Solution:

 \(2y^{2} - y + 2 = 0\)

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

Now, discriminant
= \(b^{2} - 4ac\)
= \((-\:1)^{2} - 4 \times 2 \times 2\)
= \(1 -\:16\)
= \(-\:15\)
∴ Discriminant = − 15

Solution:

 \(5m^{2} - m = 0\)
i.e. \(5m^{2} - m + 0 = 0\)

Comparing with \(am^{2} + bm + c = 0\),
 \(a = 5,\ b = -\:1,\ c = 0\)

Now, discriminant
= \(b^{2} - 4ac\)
= \((-\:1)^{2} - 4 \times 5 \times 0\)
= \(1 -\:0\)
= \(1\)
∴ Discriminant = 1

Solution:

 \(\sqrt {5}x^{2} - x - \sqrt {5}= 0\)

Comparing with \(ax^{2} + bx + c = 0\),
 \(a = \sqrt {5},\ b = -\:1,\ c = -\:\sqrt {5}\)

Now, discriminant
= \(b^{2} - 4ac\)
= \((-\:1)^{2} - 4 \times \sqrt {5} \times -\:\sqrt {5}\)
= \(1 + 4 \times 5\)
= \(1 + 20\)
= \(21\)
∴ Discriminant = 21

Solution:

\(-\:2\) is one root of the equation \(2x^{2} + kx - 2 = 0\).
∴ Substituting \(x = -\:2\) in the given equation,
 \(2x^{2} + kx - 2 = 0\)
∴ \(2 \times (-\:2)^{2} + k \times -\:2 - 2 = 0\)
∴ \(2 \times 4 -\:2k - 2 = 0\)
∴ \(8 -\:2k -\:2 = 0\)
∴ \(6 -\:2k = 0\)
∴ \(6 = 2k\)
i.e. \(2k = 6\)

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

∴ \(k = 3\)

Solution:

Let, \(\alpha = 10\) and \(\beta = -\:10\)

Now, the required equation is:
 \(x^{2} - (\alpha + \beta)x + \alpha\beta = 0\)
∴ \(x^{2} -\:(10 -\:10)x + 10 \times -\:10 = 0\)
∴ \(x^{2} -\:(0)x -\:100 = 0\)
∴ \(x^{2} -\:100 = 0\)

Solution:

Let, \(\alpha = 1 -\:3\sqrt{5}\) and \(\beta = 1 + 3\sqrt{5}\)

Now, the required equation is:
 \(x^{2} - (\alpha + \beta)x + \alpha\beta = 0\)
∴ \(x^{2} -\:(1 \cancel{-\:3\sqrt{5}} + 1 \cancel{+ 3\sqrt{5}})x + (1 -\:3\sqrt{5}) \times (1 + 3\sqrt{5}) = 0\)
∴ \(x^{2} -\:2x + (1)^2 -\:(3\sqrt{5})^2 = 0\) ... [\(\because (a - b)(a + b) = a^{2} - b^{2}\)]
∴ \(x^{2} -\:2x + 1 -\:9 \times 5 = 0\)
∴ \(x^{2} -\:2x + 1 -\:45 = 0\)
∴ \(x^{2} -\:2x -\:44 = 0\)

Solution:

Let, \(\alpha = 0\) and \(\beta = 7\)

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

Solution:

 \(3x^{2} - 5x + 7 = 0\)

Comparing with \(ax^{2} + bx + c = 0\),
 \(a = 3,\ b = -\:5,\ c = 7\)

Now, \(\Delta = b^{2} - 4ac\)
∴ \(\Delta = (- 5)^{2} - 4 \times 3 \times 7\)
∴ \(\Delta = 25 - 84\)
∴ \(\Delta = -\:59\)
i.e. \(\Delta \lt 0\)
∴ The roots are not real.

Solution:

 \(\sqrt {3}x^{2} + \sqrt {2}x -\:2\sqrt {3} = 0\)

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

Now, \(\Delta = b^{2} - 4ac\)
∴ \(\Delta = (\sqrt {2})^{2} - 4 \times \sqrt {3} \times -\:2\sqrt {3}\)
∴ \(\Delta = 2 + 8 \times 3\)
∴ \(\Delta = 2 + 24\)
∴ \(\Delta = 26\)
i.e. \(\Delta \gt 0\)
∴ The roots are real and unequal.

Solution:

 \(m^{2} -\:2m + 1 = 0\)

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

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

Solution:

\(\displaystyle \frac {1}{x + 5} = \frac {1}{x^{2}}\)

By invertendo,
  \(x + 5 = x^{2}\)
i.e. \(x^{2} = x + 5\)
∴ \(x^{2} - x - 5 = 0\)

Comparing with \(ax^{2} + bx + c = 0\),
 \(a = 1,\ b = - 1,\ c = - 5\)

Now, \(b^{2} - 4ac\)
= \((- 1)^{2} - 4 \times 1 \times - 5\)
= \(1 + 20\)
= \(21\)
∴ \(b^{2} - 4ac = 21\)

Now, using Quadratic Formula,

 \(\displaystyle x = \frac {- b \pm \sqrt {b^{2} - 4ac}}{2a}\)

∴ \(\displaystyle x = \frac {- (- 1) \pm \sqrt {21}}{2 \times 1}\)

∴ \(\displaystyle x = \frac {1 \pm \sqrt {21}}{2}\)

∴ \(\displaystyle x = \frac {1 + \sqrt {21}}{2}\) or \(\displaystyle x = \frac {1 - \sqrt {21}}{2}\)

∴ \(\displaystyle \frac {1 + \sqrt {21}}{2}\), \(\displaystyle \frac {1 - \sqrt {21}}{2}\) are the roots of the given quadratic equation.

Solution:

\(\displaystyle x^{2} - \frac {3x}{10} - \frac {1}{10} = 0\)

Multiplying both sides by 10,

 \(\displaystyle x^{2} \times 10 - \frac {3x}{\cancel {10}} \times \cancel {10} - \frac {1}{\bcancel {10}} \times \bcancel {10} = 0 \times 10\)

∴ \(\displaystyle 10x^{2} - 3x - 1 = 0\)

Comparing with \(ax^{2} + bx + c = 0\),
 \(a = 10,\ b = - 3,\ c = - 1\)

Now, \(b^{2} - 4ac\)
= \((- 3)^{2} - 4 \times 10 \times - 1\)
= \(9 + 40\)
= \(49\)
∴ \(b^{2} - 4ac = 49\)

Now, using Quadratic Formula,

 \(\displaystyle x = \frac {- b \pm \sqrt {b^{2} - 4ac}}{2a}\)

∴ \(\displaystyle x = \frac {- (- 3) \pm \sqrt {49}}{2 \times 10}\)

∴ \(\displaystyle x = \frac {3 \pm 7}{20}\)

∴ \(\displaystyle x = \frac {3 + 7}{2}\) or \(\displaystyle x = \frac {3 - 7}{2}\)

∴ \(\displaystyle x = \frac {\cancelto {1}{10}}{\cancelto {2}{20}}\) or \(\displaystyle x = \frac {\cancelto{- 1}{- 4}}{\cancelto {5}{20}}\)

∴ \(\displaystyle x = \frac {1}{2}\) or \(\displaystyle x = - \frac {1}{5}\)

∴ \(\displaystyle \frac {1}{2}\), \(\displaystyle - \frac {1}{5}\) are the roots of the given quadratic equation.

Solution:

\((2x + 3)^{2} = 25\)

Taking square roots of both sides,

 \(2x + 3 = \pm 5\)

∴ \(2x = - 3 \pm 5\)

∴ \(2x = - 3 + 5\) OR \(2x = - 3 - 5\)

∴ \(2x = 2\) OR \(2x = - 8\)

∴ \(\displaystyle x = \frac {\cancel{2}}{\cancel{2}}\) OR \(\displaystyle x = \frac {\cancelto {- 4}{- 8}}{\cancelto {1}{2}}\)

∴ \(x = 1\) OR \(x = - 4\)

∴ 1, − 4 are the roots of the given quadratic equation.

Solution:

\(m^{2} + 5m + 5 = 0\)

Comparing with \(am^{2} + bm + c = 0\),
 \(a = 1,\ b = 5,\ c = 5\)

Now, \(b^{2} - 4ac\)
= \(5^{2} - 4 \times 1 \times 5\)
= \(25 - 20\)
= \(5\)
∴ \(b^{2} - 4ac = 5\)

Now, using Quadratic Formula,

 \(\displaystyle m = \frac {- b \pm \sqrt {b^{2} - 4ac}}{2a}\)

∴ \(\displaystyle m = \frac {- 5 \pm \sqrt {5}}{2 \times 1}\)

∴ \(\displaystyle m = \frac {- 5 \pm \sqrt {5}}{20}\)

∴ \(\displaystyle m = \frac {- 5 + \sqrt {5}}{2}\) or \(\displaystyle m = \frac {- 5 - \sqrt {5}}{2}\)

∴ \(\displaystyle m = \frac {1}{2}\) or \(\displaystyle m = - \frac {1}{5}\)

Solution:

\(m^{2} + 5m + 5 = 0\)

Comparing with \(am^{2} + bm + c = 0\),
 \(a = 5,\ b = 2,\ c = 1\)

Now, \(b^{2} - 4ac\)
= \(2^{2} - 4 \times 5 \times 1\)
= \(4 - 20\)
= \(- 16\)
∴ \(\Delta = b^{2} - 4ac = - 16\)
∴ \(\Delta \lt 0\)
∴ The roots are not real and equal.

Solution:

\(x^2 - 4x - 3 = 0\)

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

Now, \(b^{2} - 4ac\)
= \((- 4)^{2} - 4 \times 1 \times - 3\)
= \(16 + 12\)
= \(28\)
∴ \(b^{2} - 4ac = 28\)

Now, using Quadratic Formula,

 \(\displaystyle x = \frac {- b \pm \sqrt {b^{2} - 4ac}}{2a}\)

∴ \(\displaystyle x = \frac {- (- 4) \pm \sqrt {28}}{2 \times 1}\)

∴ \(\displaystyle x = \frac {4 \pm \sqrt {4 \times 7}}{2}\)

∴ \(\displaystyle x = \frac {4 \pm 2\sqrt {7}}{2}\)

∴ \(\displaystyle x = \frac {\cancel{2}(2 \pm \sqrt {7})}{\cancel{2}}\)

∴ \(x = 2 \pm \sqrt {7}\)

∴ \(x = 2 + \sqrt {7}\) or \(x = 2 - \sqrt {7}\)

∴ \(2 + \sqrt {7}\), \(2 - \sqrt {7}\) are the roots of the given quadratic equation.

Solution:

 \((m - 12)x^{2} + 2(m - 12)x + 2 = 0\)

Comparing with \(ax^{2} + bx + c = 0\),
 \(a = m - 12,\ b = 2(m - 12),\ c = 2\)

Now, \(b^{2} - 4ac\)
= \([2(m - 12)]^{2} - 4 \times (m - 12) \times 2\)
= \(4(m^{2} - 24m + 144) - 8(m - 12)\)
= \(4m^{2} - 96m + 576 - 8m + 96\)
= \(4m^{2} - 104m + 672\)
∴ \(b^{2} - 4ac = 4m^{2} - 104m + 672\) ... (i)

But, the roots are real and equal. ... (Given)
∴ \(b^{2} - 4ac = 0\)
∴ \(4m^{2} - 104m + 672 = 0\)
Dividing both sides by 4,
 \(m^{2} - 26m + 168 = 0\)

∴ \(\underline {m^{2} - 14m}\ \underline {-\ 12m + 168} = 0\)

∴ \(m\underline {(m - 14)}\ - 12 \underline {(m - 14)} = 0\)

∴ \((m - 14)(m - 12) = 0\)

∴ \(m - 14 = 0 \text { OR } m - 12 = 0\)

∴ \(m = 14 \text { OR } m = 12\)

Now, if \(m = 12\), then
\(a = m - 12 = 12 - 12 = 0\)
But, in a quadratic equation, \(ax^2 + bx + c = 0, a \ne 0\)
∴ \(m \ne 12\)
∴ \(m = 14\)

Solution:

Let, \(\alpha\) and \(\beta\) be the roots of that equation.

From the given information,
 \(\alpha + \beta = 5\) ... (i) and

 \(\alpha^{3} + \beta^{3} = 35\) ... (ii)
∴ \((\alpha + \beta)^{3} - 3\alpha\beta(\alpha + \beta) = 35\)
∴ \(5^{3} - 3\alpha\beta \times 5 = 35\)
∴ \(125 - 15\alpha\beta = 35\)
∴ \(125 - 35 = 15\alpha\beta\)
∴ \(90 = 15\alpha\beta\)
i.e. \(15\alpha\beta = 90\)

∴ \(\displaystyle \alpha\beta = \frac {\cancelto {6}{90}}{\cancelto {1}{15}}\)

∴ \(\alpha\beta = 6\) ... (iii)

Now, the required equation is:
 \(x^2 - (\alpha + \beta)x + \alpha\beta = 0\)
∴ \(x^2 - 5x + 6 = 0\)

Solution:

\(2x^2 + 2(p + q)x + p^2 + q^2 = 0\)

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

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

∴ \(\displaystyle \alpha + \beta = - \frac {\cancel {2}(p + q)}{\cancel {2}}\)

∴ \(\displaystyle \alpha + \beta = - (p + q)\) ... (i)

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

∴ \(\displaystyle \alpha\beta = \frac {p^2 + q^2}{2}\) ... (ii)

Now, the roots of the required equation are \((\alpha + \beta)^2\) and \((\alpha - \beta)^2\).

Let, \(\alpha_1\) and \(\beta_1\) be the roots of the required equation.

Also, let \(\alpha_1 = (\alpha + \beta)^2\) and \(\beta_1 = (\alpha - \beta)^2\).

Now, the sum of the required roots:

= \(\alpha_1 + \beta_1\)

= \((\alpha + \beta)^2 + (\alpha - \beta)^2\)

= \((\alpha + \beta)^2 + (\alpha + \beta)^2 - 4\alpha\beta\)

= \(\displaystyle [-(p + q)^2] + [-(p + q)^2] - \cancelto {2}{4} \times \frac {p^2 + q^2}{\cancelto {1}{2}}\)

= \((p + q)^2 + (p + q)^2 - 2(p^2 + q^2)\)

= \(\cancel{p^2} + 2pq + \bcancel{q^2} + \cancel{p^2} + 2pq + \bcancel{q^2} - \cancel{2p^2} - \bcancel{2q^2}\)

= \(4pq\)

∴ \(\alpha_1 + \beta_1 = 4pq\) ... (iii)

And, the product of the required roots:

= \(\alpha_1 \beta_1\)

= \((\alpha + \beta)^2 \times (\alpha - \beta)^2\)

= \((\alpha + \beta)^2 \times [(\alpha + \beta)^2 - 4\alpha\beta]^2\)

= \(\displaystyle [- (p + q)]^2 \times \left ([- (p + q)^2] - \cancelto {2}{4} \times \frac {p^2 + q^2}{\cancelto {1}{2}}\right )\)

= \((p + q)^2 \times [(p + q)^2 - 2(p^2 + q^2)]\)

= \((p + q)^2 \times (p^2 + 2pq + q^2 - 2p^2 - 2q^2)\)

= \((p + q)^2 \times (- p^2 + 2pq - q^2)\)

= \((p + q)^2 \times - (p^2 - 2pq + q^2)\)

= \((p + q)^2 \times - (p - q)^2\)

= \(- [(p + q)^2 \times (p - q)^2]\)

= \(- [(p + q)(p - q)]^2\)

= \(- (p^2 - q^2)^2\)

∴ \(\alpha_1 \beta_1 = - (p^2 - q^2)^2\) ... (iv)

Now, the required equation is:
 \(x^2 - (\alpha + \beta)x + \alpha\beta = 0\)
∴ \(x^2 - 4pqx - (p^2 - q^2)^2 = 0\)

Solution:

Let, the amount with Sagar be ₹ \(x\).
∴ The amount with Mukund is ₹ \(x + 50\).

From the given information,

 \(x \times (x + 50) = 15000\)

∴ \(x^2 + 50x = 15000\)

 \(x^{2} + 50x - 15000 = 0\)

∴ \(\underline {x^{2} + 150x}\ \underline {-\ 100x -\ 15000} = 0\)

∴ \(x\underline {(x + 150)}\ - 100 \underline {(x + 150)} = 0\)

∴ \((x + 150)(x - 100) = 0\)

∴ \(x + 150 = 0 \text { OR } x - 100 = 0\)

∴ \(x = - 150 \text { OR } x = 100\)

But, \(x\) is the amount with Sagar.
∴ It can’t be negative.
∴ \(x = 100\) ... (i)

And \(x + 50 = 100 + 50 = 150\) ... (ii)

∴ Sagar has ₹ 100 and Mukund has ₹ 150.

Solution:

Let, the greater number be \(x\).
∴ The square of the smaller number = \(2x\).

From the given information,

 \(x^2 - 2x = 120\)

∴ \(x^2 - 2x - 120 = 0\)

∴ \(\underline {x^{2} - 12x}\ \underline {+ 10x -\ 120} = 0\)

∴ \(x\underline {(x - 12)}\ + 10 \underline {(x - 12)} = 0\)

∴ \((x - 12)(x + 10) = 0\)

∴ \(x - 12 = 0 \text { OR } x + 10 = 0\)

∴ \(x = 12 \text { OR } x = - 10\)

But, \(x\) can’t be negative.

∴ \(x = 12\) ... (i)

And the square of the smaller number = \(2x = 2 \times 12 = 24\)

∴ The smaller number = \(\pm \sqrt {24}\) ... (ii)

∴ Those numbers are 12 and \(\sqrt {24}\) or 12 and \(- \sqrt {24}\).

(This is the answer given in the textbook.)

Solution:

Let, the number of students be \(x\).

If 540 oranges are distributed equally between \(x\) students,

each student will get \(\displaystyle \frac {540}{x}\) oranges.

Now, if 30 students were more, the number of students would become \(x + 30\).

And each student would now get \(\displaystyle \frac {540}{x + 30}\) oranges.

From the given information,

  \(\displaystyle \frac {540}{x} - \frac {540}{x + 30} = 3\)

∴ \(\displaystyle \frac {540 (x + 30) - 540x}{x (x + 30)} = 3\)

∴ \(\displaystyle \frac {\cancel {540x} + 16200 - \cancel {540x}}{x^2 + 30x} = 3\)

∴ \(\displaystyle \frac {16200}{x^2 + 30x} = 3\)

∴ \(16200 = 3(x^2 + 30x)\)

∴ \(16200 = 3x^2 + 90x\)

∴ \(0 = 3x^2 + 90x - 16200\)

i.e. \(3x^2 + 90x - 16200 = 0\)

Dividing both sides by 3,

 \(x^2 + 30x - 5400 = 0\)

∴ \(\underline {x^{2} + 90x}\ \underline {-\ 60x - 5400} = 0\)

∴ \(x\underline {(x + 90)}\ - 60 \underline {(x + 90)} = 0\)

∴ \((x + 90)(x - 60) = 0\)

∴ \(x + 90 = 0 \text { OR } x - 60 = 0\)

∴ \(x = - 90 \text { OR } x = 60\)

But, \(x\) is the number of students.
∴ It can’t be negative.
∴ \(x = 60\). ... (i)

∴ The number of students is 60.

Solution:

Let, the breadth of the field be \(x\) m.
∴ It’s length = \(2x + 10\) m.

∴ Area of the farm
= length × breadth
= \((2x + 10)\times x\)
= (\(2x^2 + 10x\)) m² ... (i)

And, the side of the square shaped pond
= \(\displaystyle \frac {1}{3} x\) m.

∴ The area of the square shaped pond
= \(\displaystyle \left (\frac {1}{3}x\right)^2\)

= \(\displaystyle \frac {x^2}{9}\) m² ... (ii)

From the given information,
∴ \(\displaystyle 2x^2 + 10x = 20 \times \frac {x^2}{9}\)

Multiplying both sides by 9,
 \(\displaystyle 2x^2 \times 9 + 10x \times 9 = 20 \times \frac {x^2}{\cancel {9}} \times \cancel {9}\)
∴ \(18x^2 + 90x = 20x^2\)
∴ \(0 = 20x^2 - 18x^2 - 90x\)
∴ \(0 = 2x^2 - 90x\)
i.e. \(2x^2 - 90x = 0\)
Dividing both sides by 2,
 \(x^2 - 45x = 0\)
∴ \(x(x - 45) = 0\)
∴ \(x = 0 \text { OR }x - 45 = 0\)
∴ \(x = 0 \text { OR }x = 45\)

But, \(x\) is the breadth of the field.
∴ It cannot be 0.
∴ \(x = 45\) m ... (iii)

And, \(2x + 10 = 2 \times 45 + 10 = 90 + 10 = 100\) m ... (iv)

Also, the side of the pond
= \(\displaystyle \frac {1}{3}x\)
= \(\displaystyle \frac {1}{\cancelto {1}{3}} \times \cancelto {15}{45}\)
= \(15\) m ... (v)

∴ The length and breadth of the field is 100 m and 45 m and the side of the pond is 15 m.

Solution:

Let, the time taken by larger tap to fill the tank be \(x\) hours.
∴ The time taken by smaller tap to fill the tank = \(x + 3\) hours.

Now, in 1 hour, the larger tap fills \(\displaystyle \frac {1}{x}\) part of the tank and the smaller tap fills \(\displaystyle \frac {1}{x + 3}\) part of the tank.

But, both the taps together fill the tank in 2 hours.

 \(\displaystyle 2\left[\frac {1}{x} + \frac {1}{x + 3}\right] = 1\)

∴ \(\displaystyle \frac {1}{x} + \frac {1}{x + 3} = \frac {1}{2}\)

∴ \(\displaystyle \frac {x + 3 + x}{x (x + 3)} = \frac {1}{2}\)

∴ \(\displaystyle \frac {2x + 3}{x^2 + 3x} = \frac {1}{2}\)

∴ \(2(2x + 3) = 1(x^2 + 3x)\)

∴ \(4x + 6 = x^2 + 3x\)

∴ \(0 = x^2 + 3x - 4x - 6\)

∴ \(0 = x^2 - x - 6\)

i.e. \(x^2 - x - 6 = 0\)

∴ \(\underline {x^{2} - 3x}\ \underline {+ 2x - 6} = 0\)

∴ \(x\underline {(x - 3)}\ + 2 \underline {(x - 3)} = 0\)

∴ \((x - 3)(x + 2) = 0\)

∴ \(x - 3 = 0 \text { OR } x + 2 = 0\)

∴ \(x = 3 \text { OR } x = - 2\)

But, \(x\) is the time required to fill the tank.
∴ It cannot be negative.
∴ \(x = 3\) hours ... (i)

And, \(x + 3 = 3 + 3 = 6\) hours ... (ii)

∴ The larger tap will fill that tank in 3 hours and the smaller tap will take 6 hours to fill the tank.