Answer :
To find the non-zero value of [tex]\( k \)[/tex] for which the quadratic equation [tex]\( 2 k x^2 - 2 (k - 1) x + 1 = 0 \)[/tex] has equal roots and subsequently find the roots of the equation, we need to follow these steps:
1. Identify the standard form of a quadratic equation:
A quadratic equation is generally given by [tex]\[ ax^2 + bx + c = 0 \][/tex], where [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex] are coefficients.
2. Compare given quadratic with the standard form to identify coefficients:
Given equation: [tex]\[ 2 k x^2 - 2 (k - 1) x + 1 = 0 \][/tex]
Here: [tex]\( a = 2k \)[/tex], [tex]\( b = -2(k - 1) \)[/tex], and [tex]\( c = 1 \)[/tex].
3. Condition for equal roots:
For a quadratic equation to have equal roots, its discriminant must be zero.
The discriminant [tex]\(\Delta\)[/tex] is given by: [tex]\[ \Delta = b^2 - 4ac \][/tex]
Setting the discriminant to zero for equal roots: [tex]\[ \Delta = 0 \][/tex]
4. Calculate the discriminant:
Substitute [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex] into the discriminant formula:
[tex]\[ b = -2(k - 1) \implies b^2 = [-2(k - 1)]^2 = 4(k - 1)^2 \][/tex]
[tex]\[ a = 2k \][/tex]
[tex]\[ c = 1 \][/tex]
Therefore:
[tex]\[ \Delta = 4(k - 1)^2 - 4(2k)(1) \][/tex]
[tex]\[ \Delta = 4(k - 1)^2 - 8k \][/tex]
5. Simplify and set the discriminant to zero:
[tex]\[ 4(k - 1)^2 - 8k = 0 \][/tex]
Divide by 4 to simplify:
[tex]\[ (k - 1)^2 - 2k = 0 \][/tex]
[tex]\[ k^2 - 2k + 1 - 2k = 0 \][/tex]
[tex]\[ k^2 - 4k + 1 = 0 \][/tex]
6. Solve the quadratic equation for [tex]\( k \)[/tex]:
[tex]\[ k^2 - 4k + 1 = 0 \][/tex]
Using the quadratic formula [tex]\( k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex] where [tex]\( a = 1 \)[/tex], [tex]\( b = -4 \)[/tex], and [tex]\( c = 1 \)[/tex]:
[tex]\[ k = \frac{4 \pm \sqrt{16 - 4}}{2} \][/tex]
[tex]\[ k = \frac{4 \pm \sqrt{12}}{2} \][/tex]
[tex]\[ k = \frac{4 \pm 2\sqrt{3}}{2} \][/tex]
[tex]\[ k = 2 \pm \sqrt{3} \][/tex]
So, the values of [tex]\( k \)[/tex] are [tex]\( k = 2 + \sqrt{3} \)[/tex] and [tex]\( k = 2 - \sqrt{3} \)[/tex].
7. Select the non-zero value of [tex]\( k \)[/tex]:
Both solutions are non-zero; let's pick [tex]\( k = 2 - \sqrt{3} \)[/tex] for subsequent steps.
8. Substitute [tex]\( k \)[/tex] back into the quadratic equation:
Given [tex]\( k = 2 - \sqrt{3} \)[/tex]:
[tex]\[ 2(2 - \sqrt{3})x^2 - 2((2 - \sqrt{3}) - 1)x + 1 = 0 \][/tex]
[tex]\[ (4 - 2\sqrt{3})x^2 - 2(1 - \sqrt{3})x + 1 = 0 \][/tex]
[tex]\[ (4 - 2\sqrt{3})x^2 - (2 - 2\sqrt{3})x + 1 = 0 \][/tex]
9. Solve for the roots of the quadratic equation with the chosen [tex]\( k \)[/tex]:
For the quadratic equation to have equal roots, the roots can be found from:
[tex]\[ x = \frac{-b}{2a} \][/tex]
Here:
[tex]\[ a = 4 - 2\sqrt{3} \][/tex]
[tex]\[ b = -(2 - 2\sqrt{3}) = -2 + 2\sqrt{3} \][/tex]
Therefore:
[tex]\[ x = \frac{-(-2 + 2\sqrt{3})}{2(4 - 2\sqrt{3})} = \frac{2 - 2\sqrt{3}}{2(4 - 2\sqrt{3})} \][/tex]
Simplifying:
[tex]\[ x = \frac{2(1 - \sqrt{3})}{2 \cdot 2(2 - \sqrt{3})} = \frac{1 - \sqrt{3}}{4 - 2\sqrt{3}} \][/tex]
10. Further simplification (if necessary):
Rationalize the denominator:
[tex]\[ x = \frac{1 - \sqrt{3}}{4 - 2\sqrt{3}} \cdot \frac{4 + 2\sqrt{3}}{4 + 2\sqrt{3}} = \frac{(1 - \sqrt{3})(4 + 2\sqrt{3})}{16 - 12} = \frac{4 + 2\sqrt{3} - 4\sqrt{3} - 6}{4} = \frac{-2 - 2\sqrt{3}}{4} = -\frac{1 + \sqrt{3}}{2} \][/tex]
Therefore, the roots of the quadratic equation when [tex]\( k = 2 - \sqrt{3} \)[/tex] are [tex]\( -\frac{1 + \sqrt{3}}{2} \)[/tex].
1. Identify the standard form of a quadratic equation:
A quadratic equation is generally given by [tex]\[ ax^2 + bx + c = 0 \][/tex], where [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex] are coefficients.
2. Compare given quadratic with the standard form to identify coefficients:
Given equation: [tex]\[ 2 k x^2 - 2 (k - 1) x + 1 = 0 \][/tex]
Here: [tex]\( a = 2k \)[/tex], [tex]\( b = -2(k - 1) \)[/tex], and [tex]\( c = 1 \)[/tex].
3. Condition for equal roots:
For a quadratic equation to have equal roots, its discriminant must be zero.
The discriminant [tex]\(\Delta\)[/tex] is given by: [tex]\[ \Delta = b^2 - 4ac \][/tex]
Setting the discriminant to zero for equal roots: [tex]\[ \Delta = 0 \][/tex]
4. Calculate the discriminant:
Substitute [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex] into the discriminant formula:
[tex]\[ b = -2(k - 1) \implies b^2 = [-2(k - 1)]^2 = 4(k - 1)^2 \][/tex]
[tex]\[ a = 2k \][/tex]
[tex]\[ c = 1 \][/tex]
Therefore:
[tex]\[ \Delta = 4(k - 1)^2 - 4(2k)(1) \][/tex]
[tex]\[ \Delta = 4(k - 1)^2 - 8k \][/tex]
5. Simplify and set the discriminant to zero:
[tex]\[ 4(k - 1)^2 - 8k = 0 \][/tex]
Divide by 4 to simplify:
[tex]\[ (k - 1)^2 - 2k = 0 \][/tex]
[tex]\[ k^2 - 2k + 1 - 2k = 0 \][/tex]
[tex]\[ k^2 - 4k + 1 = 0 \][/tex]
6. Solve the quadratic equation for [tex]\( k \)[/tex]:
[tex]\[ k^2 - 4k + 1 = 0 \][/tex]
Using the quadratic formula [tex]\( k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex] where [tex]\( a = 1 \)[/tex], [tex]\( b = -4 \)[/tex], and [tex]\( c = 1 \)[/tex]:
[tex]\[ k = \frac{4 \pm \sqrt{16 - 4}}{2} \][/tex]
[tex]\[ k = \frac{4 \pm \sqrt{12}}{2} \][/tex]
[tex]\[ k = \frac{4 \pm 2\sqrt{3}}{2} \][/tex]
[tex]\[ k = 2 \pm \sqrt{3} \][/tex]
So, the values of [tex]\( k \)[/tex] are [tex]\( k = 2 + \sqrt{3} \)[/tex] and [tex]\( k = 2 - \sqrt{3} \)[/tex].
7. Select the non-zero value of [tex]\( k \)[/tex]:
Both solutions are non-zero; let's pick [tex]\( k = 2 - \sqrt{3} \)[/tex] for subsequent steps.
8. Substitute [tex]\( k \)[/tex] back into the quadratic equation:
Given [tex]\( k = 2 - \sqrt{3} \)[/tex]:
[tex]\[ 2(2 - \sqrt{3})x^2 - 2((2 - \sqrt{3}) - 1)x + 1 = 0 \][/tex]
[tex]\[ (4 - 2\sqrt{3})x^2 - 2(1 - \sqrt{3})x + 1 = 0 \][/tex]
[tex]\[ (4 - 2\sqrt{3})x^2 - (2 - 2\sqrt{3})x + 1 = 0 \][/tex]
9. Solve for the roots of the quadratic equation with the chosen [tex]\( k \)[/tex]:
For the quadratic equation to have equal roots, the roots can be found from:
[tex]\[ x = \frac{-b}{2a} \][/tex]
Here:
[tex]\[ a = 4 - 2\sqrt{3} \][/tex]
[tex]\[ b = -(2 - 2\sqrt{3}) = -2 + 2\sqrt{3} \][/tex]
Therefore:
[tex]\[ x = \frac{-(-2 + 2\sqrt{3})}{2(4 - 2\sqrt{3})} = \frac{2 - 2\sqrt{3}}{2(4 - 2\sqrt{3})} \][/tex]
Simplifying:
[tex]\[ x = \frac{2(1 - \sqrt{3})}{2 \cdot 2(2 - \sqrt{3})} = \frac{1 - \sqrt{3}}{4 - 2\sqrt{3}} \][/tex]
10. Further simplification (if necessary):
Rationalize the denominator:
[tex]\[ x = \frac{1 - \sqrt{3}}{4 - 2\sqrt{3}} \cdot \frac{4 + 2\sqrt{3}}{4 + 2\sqrt{3}} = \frac{(1 - \sqrt{3})(4 + 2\sqrt{3})}{16 - 12} = \frac{4 + 2\sqrt{3} - 4\sqrt{3} - 6}{4} = \frac{-2 - 2\sqrt{3}}{4} = -\frac{1 + \sqrt{3}}{2} \][/tex]
Therefore, the roots of the quadratic equation when [tex]\( k = 2 - \sqrt{3} \)[/tex] are [tex]\( -\frac{1 + \sqrt{3}}{2} \)[/tex].
