a x² + b x + c= 0
2 x² - 20 x + 50 = 0
In this case:
a = 2 , b = - 20 , c = 50
The discriminant:
d = b² - 4 a c
d = (- 20)² - 4 ∙ 2 ∙ 50
d = 400 - 400 = 0
If d < 0 there are no real root
If d = 0 the roots are real and equal ( one real root )
If d > 0 the roots are real and unequal ( two distinct real roots )
In this case d = 0 so one real root , 1 solution.
Answer A
how many real number solutions does the equation have?
0=2x^2-20x+50
A. 1 solution
B. 2 solutions
C. no solutions**
D. infinite solutions
4 answers
nope. why did you just guess?
the discriminant is 20^2 - 4*2*50 = 400 - 400 = 0
In fact,
2x^2-20x+50 = 2(x^2-10x+25) = 2(x-5)^2
so both roots are x=5
the discriminant is 20^2 - 4*2*50 = 400 - 400 = 0
In fact,
2x^2-20x+50 = 2(x^2-10x+25) = 2(x-5)^2
so both roots are x=5
What type of solutions does this equation have?
64t2–50=0
64t2–50=0
oobleck shut it you nerd
1 answer