Form a quadratic equation ax^2+bx+c=0 with a=1 that has the given roots:
1± 2
Nice explanation Reiny...
I wondered why the +- would be in the question.
No worries ... she has more of them for you to work on : )
Yours in Mathematical fun, Ms Pi
I assume the 1+ - 2 comes from the quadratic formula?
Please advise...
A quadratic can have at most 2 roots.
1± 2 to me means 3 or -1
so the quadratic could b a(x-3)(x+1) = 0
but you said that a=1 , so
(x-3)(x+1) = 0
x^2 - 2x - 3 = 0
To form a quadratic equation with the given roots, we know that the roots are 1+2 and 1-2.
The general form of a quadratic equation is ax^2 + bx + c = 0, with a, b, and c being constants. In this case, we are given that a = 1, so we can substitute that value into the equation:
1*x^2 + bx + c = 0
Now, let's find the values of b and c using the given roots:
When x = 1 + 2:
1*(1 + 2)^2 + b*(1 + 2) + c = 0
1*9 + 3b + c = 0
9 + 3b + c = 0 ➊
When x = 1 - 2:
1*(1 - 2)^2 + b*(1 - 2) + c = 0
1*1 - 2b + c = 0
1 - 2b + c = 0 ➋
Now we have two equations (equation ➊ and ➋) with two variables (b and c). We can solve this system of equations to find the values of b and c.
Subtract equation ➋ from equation ➊:
(9 + 3b + c) - (1 - 2b + c) = 0 - 0
9 + 3b + c - 1 + 2b - c = 0
(3b + 2b) + (c - c) + (9 - 1) = 0
5b + 8 = 0
We can solve this equation for b:
5b = -8
b = -8/5
Now substitute the value of b into equation ➊ to solve for c:
9 + 3(-8/5) + c = 0
9 - 24/5 + c = 0
45/5 - 24/5 + c = 0
(45 - 24)/5 + c = 0
21/5 + c = 0
c = -21/5
Therefore, the quadratic equation with the given roots 1+2 and 1-2 is:
x^2 - (8/5)x - (21/5) = 0