Let the roots of the quadratic equation [tex]\( p x^2 + q = 0 \)[/tex] be denoted by [tex]\( r_1 \)[/tex] and [tex]\( r_2 \)[/tex]. Given that these roots differ by 1, we can express them as [tex]\( r_1 = r \)[/tex] and [tex]\( r_2 = r + 1 \)[/tex].
### Step 1: Sum of Roots
The sum of roots of a quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex] is given by [tex]\( -\frac{b}{a} \)[/tex]. In our equation [tex]\( p x^2 + q = 0 \)[/tex], the coefficient of [tex]\( x \)[/tex] (i.e., [tex]\( b \)[/tex]) is 0. Therefore, the sum of the roots is:
[tex]\[ r_1 + r_2 = -\frac{0}{p} = 0 \][/tex]
As [tex]\( r_1 \)[/tex] and [tex]\( r_2 \)[/tex] are [tex]\( r \)[/tex] and [tex]\( r + 1 \)[/tex] respectively, we have:
[tex]\[ r + (r + 1) = 0 \][/tex]
[tex]\[ 2r + 1 = 0 \][/tex]
Solving for [tex]\( r \)[/tex]:
[tex]\[ 2r = -1 \][/tex]
[tex]\[ r = -\frac{1}{2} \][/tex]
### Step 2: Product of Roots
The product of the roots of a quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex] is given by [tex]\( \frac{c}{a} \)[/tex]. In our equation [tex]\( p x^2 + q = 0 \)[/tex], the constant term (i.e., [tex]\( c \)[/tex]) is [tex]\( q \)[/tex] and the coefficient of [tex]\( x^2 \)[/tex] (i.e., [tex]\( a \)[/tex]) is [tex]\( p \)[/tex]. Therefore, the product of the roots is:
[tex]\[ r_1 \cdot r_2 = \frac{q}{p} \][/tex]
Substituting [tex]\( r_1 = -\frac{1}{2} \)[/tex] and [tex]\( r_2 = -\frac{1}{2} + 1 \)[/tex]:
[tex]\[ r_1 \cdot r_2 = -\frac{1}{2} \cdot \left( -\frac{1}{2} + 1 \right) \][/tex]
[tex]\[ r_1 \cdot r_2 = -\frac{1}{2} \cdot \frac{1}{2} \][/tex]
[tex]\[ r_1 \cdot r_2 = -\frac{1}{4} \][/tex]
### Step 3: Relationship Between [tex]\( p \)[/tex] and [tex]\( q \)[/tex]
We know that the product of the roots is:
[tex]\[ \frac{q}{p} = -\frac{1}{4} \][/tex]
Multiplying both sides by [tex]\( p \)[/tex]:
[tex]\[ q = -\frac{1}{4} p \][/tex]
Thus, the relationship between [tex]\( p \)[/tex] and [tex]\( q \)[/tex] is:
[tex]\[ q = -\frac{1}{4} p \][/tex]
