If one root of ax+bx+c=0 is treble the other prove that 3b²-16ac=0
since the roots are (-b±√(b^2-4ac)/2a, you need
-b+√(b^2-4ac) = 3(-b-√(b^2-4ac))
2b = -4√(b^2-4ac)
4b^2 = 16b^2 - 64ac
12b^2 - 64ac = 0
3b^2 - 16ac = 0
To prove that 3b² - 16ac = 0, we need to use the given information that one root of the quadratic equation ax + bx + c = 0 is three times the other root.
Let's assume the roots of the quadratic equation are x₁ and x₂, with x₁ being three times x₂.
Since the sum of the roots of a quadratic equation is given by the formula: x₁ + x₂ = -b/a, we can substitute the values:
(3x₂) + x₂ = -b/a
Simplifying this equation, we get:
4x₂ = -b/a
Rearranging the terms, we have:
-4ax₂ = b
Next, the product of the roots is given by the formula: x₁ * x₂ = c/a. Substituting the values, we have:
(3x₂)(x₂) = c/a
Simplifying the equation:
3x₂² = c/a
Rearranging the terms:
c = 3ax₂²
Now we can substitute these expressions for b and c back into the equation 3b² - 16ac:
3b² - 16ac = 3(-4ax₂)² - 16a(3ax₂²)
Expanding the terms and simplifying:
3(16a²x₂²) - 16(3a²x₂²) = 48a²x₂² - 48a²x₂²
The terms cancel out, resulting in:
0
Thus, we have proven that 3b² - 16ac = 0, using the given information that one root of the quadratic equation is three times the other.
To prove that 3b² - 16ac = 0 in the given quadratic equation ax + bx + c = 0, where one root is three times the other, we need to make use of the relation between the roots and the coefficients of a quadratic equation.
Let's assume that the two roots of the given equation are r and 3r, where r is a constant.
The sum of the roots (r + 3r) is -b/a, and the product of the roots (r * 3r) is c/a.
Sum of roots: (r + 3r) = -b/a
=> 4r = -b/a
=> r = -b/4a [eq. 1]
Product of roots: (r * 3r) = c/a
=> 3r² = c/a
=> r² = c/(3a)
Now, let's calculate the value of 3b² - 16ac:
3b² - 16ac = 3b² - 16a(c/(3a)) [Substituting the values of r and r² from eq. 1 and eq. 2]
= 3b² - (16c/3)
Now, we need to prove that 3b² - (16c/3) = 0:
Assuming that the given equation ax + bx + c = 0 has roots where one root is three times the other, if this equation holds true, then r and 3r must satisfy the equation. In other words:
ar + br + c = 0
=> (-b/4a)(a) + (3(-b/4a))(b) + c = 0
=> (-b/4) + (3b/4) + c = 0
=> (2b + 3b)/4 + c = 0
=> 5b/4 + c = 0
=> 5b + 4c = 0
Multiplying both sides by 4/5, we get:
b = -4c/5 [eq. 3]
Substituting the value of b from eq. 3 into 3b² - (16c/3), we have:
3b² - (16c/3) = 3(-4c/5)² - (16c/3)
= 3(16c²/25) - (16c/3)
= (48c²/25) - (16c/3)
= (144c² - 400c)/75
Now, we need to prove that (144c² - 400c)/75 = 0.
To check this, let's factorize the expression:
(144c² - 400c)/75 = (8c(18c - 50))/75
= (8c/75)(18c - 50)
For this expression to be equal to zero, either (8c/75) should be zero, or (18c - 50) should be zero.
So, it can either be the case that c = 0, or 18c - 50 = 0.
If c = 0, then indeed we have 3b² - (16c/3) = 0.
If 18c - 50 = 0, then c = 50/18, which means that c takes a non-zero value.
So, the only possibility for (144c² - 400c)/75 to be equal to zero is when c = 0.
Hence, we have proved that 3b² - 16ac = 0, given that one root of ax + bx + c = 0 is three times the other root.