Question about Sharp EL-531VB Calculator

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How do I use sharp EL 531W to solve quadratic equations?

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  • Anonymous May 24, 2009

    fregre

  • Anonymous Mar 16, 2014

    how to calculate the quadratic equation

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Hello,
This calculator does not have a program to which you give the equation, and which returns the solutions. Sorry. You have to use the formulas you learned.
Let the equation be ax^2 +bx+c=0
You compute the discriminant Callit disc.
disc= (b^2-4*a*c)
1. If discriminant is larger than zero you have two solutions. Let us call them x1 and x2.
Then x1=(-b+square root of disc)/(2*a)
and x2= (-b-square root of disc)/(2*a)

2. If disc=0 , square root of disc =0 and x1=x2=-b/(2*a)
3. If disc is negative, there are no real solutions.

To find the solutions you replace a, b, and c by their numerical values, to calculate disc, its square root, and calculate X1 and X2.

Hope it helps.



Posted on Sep 17, 2009

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Posted on Jan 02, 2017

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422b55e.jpg
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Hello,
This calculator does not have a program to which you give the equation, and which returns the solutions. Sorry. You have to use the formulas you learned.
Let the equation be ax^2 +bx+c=0
You compute the discriminant Call it disc.
disc= (b^2-4*a*c)
1. If discriminant is larger than zero you have two solutions. Let us call them x1 and x2.
Then x1=(-b+square root of disc)/(2*a)
and x2= (-b-square root of disc)/(2*a)

2. If disc=0 , square root of disc =0 and x1=x2=-b/(2*a)
3. If disc is negative, there are no real solutions.

To find the solutions you replace a, b, and c by their numerical values, to calculate disc, its square root, and calculate X1 and X2.

Hope it helps.

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Hello,
This calculator does not have a program to which you give the equation, and which returns the solutions. Sorry. You have to use the formulas you learned.
Let the equation be ax^2 +bx+c=0
You compute the discriminant Call it disc.
disc= (b^2-4*a*c)
1. If discriminant is larger than zero you have two solutions. Let us call them x1 and x2.
Then x1=(-b+square root of disc)/(2*a)
and x2= (-b-square root of disc)/(2*a)

2. If disc=0 , square root of disc =0 and x1=x2=-b/(2*a)
3. If disc is negative, there are no real solutions.

To find the solutions you replace a, b, and c by their numerical values, to calculate disc, its square root, and calculate X1 and X2.

Hope it helps.

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