math
posted by kim .
Having a lil problem
Prove that the roots of
ax^2 + (a + b)x+b are real for all values of k
note the "x"s aren't multiplication signs.
a x^2 + bx + c has the discriminant of
D = b^2  4ac.
If D is nonnegative then the function has real roots.
In this case you have
D = (a + b)^2  4 a b = (ab)^2
which is larger than or equal to zero because it is a square.
Respond to this Question
Similar Questions

can someone please help me?
simplify the following expressions involving complex numbers. (3+2i)(7i) 10 + 3i how did you do that? 
math
Could you please solve so I can double check my answers for the practice quiz? 
math
I HAVE THESE ANSWERS FOR THE PROBLEMS. COULD YOU DOUBLE CHECK PLEASE, THIS IS A PRACTICE QUIZ WHICH ISN'T A GRADE IT JUST HELPS ME GET READY FOR THE TEST. 1) a 2) b 3) d 4) a 5) d 1. Solve x^3 + 6x^2 + 13x + 10 = 0. a) –2 + 2i, –2 … 
mathematics
Use the discriminant to determine the number of real roots the equation has. 3x2 – 5x + 1 =0 A. One real root (a double root) B. Two distinct real roots C. Three real roots D. None (two imaginary roots) 
maths2
Use the discriminant to determine the number of real roots the equation has. 7x2 + 3x + 1 =0 A. One real root (a double root) B. Two distinct real roots C. Three real roots D. None (two imaginary roots) 
algebra
if a quadratic equation with real coefficents has a discriminant of 10, then what type of roots does it have? 
Math
The equation x^2 + k = 6 x has two distinct real roots. Find the range of values of k. I know that you do 36 4k > 0 4k > 36 k < 9 But I don't get why you switch the signs. So. When do u switch signs 
math
Ax^2+bx=bx^2+a prove that it's roots are rational for all real values of a and b 
Math
Find the discriminant for the quadratic equation f(x) = 5x^2  2x + 7 and describe the nature of the roots. discriminant is 144, one real root discriminant is 136, two complex roots <? 
MathematicsIntegration
Question: f(x)=ax^2 + bx + c {a,b,c €R} When 0<=x<=1 , f(x)<=1 Show that a+b+c <= 17 I don't see a way to even start this. I know we can take the discriminant of this function as [(b^2)4ac] This function will …