Question

explain me the completing square method in ur own words
no cheating guys
no spamming​

Answer 1

Answer:

See square just means to multiply the given no twice like .

14²=14*14=196

you just need to multiply.

And

for square root....

You can take by two methods either by long division method or prime factorization:

See the attachment make pair of two and solve to get the no whose square has been given.

See further. ..........

There comes completing the square here......

Say we have a simple expression like x2 + bx. Having x twice in the same expression can make life hard. What can we do?

Well, with a little inspiration from Geometry we can convert it, like this:

Completing the Square Geometry

As you can see x2 + bx can be rearranged nearly into a square ...

... and we can complete the square with (b/2)2

In Algebra it looks like this:

x2 + bx + (b/2)2 = (x+b/2)2

"Complete the Square"

So, by adding (b/2)2 we can complete the square.

And (x+b/2)2 has x only once, which is easier to use.

Keeping the Balance

Now ... we can't just add (b/2)2 without also subtracting it too! Otherwise the whole value changes.

So let's see how to do it properly with an example:

Start with: x^2 + 6x + 7

("b" is 6 in this case)

Complete the Square:

x^2 + 6x + (6/2)^2 + 7 - (6/2)^2

Also subtract the new term

Simplify it and we are done.

simplifies to (x+3)^2

The result:

x2 + 6x + 7 = (x+3)2 − 2

And now x only appears once, and our job is done!

Answer 2

$\sf\red{\underline{\underline{Answer:}}}$

$\sf\blue{Explanation:}$

$\sf{Completing \ the \ square \ method \ works}$

$\sf{on \ the \ formulas}$

$\sf{1.(a+b)^{2}}$

$\sf{2.(a-b)^{2}}$

$\sf\green{\underline{\underline{Detailed \ explanation}}}$

$\sf{General \ form \ of \ quadratic \ equation \ is}$

$\sf{\implies{ax^{2}+bx+c=0}}$

$\sf{Here \ a, \ b \ and \ c \ can \ be \ any \ integers.}$

$\sf{\implies{ax^{2}+bx+c=0}}$

$\sf{First \ step: \ Making \ the \ coefficient \ of \ x^{2} \ as \ 1.}$

$\sf{Here, \ a \ is \ any \ integer.}$

$\sf{\implies{\red{a}x^{2}+bx+c=0}}$

$\sf{Divide \ the \ equation \ through \ by \ \red{a}}$

$\sf{\implies{\frac{\red{a}x^{2}}{\red{a}}+\frac{bx}{\red{a}}+\frac{c}{\red{a}}=0}}$

$\sf{\implies{x^{2}+\frac{bx}{a}+\frac{c}{a}=0}}$

$\sf{Second \ step: \ Shifting \ the \ constant}$

$\sf{term \ \frac{c}{a} \ on \ the \ L.H.S.}$

$\sf{\implies{x^{2}+\frac{bx}{a}=\frac{-c}{a}}}$

$\sf{Now \ here, \ comparing \ with \ a^{2}+2ab+b^{2}}$

$\sf{x^{2}=a^{2},}$

$\sf{\frac{bx}{a}=2ab.}$

$\sf{He \ have \ to \ make \ a \ term \ for \ b^{2}}$

____________________________________

$\sf{\underline{\underline{Working:}}}$

$\sf{In \ a^{2}+2ab+b^{2} \ a=1}$

$\sf{\therefore{1+2b+b^{2}}}$

$\sf{we \ can \ form \ a \ term \ for \ b^{2}}$

$\sf{by \ Multipyling \ coefficient \ of \ x }$

$\sf{i.e \ \frac{b}{a} \ by \ half \ and \ taking \ it's \ square.}$

___________________________________

$\sf{Third \ step:: \ Multiply \ the \ coefficient \ of \ x}$

$\sf{by \ \frac{1}{2} \ and \ taking \ it's \ square.}$

$\sf{=>(Coefficient \ of \ x\times\frac{1}{2})^{2}}$

$\sf{=>(\frac{b}{a}\times\frac{1}{2})^{2}}$

$\sf{=>\frac{b^{2}}{4a^{2}}}$

___________________________________

$\sf{Forth \ step: \ Add \ \frac{b^{2}}{4a^{2}} \ on \ both \ sides \ of \ equation}$

$\sf{\implies{x^{2}+\frac{bx}{a}+\frac{b^{2}}{4a^{2}}=\frac{-c}{a}+\frac{b^{2}}{4a^{2}}}}$

$\sf\pink{According \ to \ identity}$

$\sf\pink{a^{2}+2ab+b^{2}=(a+b)^{2}}$

$\sf{\implies{(x+\frac{b}{2a})^{2}=\frac{-4ac+b^{2}}{4a^{2}}}}$

$\sf\pink{On \ taking \ square \ root \ of \ both \ sides}$

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

$\sf{\implies{x=\frac{\sqrt{b^{2}-4ac}}{2a}-\frac{b}{2a} \ or \ -\frac{\sqrt{b^{2}-4ac}}{2a}-\frac{b}{2a}}}$

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

$\sf{This \ is \ how \ completing \ the \ square \ method \ works}$

____________________________________

$\sf{Same \ result \ is \ obtained \ by \ formula \ method.}$

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