One solution of the equation 3x^2-16x+5=0 is 1/3. What is the other solution?
If "a" is a solution of a polynomial equation, then x - a is a factor of the polynomial. Let's call one solution "a" and the other "b." Then (x - a)(x - b) = 0 Here you know one solution is 1/3. Let's say a = 1/3. That means (x - 1/3)(x - b) = 0 Multiply both sides by 3 to get: (3x - 1)(x - b) = 0 Now you know that one of the factors of the given polynomial is 3x - 1. You need to find the other one. 3x^2-16x+5 = 0 (3x - 1)( ? ) = 0
3x-5
right?????
I'm supposed to give a numerical answer
no, because you would end up with 9x^2 as the leading term...
No. Be careful. 3x^2-16x+5 = 0 In the second set of parentheses, youneed an x, since 3x * x = 3x^2. So place an x inside the second parentheses. (3x - 1)(x ) = 0
Now for the number, notice that in the polynomial you have +5. Since you already have -1 in the first set of parentheses, what times -1 equals +5? It's -5. So place a -5 inside the second set of parentheses: (3x - 1)(x - 5) = 0
Now that the polynomial is factored, use the other binomial to find the second solution. x - 5 = 0 Solve for x.
5
in a more complicated polynomial with more than two solutions (remember, the number of solutions equals the highest exponent of the variable of the polynomial) you might not be able to factor in this way very conveniently. What you can do is divide off each factor/solution as you find it, simplifying the polynomial as you go. in this case, we could have divided 3x^2-16x+5 by (3x-1) to get a simpler polynomial whose root we could find by inspection or trivial algebra.
Is 5 correct?
plz help
yes
x-5=0 x - 5 + 5 = 0 + 5 x = 5 yes
x - 5 = 0 Add 5 to both sides: x = 5
Thanks you BTW i became your fan
also, plugging the value into the polynomial should evaluate to 0 if it is a root/zero: 3x^2-16x+5 3(5)^2-16(5) + 5 = 3*25 - 80 + 5 = 0 so 5 is a root
more tedious to do the other one, 3x - 1 = 0, 3x = 1, x = 1/3 \[3(1/3)^2 - 16(1/3) + 5 = 3(1/9)-16(1/3) + 5 =\]\[= 1/3 - 16/3 + 5 = -15/3 + 5 = -5 + 5 = 0\]
Join our real-time social learning platform and learn together with your friends!