Certainly! Let's tackle the problem step-by-step.
Given the quadratic equation:
[tex]\[ x^2 - kx - 28 = 0 \][/tex]
We know one of the roots is [tex]\( -7 \)[/tex]. Let's denote the roots of the quadratic equation by [tex]\( r_1 \)[/tex] and [tex]\( r_2 \)[/tex].
Step (i): Find the value of [tex]\( k \)[/tex]
First, we know from Vieta's formulas that for a quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex]:
1. The sum of the roots [tex]\( r_1 + r_2 = -\frac{b}{a} \)[/tex]
2. The product of the roots [tex]\( r_1 \cdot r_2 = \frac{c}{a} \)[/tex]
For our given equation [tex]\( x^2 - kx - 28 = 0 \)[/tex]:
- [tex]\( a = 1 \)[/tex]
- [tex]\( b = -k \)[/tex]
- [tex]\( c = -28 \)[/tex]
We are given [tex]\( r_1 = -7 \)[/tex]. Let's denote [tex]\( r_2 \)[/tex] as the other root.
By Vieta's formulas:
[tex]\[ r_1 \cdot r_2 = -28 \][/tex]
Substitute [tex]\( r_1 = -7 \)[/tex]:
[tex]\[ -7 \cdot r_2 = -28 \][/tex]
Solve for [tex]\( r_2 \)[/tex]:
[tex]\[ r_2 = \frac{-28}{-7} = 4 \][/tex]
Now, using the sum of the roots:
[tex]\[ r_1 + r_2 = -\frac{b}{a} \][/tex]
Substitute [tex]\( r_1 = -7 \)[/tex] and [tex]\( r_2 = 4 \)[/tex]:
[tex]\[ -7 + 4 = -\frac{-k}{1} \][/tex]
[tex]\[ -3 = -k \][/tex]
[tex]\[ k = 3 \][/tex]
So, the value of [tex]\( k \)[/tex] is [tex]\( 3 \)[/tex].
Step (ii): Find the other root
We already found in the previous calculation that the other root [tex]\( r_2 \)[/tex] is:
[tex]\[ r_2 = 4 \][/tex]
Summary:
- The value of [tex]\( k \)[/tex] is [tex]\( 3 \)[/tex].
- The other root of the equation is [tex]\( 4 \)[/tex].
