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Question

Use the discriminant to determine the number and type of solutions of the equation.

 

A) 2a^2+4a-5=0

B) 2b-b^2=1

Answer 1

For a quadratic equation 

 

Ax^2+Bx+C=0, we use the quadratic formula to find solutions for x:

 

x= [-B +/- \sqrt(B^2-4AC)]/2A

 

 

 

The term inside the square root, B^2-4AC is called the discriminant.  It discriminates what type of solutions we have for the quadratic equation:

 

If B^2-4AC =0 we see that the +/- in the formula both lead to the same SINGLE solution for x.

 

If B^2-4AC is positive, the square root here is a real number, and thus we get two REAL solutions for the quadratic equation.

 

If B^2-4AC is negative, we cannot take the square root (over the real numbers) and get NO REAL SOLUTIONS (there are still two solutions to the quadratic, but they are both necessarily complex conjugates in this case).

 

 

 

All we need to do for your problem is consider the coefficients of these quadratic equations...

 

A). 2a^2+4a-5 has coefficients 2, 4, -5, thus the discriminant is 4^2-4*2*(-5) = 16+40=56 >0...so we have two real solutions

 

B). 2b-b^2=1 is equivalent to 0=b^2-2b+1, so the coefficients are 1, -2, 1, and the discriminant is (-2)^2-4(1)(1)=0, and so we have a unique real solution (sometimes this is called a repeated root, or a root of degree two).

 

We need to be careful in B) that we first write an equation where 0=quadratic polynomial, to be sure we are using the correct coefficients for the quadratic formula...this was done above by moving all the terms to one side of the equation (by addition or subtraction).

 

Hope this helps.