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implicit double precision(a-h,o-z)

write(*,*) "please provide the a,b,c coeff"

read(*,*) A,B,C

D=B*B-4*A*C

if(D.GT.0) then

root1=(-B/(2*A))+(SQRT(D))/(2*A)

root2=(-B/(2*A))-(SQRT(D))/(2*A)

write(*,*) root1,root2

elseif(D.EQ.0) then

root1=(-B/(2*A))

root2=root1

write(*,*) root1,root2

else

root1=(-B/(2*A))+(SQRT(-D))/(2*A)

root2=(-B/(2*A))-(SQRT(-D))/(2*A)

a=(root1+root2)/2

b=(root1-root2)/2

write(*,*) 'realpartroot=',a, 'complexpartroot=',b

endif

stop

END

Q: Write a fortran code to calculate the root of quadratic equation?

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ax2 + bx + c

computer scince

Write the quadratic equation in the form ax2 + bx + c = 0 The roots are equal if and only if b2 - 4ac = 0. The expression, b2-4ac is called the [quadratic] discriminant.

First, write the equation in standard form, i.e., put zero on the right. Then, depending on the case, you may have the following options:Factor the polynomialComplete the squareUse the quadratic formula

You do not need to write anything to calculate things mentally.

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Write the quadratic equation in the form ax2 + bx + c = 0 then the roots (solutions) of the equation are: [-b ± √(b2 - 4*a*c)]/(2*a)

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ax2 + bx + c

Write the quadratic equation in the form ax2 + bx + c = 0 The roots are equal if and only if b2 - 4ac = 0. The expression, b2-4ac is called the [quadratic] discriminant.

Yes. The FORTRAN write command output's data to a file.

First, write the equation in standard form, i.e., put zero on the right. Then, depending on the case, you may have the following options:Factor the polynomialComplete the squareUse the quadratic formula

The easiest way to write a generic algorithm is to simply use the quadratic formula. If it is a computer program, ask the user for the coefficients a, b, and c of the generic equation ax2 + bx + c = 0, then just replace them in the quadratic formula.