: Yeh bhool jaate hain ki negative $c$ se multiply karne par sign change hota hai.

Fix yeh hai: $-4ac=-4\cdot 2\cdot (-4)=-8\cdot (-4)=+32$. Pehle $4ac$ compute karo, phir use subtract karo: $49-(-32)=49+32=81$.

Connections

• Completing the square technique
• Perfect square trinomials
• Discriminant and nature of roots
• Factoring quadratics
• Complex numbers and quadratic equations
• Parabola and its vertex form
• Vieta's formulas for sum and product of roots
• Quadratic inequalities

Ya visually: formula ek fraction hai jisme neeche $2a$ hai, aur upar "$b$ ka opposite" plus/minus ek square root hai jisme "$b$ squared minus four-$a$-$c$" hai.

Recall Ek 12-saal ke bachche ko samjhao

Socho tumhare paas ek mystery number $x$ hai, aur koi tumhe bolta hai: "Agar tum isse square karo, kisi number se multiply karo, phir iska ek aur multiple add karo, aur ek constant add karo, toh zero milta hai." Yeh complicated lagta hai! Lekin yahan ek magic trick hai:

Hum equation ko completing the square se reshape karte hain — yeh aisa hai jaise puzzle pieces ko rearrange karo taaki woh ek perfect square shape mein fit ho jayein, jaise $(x+\text{kuch}{)}^2$. Jab ek baar yeh mil jaaye, toh hum square root lekar square ko "undo" kar sakte hain, jo $x$ guess karne se kaafi aasaan hai.

Quadratic formula is puzzle ka final answer hai. Tum bas apni equation se teen numbers ($a$, $b$, $c$) plug in karo, aur yeh mystery number $x$ de deta hai. Kabhi kabhi do answers milte hain (parabola $x$-axis ko do baar cross karta hai), kabhi ek (sirf touch karta hai), aur kabhi regular numbers mein koi nahi (kabhi cross nahi karta, lekin tum imaginary numbers use kar sakte ho).

Sabse cool baat? Yeh formula kisi bhi quadratic ke liye kaam karta hai, chahe numbers kitni bhi messy ho!

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