Question

If the quadratic equation \(ax^2 + bx + c = 0\) has roots that differ by 4, and \(a + b + c = 12\), what is the value of \(b^2 - 4ac\)?

Hint:

The square of the difference of the roots of \(ax^2+bx+c=0\) is directly related to the discriminant (D) by \((\alpha - \beta)^2 = D/a^2\). This is a very useful shortcut.

Answer 1

Correct Answer: 16

Step 1: Understanding the Question:
We are asked to find the value of the discriminant (\(D = b^2 - 4ac\)) of a quadratic equation. We are given two conditions: the difference between its roots and the sum of its coefficients.
Step 2: Key Formula or Approach:
Let the roots of the equation be \(\alpha\) and \(\beta\). We will use Vieta's formulas and the relationship between the difference of roots and the discriminant.
1. Sum of roots: \(\alpha + \beta = -b/a\).
2. Product of roots: \(\alpha\beta = c/a\).
3. Difference of roots: \((\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta\).
The discriminant \(D = b^2 - 4ac\). The relationship is \((\alpha - \beta)^2 = D/a^2\).
The sum of coefficients \(a+b+c\) is the value of the polynomial at \(x=1\).
Step 3: Detailed Explanation:
Part A: Using the Difference of Roots
We are given that the roots differ by 4, so \(|\alpha - \beta| = 4\). Squaring this gives \((\alpha - \beta)^2 = 16\).
Using the formula that relates the difference of roots to the discriminant:
\[ (\alpha - \beta)^2 = \frac{b^2 - 4ac}{a^2} \] \[ 16 = \frac{b^2 - 4ac}{a^2} \] \[ b^2 - 4ac = 16a^2 \] This shows that the value we need depends on 'a'. We must find 'a' using the second condition.
Part B: Using the Sum of Coefficients
The condition \(a + b + c = 12\) means the value of the polynomial \(f(x) = ax^2 + bx + c\) at \(x=1\) is 12.
\[ f(1) = a(1)^2 + b(1) + c = a + b + c = 12 \] Since \(\alpha\) and \(\beta\) are roots, the polynomial can also be written as \(f(x) = a(x - \alpha)(x - \beta)\).
Therefore, \(f(1) = a(1 - \alpha)(1 - \beta) = 12\).
Let the roots be \(k\) and \(k+4\). Substituting these for \(\alpha\) and \(\beta\):
\[ a(1 - k)(1 - (k+4)) = 12 \] \[ a(1 - k)(-k - 3) = 12 \] \[ a(k - 1)(k + 3) = 12 \] \[ a(k^2 + 2k - 3) = 12 \] If 'a' and 'k' are integers, then 'a' must be a divisor of 12. Let's test integer values for 'a'.
- If \(a=1\): \((k+3)(k-1) = 12 \implies k^2+2k-3=12 \implies k^2+2k-15=0 \implies (k+5)(k-3)=0\). This gives integer solutions \(k=3\) or \(k=-5\). This is a valid scenario.
- If \(a=2\): \((k+3)(k-1) = 6 \implies k^2+2k-9=0\). This does not give integer solutions for k.
- For other integer values of 'a' (3, 4, 6, 12), we also find no integer solutions for 'k'.
Thus, the only possibility under the assumption of integer coefficients and roots is \(a=1\).
Part C: Final Calculation
Using \(a=1\) in the result from Part A:
\[ b^2 - 4ac = 16a^2 = 16(1)^2 = 16 \] Step 4: Final Answer:
The value of \(b^2 - 4ac\) is 16.

Answer 2

Roots-and-relations method: Let the two roots be \( p \) and \( p+4 \) (they differ by 4). For the monic form \( x^2+bx+c=0 \) (taking \( a=1 \), the natural reading when a leading coefficient is not otherwise fixed), the sum of roots is \( 2p+4=-b \) and the product is \( p(p+4)=c \).

Using \( a+b+c=12 \) with \( a=1 \): \( 1+b+c=12 \Rightarrow b+c=11 \). Substituting \( b=-(2p+4) \) and \( c=p(p+4) \):
\[ -(2p+4)+p(p+4)=11 \;\Rightarrow\; p^2+2p-15=0 \;\Rightarrow\; (p+5)(p-3)=0. \] So \( p=3 \) (roots \( 3,7 \)) or \( p=-5 \) (roots \( -5,-1 \)). Either pair gives a valid quadratic satisfying every condition.

For roots that differ by 4 in a monic quadratic, the discriminant is always \( b^2-4ac=(\text{root difference})^2=4^2=16 \): checking directly, with \( b=-10,c=21 \), \( b^2-4ac=100-84=16 \); with \( b=6,c=5 \), \( b^2-4ac=36-20=16 \). Both cases agree.

\[ \boxed{b^2-4ac=16} \]