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Posted on Jun 26, 2009
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How do I use sharp EL W531 to solve quadratic equations?

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k24674

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  • Sharp Master 8,093 Answers
  • Posted on Sep 17, 2009
k24674
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Joined: Aug 17, 2009
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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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You cannot solve linear equations on a SHARP EL531. If you would like a calculator that solves equations, you need to get a SHARP EL W506
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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.



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How to solve quadratic equation in SHARP EL-531WH

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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