These questions are pertaining to College Algebra

Question

a) A quadratic equation ax^2+bx+c=0 has 5-the square root of 26 and 5+ the square root of 26 as solutions. Find the values of b and c if the value of a is 1.

 

b) Determine the descriminant of the quadratic equation. Use the value of the descriminant to determine whether the quadratic equation has two rational solutions, one repeated real solution, or two complex solutions that are not real.

                                                                                 3z^2+3z+5=0

Answer 1

a)  x = 5 ±√26
 
x = [-b ±√(b²-4ac)]/2a = 5±√26
 
First... this solution started from
[10±2√26]/2   since we are given a=1
 
so...   a= 1, b=-10
 
From the 2√26 part I can figure out b²-4ac
2√26 = √[4(26)] = √104
 
b²-4ac = 104
 
we already know b=-10, a = 1
 
(-10)²-4(1)(c) = 104
100-4c = 104
c = -1
 
a=1, b=-10, c = -1
 
Your equation is x²-10x-1=0

 

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3z^2+3z+5=0

 

discriminant is the part under the square root

√(b²-4ac)

 

If b²-4ac > 0  there are 2 real roots

If b²-4ac = 0 there is one real root of multiplicity 2

if b²-4ac < 0 There are two complex roots in a conjugate pair

 

a=3, b=3, c=5

 

b²-4ac = 3²-4(3)(5) = 9-60=-51

 

Since the discriminant is negative there are 2 complex roots

 

 

Answer 2

b) Discriminant = b²-4ac = 3² - 4(3)(5) which is a negative quantity, therefore there are two complex roots (a conjugate pair)--not Real.

 

a) You know that -b = 5 and a=1, and b²-4ac = 26; substituting, we get

25 - 4(c) = 26  which means 4c = -1  so c = -¼