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
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
What type of solutions does this equation have?
64t2–50=0
oobleck shut it you nerd
Well, let me analyze this equation for you. It's like trying to find the least interesting person at a party - we need to see how many solutions it has!
If we take a closer look at the equation 0=2x^2-20x+50, we notice that it's a quadratic equation. And just like trying to find a good clown, sometimes it's not easy.
So, to find the number of real solutions, we can use the almighty Discriminant, which is calculated as b^2 - 4ac. In this equation, a=2, b=-20, and c=50.
Now, let's plug these values into the Discriminant formula: (-20)^2 - 4(2)(50). After doing some magic calculations, the Discriminant reveals itself to be -300. Ouch!
Since the Discriminant is negative, it means that the equation has no real solutions. So the answer is C, no solutions. Just like trying to find a serious clown - they're simply nowhere to be found!
To determine the number of real number solutions for the equation 0=2x^2-20x+50, we can use the discriminant formula. The discriminant (denoted as Δ) is calculated as Δ = b^2 - 4ac, where a, b, and c are the coefficients of the quadratic equation in the form ax^2 + bx + c = 0.
In this case, a = 2, b = -20, and c = 50. Substituting the values into the discriminant formula, we have:
Δ = (-20)^2 - 4(2)(50)
= 400 - 400
= 0
Since the discriminant is equal to zero (Δ = 0), we conclude that the equation has:
A. 1 real number solution
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