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bezimeni [28]
4 years ago
5

Use complete sentences to explain how the quadratic formula is related to the process of completing the square.

Mathematics
2 answers:
Vitek1552 [10]4 years ago
5 0

Answer:

Quadratic formula can be derived from completing the square method .

Step-by-step explanation:

Consider a quadratic equation : ax^2+bx+c=0 where a , b , c are variables and a\neq 0

On solving this equation by completing the square , we get

ax^2+bx+c=0\\x^2+\frac{bx}{a}+\frac{c}{a}=0\\x^2+2\left ( \frac{b}{2a} \right )x+\frac{c}{a}=0\\x^2+2\left ( \frac{b}{2a} \right )x+\left ( \frac{b}{2a} \right )^2-\left ( \frac{b}{2a} \right )^2+\frac{c}{a}=0

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

On taking square root on both sides, we get

\left ( x+\frac{b}{2a} \right )=\pm \sqrt{\frac{b^2-4ac}{4a^2}}\\\left ( x+\frac{b}{2a} \right )=\pm \sqrt{\frac{D}{4a^2}}\,\,;\,\,D=b^2-4ac\\\left ( x+\frac{b}{2a} \right )=\pm \frac{\sqrt{D}}{2a}\\x=\frac{-b\pm \sqrt{D}}{2a}

which is basically a quadratic formula .

nadya68 [22]4 years ago
4 0
Quadratic formula is derived from completing the square: 

<span>ax² + bx + c = 0 </span>
<span>ax² + bx = −c </span>
<span>x² + b/a x = −c/a </span>

<span>Complete square on left side by adding (b/(2a))² to both sides: </span>

<span>x² + b/a x + (b/(2a))² = (b/(2a))² − c/a </span>
<span>(x + b/(2a))² = (b²−4ac)/(2a)² </span>
<span>x + b/(2a) = ± √(b²−4ac)/(2a) </span>
<span>x = −b/(2a) ± √(b²−4ac)/(2a) </span>
<span>x = (−b ± √(b²−4ac)) / (2a) </span>

<span>- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - </span>

<span>or </span>

<span>ax² + bx + c = 0 </span>
<span>ax² + bx = −c </span>
<span>4a (ax² + bx) = −4ac </span>
<span>4a²x² + 4abx = −4ac </span>

<span>Complete the square on left side by adding b² to both sides </span>

<span>4a²x² + 4abx + b² = b²−4ac </span>
<span>(2ax + b)² = b²−4ac </span>
<span>2ax + b = ± √(b²−4ac) </span>
<span>2ax = −b ± √(b²−4ac) </span>
<span>x = (−b ± √(b²−4ac)) / (2a)</span>
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