Question

derive the quadratic formula

Answer 1

ʜᴇy ᴜꜱᴇʀ

ʜᴇʀᴇ ɪꜱ yᴏᴜʀ ᴀɴꜱᴡᴇʀ :-

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$a {x}^{2} + bx + c = 0 \\ \\ {x}^{2} + \frac{b}{a} x + \frac{c}{a} = 0 \: \: \: \: \: \: \: \: dividing \: by \: a \: \\ \\ {x}^{2} + \frac{b}{a} x + { (\frac{b}{2a} )}^{2} - {( \frac{b}{2a}) }^{2} + \frac{c}{a} = 0 \\ \\ {(x + \frac{b}{2a} }^{2} - \frac{ {b}^{2} }{4 {a}^{2} } + \frac{c}{a} \\ \\ {(x + \frac{b}{2a} }^{2} - \frac{ {b}^{2} - 4ac }{4 {a}^{2} } = 0 \\ \\ x + \frac{b}{2a} = \sqrt{ \frac{ {b}^{2} - 4ac }{4 {a}^{2} } } \: \: \: \: \: \: \: \: or \: \: \: \: \: \: \: x + \frac{b}{2a} = - \sqrt{ \frac{ {b}^{2} - 4ac }{4 {a}^{2} } } \\ \\ x = - \frac{b}{2a} + \frac{ \sqrt{ {b}^{2} - 4ac } }{2a} \: \: \: \: \: \: \: or \: \: \: \: \: \: x = - \frac{b}{2a} - \frac{ \sqrt{ {b}^{2} - 4ac} }{2a} \\ \\ x = \frac{ - b + - \sqrt{ {b}^{2} - 4ac } }{2a}$

ʜᴀᴠᴇ ᴀ ɴɪᴄᴇ ᴅᴀy

Answer 2

♥️Hi there ♥️

1.Write the quadratic eqn in its standard form ;

${ax}^{2} + bx + c = 0 \:$
where a ≠ 0

2.Divide it by 'a'
$\frac{ {ax}^{2} }{a} + \frac{b}{a} x + \frac{c}{a} = 0$

So it will become
${x}^{2} + \frac{b}{a} x + \frac{c}{ \:a} = 0$

3.Now take coefficient of x that is b/a and multiply b/a by 1/2 so will get b/2a .

4.Add and subtract (b/2a)² that is b²/4a² in x²+b/a.x + c/a to convert in the form of :-

${x}^{2} + \frac{b}{a} x + \frac{c}{a} + \frac{ {b}^{2} }{4 {a}^{2} } - \frac{ {b}^{2} }{4 {a}^{2} } = 0 -(1)$

5. Now re write :- eq (1) in the form of ;-

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

6 .Simplify it to (x+ b/ 2a)² = b²-4ac / 4a².

7.Take a square root on both the sides to get.

$\sqrt{(x + \frac{ {b} }{2a} ) {}^{2} }$

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

= ± √b^2- 4ac /2a

8.Finally we get two values that is quadratic equation.

Thanks ✋

Hope it's right.