Having a lil problem
Prove that the roots of
ax^2 + (a + b)x+b are real for all values of k
note the "x"s aren't multiplication signs.
a x^2 + bx + c has the discriminant of
D = b^2 - 4ac.
If D is nonnegative then the function has real roots.
In this case you have
D = (a + b)^2 - 4 a b = (a-b)^2
which is larger than or equal to zero because it is a square.
To prove that the roots of the quadratic equation ax^2 + (a + b)x + b are real for all values of k, we need to show that the discriminant D is nonnegative.
The discriminant of a quadratic equation ax^2 + bx + c is given by the formula D = b^2 - 4ac. If D is nonnegative, it means that the quadratic equation has real roots.
In our case, the quadratic equation is ax^2 + (a + b)x + b. Comparing it with the standard form ax^2 + bx + c, we can see that a = a, b = (a + b), and c = b.
Applying the formula for the discriminant, we have:
D = (a + b)^2 - 4a * b.
Simplifying further:
D = a^2 + 2ab + b^2 - 4ab,
D = a^2 - 2ab + b^2.
Rearranging the terms:
D = (a - b)^2.
Since (a - b)^2 is a square, it is always nonnegative. Therefore, the discriminant D is nonnegative for all values of k, which means that the quadratic equation ax^2 + (a + b)x + b has real roots for all values of k.