Algebra Quadric Equations help?

Question

Solve using quadric equations:
1. 2x^2 + x - 2= 0
2. 2x^2 +5x= 1

I ment quadric formula

Answer 1

1.)
2x^2 + x - 2 = 0

x = (-b ± sqrt(b^2 - 4ac))/(2a)

x = (-1 ± sqrt(1^2 - 4(2)(-2)))/(2(2))
x = (-1 ± sqrt(1 + 16))/4
x = (-1 ± sqrt(17))/4

x = (1/4)(-1 ± sqrt(17))

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2.)
2x^2 + 5x = 1
2x^2 + 5x - 1 = 0

x = (-b ± sqrt(b^2 - 4ac))/(2a)

x = (-5 ± sqrt(5^2 - 4(2)(-1)))/(2(2))
x = (-5 ± sqrt(25 + 8))/4
x = (-5 ± sqrt(33))/4

ANS :

(1/4)(-5 ± sqrt(33))

Answer 2

The general form of a quadratic equation in x is:

ax^2 +bx +c = 0 where a,b,c are constants and the solutions for x are given by the quadratic formula:

x = (-b +(or -) sqrt(b^2 -4ac))/2a

1. a=2, b=1, c= -2 so

x = (-1+(or -) sqrt(1-4*1*-2))/2
= (-1+(or-) sqrt(9))/2
= (-1+3)/2 = 1 or (-1-3)/2 = -2

Just be careful with the signs of the constants and of the product 4ac.

I'll leave 2 with you.

Answer 3

Do you mean 'quadratic formula'?

It give the two solutions of a 2nd order equation

a is the coefficient of the squared term
b is the coef of the linear term
c is the constant

Note that the equation must be written to equal zero. Thus for eqn 2 you must move the 1 to the left side.
You just plug the a,b, and c into the quadratic equation and simplify.

Answer 4

If ax^2+bx+c=0
x={-b+-sqrt[b^2-4ac]}/2a
we will use these formula to solve
qn1
x={-2+-sqrt[1+16]}/4
=-1/2+-1/4sqrt17
qn2
x={-5+-sqrt[25-8]}/4
=1/4[-5-sqrt17]

Answer 5

You mean quadratic equations, right?

Use the quadratic formula to simplify.

If you don't know how to use the quadratic formula, go to:

purplemath.com and search for the quadratic formula lesson.

Purplemath.com will show you how easy it is to use the formula to solve your questions.

Guido

Answer 6

the standard form of the equation is:
ax^2+bx+c
quadratic formula is:
x=(-b+_sq.root of(b^2-4ac))/2a

plug everything in and you will find the answers

(+_)=plus minus