(1)

(2)

(3) If α, β are roots of quadratic equation

Solution:

 \(x^{2} + 7x - 1 = 0\)

Comparing with \(ax^{2} + bx + c = 0\),
 \(a = 1,\ b = 7,\ c = -\:1\)

Now, discriminant
= \(b^{2} - 4ac\)
= \(7^{2} - 4 \times 1 \times -\:1\)
= \(49 + 4\)
= \(53\)
∴ Discriminant = 53

Solution:

 \(2y^{2} - 5y + 10 = 0\)

Comparing with \(ay^{2} + by + c = 0\),
 \(a = 2,\ b = -\:5,\ c = 10\)

Now, discriminant
= \(b^{2} - 4ac\)
= \((-\:5)^{2} - 4 \times 2 \times 10\)
= \(25 -\:80\)
= \(-\:55\)
∴ Discriminant = − 55

Solution:

 \(\sqrt {2}x^{2} + 4x + 2\sqrt {2} = 0\)

Comparing with \(ax^{2} + bx + c = 0\),
 \(a = \sqrt {2},\ b = 4,\ c = 2\sqrt {2}\)

Now, discriminat
= \(b^{2} - 4ac\)
= \(4^{2} - 4 \times \sqrt {2} \times 2\sqrt {2}\)
= \(16 - 8 \times 2\)
= \(16 - 16\)
= \(0\)

∴ Discriminant = 0

Solution:

 \(x^{2} - 4x + 4 = 0\)

Comparing with \(ax^{2} + bx + c = 0\),
 \(a = 1,\ b = - 4,\ c = 4\)

Now, \(\Delta = b^{2} - 4ac\)
∴ \(\Delta = (- 4)^{2} - 4 \times 1 \times 4\)
∴ \(\Delta = 16 - 16\)
∴ \(\Delta = 0\)
∴ The roots are real and equal.

Solution:

 \(2y^{2} - 7y + 2 = 0\)

Comparing with \(ay^{2} + by + c = 0\),
 \(a = 2,\ b = -\:7,\ c = 2\)

Now, \(\Delta = b^{2} - 4ac\)
∴ \(\Delta = (- 7)^{2} - 4 \times 2 \times 2\)
∴ \(\Delta = 49 - 16\)
∴ \(\Delta = 33\)
∴ \(\Delta \gt 0\)
Since, \(\Delta \gt 0\), the roots are real and unequal.

Solution:

 \(m^{2} + 2m + 9 = 0\)

Comparing with \(am^{2} + bm + c = 0\),
 \(a = 1,\ b = 2,\ c = 9\)

Now, \(\Delta = b^{2} - 4ac\)
∴ \(\Delta = 2^{2} - 4 \times 1 \times 9\)
∴ \(\Delta = 4 -\:36\)
∴ \(\Delta = -\:32\)
∴ \(\Delta \lt 0\)
Since, \(\Delta \lt 0\), the roots are not real.

Solution:

 Let, \(\alpha = 0 \text { and } \beta = 4\)

Now, the required equation:
 \(x^{2} - (\alpha + \beta)x + \alpha \beta = 0\)
∴ \(x^{2} - (0 + 4)x + 0 \times 4 = 0\)
∴ \(x^{2} - 4x + 0 = 0\)
∴ \(x^{2} - 4x = 0\)
∴ The required quadratic equation is \(x^{2} - 4x = 0\).

Solution:

 Let, \(\alpha = 3 \text { and } \beta = -\:10\)

Now, the required equation:
 \(x^{2} - (\alpha + \beta)x + \alpha \beta = 0\)
∴ \(x^{2} - (3 -\:10)x + 3 \times -\:10 = 0\)
∴ \(x^{2} - (-\:7)x -\:30 = 0\)
∴ \(x^{2} + 7x -\:30 = 0\)
∴ The required quadratic equation is \(x^{2} + 7x -\:30 = 0\).

Solution:

 Let, \(\displaystyle \alpha = \frac {1}{2} \text { and } \beta = -\:\frac {1}{2}\)

Now, the required equation:
 \(x^{2} - (\alpha + \beta)x + \alpha \beta = 0\)

∴ \(\displaystyle x^{2} - \left(\cancel {\frac {1}{2}} -\:\cancel {\frac {1}{2}}\right)x + \frac {1}{2} \times -\:\frac {1}{2} = 0\)

∴ \(\displaystyle x^{2} - 0x - \frac {1}{4} = 0\)

∴ \(\displaystyle x^{2} - \frac {1}{4} = 0\)

∴ The required quadratic equation is \(\displaystyle x^{2} - \frac {1}{4} = 0\).

Solution:

 Let, \(2 - \sqrt{5} \text { and } \beta = 2 + \sqrt{5}\)

Now, the required equation:
 \(x^{2} - (\alpha + \beta)x + \alpha \beta = 0\)
∴ \(x^{2} - (2 - \cancel {\sqrt{5}} + 2 + \cancel {\sqrt{5}})x + (2 - \sqrt{5})(2 + \sqrt{5}) = 0\)
∴ \(x^{2} - 4x + 4 - 5 = 0\)
∴ \(x^{2} - 4x - 1 = 0\)
∴ The required quadratic equation is \(x^{2} - 4x - 1 = 0\).

Solution:

 \(x^{2} - 4kx + k + 3 = 0\)

Comparing with \(ax^{2} + bx + c = 0\),
 \(a = 1,\ b = -\:4k,\ c = k + 3\)

Now, \(\displaystyle \alpha + \beta = - \frac {b}{a}\)

∴ \(\displaystyle \alpha + \beta = - \frac {-\:4k}{1}\)

∴ \(\displaystyle \alpha + \beta = 4k\) ... (i)

And, \(\displaystyle \alpha \beta = \frac {c}{a}\)

∴ \(\displaystyle \alpha \beta = \frac {k + 3}{1}\)

∴ \(\displaystyle \alpha \beta = k + 3\) ... (ii)

Also, from the given condition,
 \(\alpha + \beta = 2\alpha \beta\)
∴ \(4k = 2(k + 3)\) ... [From (i) and (ii)]
∴ \(4k = 2k + 6\)
∴ \(4k - 2k = 6\)
∴ \(2k = 6\)

∴ \(\displaystyle k = \frac {\cancelto {3}{6}}{\cancelto {1}{2}}\)

∴ \(k = 3\)

Solution:

 \(y^{2} - 2y - 7 = 0\)

Comparing with \(ay^{2} + by + c = 0\),
 \(a = 1,\ b = -\:2,\ c = -\:7\)

Now, \(\displaystyle \alpha + \beta = - \frac {b}{a}\)

∴ \(\displaystyle \alpha + \beta = - \frac {-\:2}{1}\)

∴ \(\displaystyle \alpha + \beta = 2\) ... (i)

And, \(\displaystyle \alpha \beta = \frac {c}{a}\)

∴ \(\displaystyle \alpha \beta = \frac {-\:7}{1}\)

∴ \(\displaystyle \alpha \beta = -\:7\) ... (ii)

Now, \(\alpha^{2} + \beta^{2} = (\alpha + \beta)^{2} - 2\alpha\beta\)
∴ \(\alpha^{2} + \beta^{2} = (2)^{2} - 2\times -\:7\) ... [From (i) and (ii)]
∴ \(\alpha^{2} + \beta^{2} = 4 + 14\)
∴ \(\alpha^{2} + \beta^{2} = 18\) ... (iii)

Solution:

Now, \(\alpha^{3} + \beta^{3} = (\alpha + \beta)(\alpha^{2} + \beta^{2} - \alpha \beta)\)

∴ \(\alpha^{3} + \beta^{3} = 2[18 -\:(-\:7)]\) ... [From (i) (ii) and (iii)]

∴ \(\alpha^{3} + \beta^{3} = 2(18 + 7)\)

∴ \(\alpha^{3} + \beta^{3} = 2(25)\)

∴ \(\alpha^{3} + \beta^{3} = 50\) ... (iv)

Solution:

 \(3y^{2} + ky + 12 = 0\)

Comparing with \(ay^{2} + by + c = 0\),
 \(a = 3,\ b = k,\ c = 12\)

Now, the roots are real and equal. ... (Given)
∴ \(b^{2} - 4ac = 0\)
∴ \(k^{2} - 4 \times 3 \times 12 = 0\)
∴ \(k^{2} - 144 = 0\)
∴ \(k^{2} = 144\)
∴ \(k = \sqrt {144}\)
∴ \(k = \pm 12\)
∴ \(k = + 12\) or \(k = -\:12\).

Solution:

 \(kx(x - 2) + 6 = 0\)
∴ \(kx^{2} - 2kx + 6 = 0\)

Comparing with \(ax^{2} + bx + c = 0\),
 \(a = k,\ b = - 2k,\ c = 6\)

Now, the roots are real and equal. ... (Given)
∴ \(b^{2} - 4ac = 0\)
∴ \((- 2k)^{2} - 4 \times k \times 6 = 0\)
∴ \(4k^{2} - 24k = 0\)
∴ \(4k(k - 6) = 0\)
∴ \(4k = 0\) or \(k - 6 = 0\)
∴ \(k = 0\) or \(k = 6\).
But, \(a = k\) and in a quadratic equation \(ax^{2} + bx + c = 0\), the value of \(a\) can not be \(0\).
∴ \(k = 6\)