Sure, let's solve the given quadratic equation for [tex]\( x \)[/tex] in terms of [tex]\( k \)[/tex].
The quadratic equation given is:
[tex]\[ x^2 + (8k - 3)x + (16k^2 - 15k) = 0 \][/tex]
To solve a quadratic equation of the form [tex]\( ax^2 + bx + c = 0 \)[/tex], we can use the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Here, [tex]\( a = 1 \)[/tex], [tex]\( b = 8k - 3 \)[/tex], and [tex]\( c = 16k^2 - 15k \)[/tex].
Let's apply the quadratic formula step by step:
1. Identify the coefficients:
[tex]\[ a = 1 \][/tex]
[tex]\[ b = 8k - 3 \][/tex]
[tex]\[ c = 16k^2 - 15k \][/tex]
2. Calculate the discriminant:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
Substitute the known values:
[tex]\[ \Delta = (8k - 3)^2 - 4 \cdot 1 \cdot (16k^2 - 15k) \][/tex]
[tex]\[ \Delta = (8k - 3)^2 - 4(16k^2 - 15k) \][/tex]
[tex]\[ \Delta = 64k^2 - 48k + 9 - 64k^2 + 60k \][/tex]
[tex]\[ \Delta = 12k + 9 \][/tex]
3. Take the square root of the discriminant:
[tex]\[ \sqrt{\Delta} = \sqrt{12k + 9} \][/tex]
4. Apply the quadratic formula:
[tex]\[ x = \frac{-(8k - 3) \pm \sqrt{12k + 9}}{2} \][/tex]
5. Simplify the expression:
[tex]\[ x = \frac{-8k + 3 \pm \sqrt{12k + 9}}{2} \][/tex]
[tex]\[ x = \frac{-8k + 3}{2} \pm \frac{\sqrt{12k + 9}}{2} \][/tex]
[tex]\[ x = -4k + \frac{3}{2} \pm \frac{\sqrt{12k + 9}}{2} \][/tex]
Therefore, the solutions to the quadratic equation [tex]\( x^2 + (8k - 3)x + (16k^2 - 15k) = 0 \)[/tex] are:
[tex]\[ x = -4k + \frac{3}{2} - \frac{\sqrt{12k + 9}}{2} \][/tex]
and
[tex]\[ x = -4k + \frac{3}{2} + \frac{\sqrt{12k + 9}}{2} \][/tex]
