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July 25, 2014

July 25, 2014

Posted by **Lisa D** on Sunday, June 27, 2010 at 7:58pm.

Use the discriminant to determine how many real-number solutions the equation has

A.2

B.1

C.0

- Equation -
**Jen**, Sunday, June 27, 2010 at 8:10pmThe discriminant is b^2-4ac. For the equation you posted, a=1 (the coefficient on w^2), b=-2, and c=2. Plug those numbers in. If the number you get is positive, there are 2 real number solutions. If you get zero, there is 1 solution. If you get a negative number, there are no solutions. Post your answer if you want it checked.

- Equation -
**bobpursley**, Sunday, June 27, 2010 at 8:24pmActually, if you get a negative discriminant, you get no

**real**solutions, but insteady, you get complex (in the imaginary domain) solutions, that in general, are valid.

eg. x^2+0x+1=0

the discriminate is sqrt(-4)

and in fact, then the solution is

x= +- i

- Equation -
**Lisa D**, Sunday, June 27, 2010 at 9:11pmIs the answer 0 not sure if this is correct.

- Equation -
**Jen**, Monday, June 28, 2010 at 1:02amShow me your work. You have to plug in the numbers.

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