The quadratic equation x2 −2kx+(k−1) = 0 has roots a and b such that a2 +b2 = 4. Without solving the equation, find the possible values of the real number k
If the roots are a and b, then the equation is
(x-a)(x-b) = 0
x^2 -ax - bx + ab = 0
x^2 - (a+b)x + ab = 0, but we also know that
x^2 - 2kx + k+1 = 0
by comparison:
ab = k-1
a+b = 2k
We know that (a+b)^2 = a^2 + b^2 + 2ab
(a+b)^2 = 4 + 2ab
subbing in our given,
4k^2 = 4 + 2(k-1)
4k^2 = 4 + 2k - 2
4k^2 - 2k -2 = 0
2k^2 - k - 1 = 0
(2k + 1)(k - 1) = 0
k = -1/2 or k = 1
Well, well, well, let's juggle with some numbers here! We have the quadratic equation x^2 - 2kx + (k-1) = 0, and we're told that the sum of the squares of the roots, a and b, is 4. No need to panic, I've got this!
We know that the sum of the roots is given by the formula -b/a (yep, b divided by a, don't get too dizzy!). In our case, we have a + b = 2k.
Now, let's make use of the sum of squares of the roots, which is a^2 + b^2. In this case, we're told that a^2 + b^2 equals 4. So, that means (a + b)^2 - 2ab is equal to 4.
Substituting the value we found for a + b, we get:
(2k)^2 - 2ab = 4
4k^2 - 2ab = 4
Hold on tight, we're almost there! We're given that a^2 + b^2 equals 4. But remember, a^2 + b^2 is also equal to (a + b)^2 - 2ab (I hope you're still following along with this mathematical circus act!).
Substituting the values we found:
4 = (2k)^2 - 2ab
4 = 4k^2 - 2ab
Now, let's bring all these equations together and put on a grand finale! We have:
4k^2 - 2ab = 4
4 = 4k^2 - 2ab
By comparing these two equations, we can see that 4k^2 - 2ab must equal 4. So, we have the equation:
4k^2 - 2ab = 4
Now, if we draw back the curtains and solve this equation, we'll find the possible values for the real number k. So, grab your calculators, ladies and gentlemen, it's showtime!
To find the possible values of k, we can use Vieta's formulas.
Vieta's formulas state that the sum of the roots of a quadratic equation ax^2 + bx + c = 0 is equal to -b/a, and the product of the roots is equal to c/a.
In this case, we have:
Sum of the roots (a + b) = 2k
Product of the roots (ab) = k - 1
We are given that a^2 + b^2 = 4, which implies that (a + b)^2 - 2ab = 4.
Substituting the values we've obtained from Vieta's formulas, we have:
(2k)^2 - 2(k - 1) = 4
Expanding and simplifying the equation, we get:
4k^2 - 2k + 2 = 4
Moving all the terms to one side, we have:
4k^2 - 2k - 2 = 0
Now we can use the quadratic formula to solve for k:
k = (-b ± √(b^2 - 4ac)) / (2a)
Plugging in the values, we have:
k = (2 ± √((-2)^2 - 4(4)(-2))) / (2(4))
k = (2 ± √(4 + 32)) / 8
k = (2 ± √36) / 8
k = (2 ± 6) / 8
By simplifying further, we get two possible values for k:
k1 = (2 + 6) / 8 = 8/8 = 1
k2 = (2 - 6) / 8 = -4/8 = -1/2
Therefore, the possible values of the real number k are 1 and -1/2.
To find the possible values of the real number k without solving the equation, we can use the Vieta's formulas.
For a quadratic equation in the form ax^2 + bx + c = 0, the sum of the roots is given by -b/a, and the product of the roots is given by c/a.
In this case, the given quadratic equation is x^2 - 2kx + (k - 1) = 0. According to Vieta's formulas, the sum of the roots (a + b) is equal to 2k, and the product of the roots (ab) is equal to k - 1.
We know that a^2 + b^2 = 4. Let's express this equation in terms of a and b using the sum and product of roots:
(a + b)^2 - 2ab = 4
Expanding and substituting the known values from Vieta's formulas, we get:
(2k)^2 - 2(k - 1) = 4
4k^2 - 2k + 2 = 4
4k^2 - 2k - 2 = 0
Dividing through by 2:
2k^2 - k - 1 = 0
To find the possible values of k, we can use the quadratic formula:
k = (-b ± √(b^2 - 4ac))/(2a)
In this case, a = 2, b = -1, and c = -1. Plugging in these values, we get:
k = (-(-1) ± √((-1)^2 - 4(2)(-1)))/(2(2))
k = (1 ± √(1 + 8))/4
k = (1 ± √9)/4
k = (1 ± 3)/4
So, the possible values of k are:
k = (1 + 3)/4 = 4/4 = 1
k = (1 - 3)/4 = -2/4 = -1/2
Therefore, the possible values of the real number k are 1 and -1/2.