Question

Hello everyone!


Find the value of x in the above equation.

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Answer 1

Step-by-step explanation:

(x+b/2a)² + c/a - (b/2a)² = 0

x² + bx/a + (b/2a)² + c/a - (b/2a)²=0

x² + bx/a + c/a = 0

=> x =[ -(b/a) + √((b²/a² - 4c/a)]/2

=> x = [(-b/a) + √(b²-4ac)/a²)]/2

=> x = [ -b + √(b²-4ac) ]/2a

2nd x = [-b - √(b²-4ac)]/2a

Answer 2

Aɴꜱᴡᴇʀ

$\large \orange{\sf{}x = \sqrt{ - \frac{bx + c}{a} } }$

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Gɪᴠᴇɴ

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

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ᴛᴏ ꜰɪɴᴅ

The value of X?

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Sᴛᴇᴘꜱ

$\large \tt{} (x + \frac{b}{2a} {)}^{2} + \frac{c}{a} - ( \frac{b}{2a} {)}^{2} = 0 \\ \\ \tt{} \: with \: the \: help \: of \: \red{ ( {x + {y)}^{2} } = {x}^{2} + 2xy + {y}^{2} } \\ \\ \leadsto\tt{} {x}^{2} + \frac{ bx}{a} +{ \cancel { \frac{b}{ {4a}^{2} } }^{2} }+ \frac{c}{a} - { \cancel{ \frac{ {b}^{2} }{ {4a}^{2} } }} = 0 \\ \\ \leadsto\tt{} {x}^{2} + \frac{bx + c}{a} = 0 \\ \\ \leadsto\tt{} {x}^{2} = - \frac{bx + c}{a} \\ \\ \pink {\tt\dashrightarrow{}x = \sqrt{ - \frac{bx + c}{a} } }$

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$\huge{\mathfrak{\purple{hope\; it \;helps}}}$