Certainly! Let's begin by expressing the roots of the quadratic equation and see how we can establish the given relationship between the coefficients.
### Step-by-Step Solution
Given the quadratic equation:
[tex]\[ ax^2 + bx + c = a \][/tex]
Suppose the roots of this equation are [tex]\( r \)[/tex] and [tex]\( 4r \)[/tex]. By Vieta's formulas, the sum and product of the roots of the quadratic equation [tex]\( ax^2 + bx + c = a \)[/tex] help us establish relationships between [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex].
1. Sum of the roots:
[tex]\[ r + 4r = 5r \][/tex]
According to Vieta's formulas:
[tex]\[ \frac{-b}{a} = r + 4r \][/tex]
[tex]\[ \frac{-b}{a} = 5r \][/tex]
[tex]\[ b = -5ar \][/tex]
2. Product of the roots:
[tex]\[ r \times 4r = 4r^2 \][/tex]
According to Vieta's formulas:
[tex]\[ \frac{c}{a} = 4r^2 \][/tex]
[tex]\[ c = 4ar^2 \][/tex]
Now, the equation given in the problem is:
[tex]\[ 4b^2 = 25ac \][/tex]
Let's substitute the expressions we have for [tex]\( b \)[/tex] and [tex]\( c \)[/tex]:
3. Substitute [tex]\( b = -5ar \)[/tex]:
[tex]\[ 4b^2 = 4(-5ar)^2 \][/tex]
[tex]\[ 4b^2 = 4 \times 25a^2r^2 \][/tex]
[tex]\[ 4b^2 = 100a^2r^2 \][/tex]
4. Now, substitute [tex]\( c = 4ar^2 \)[/tex]:
[tex]\[ 25ac = 25a \times 4ar^2 \][/tex]
[tex]\[ 25ac = 25 \times 4a^2r^2 \][/tex]
[tex]\[ 25ac = 100a^2r^2 \][/tex]
We see that:
[tex]\[ 4b^2 = 100a^2r^2 \][/tex]
[tex]\[ 25ac = 100a^2r^2 \][/tex]
Since both expressions are equal, we have:
[tex]\[ 4b^2 = 25ac \][/tex]
Therefore, we have successfully demonstrated that [tex]\( 4b^2 = 25ac \)[/tex] when one root of the equation [tex]\( ax^2 + bx + c = a \)[/tex] is four times the other root.
