Solution:

1. \(m^{2} + 5m + 3 = 0\)
2. \(y^{2} - 3 = 0\)

Solution:

In this equation, \(x\) is the only variable and the maximum index of the variable is 2.

∴ This is a quadratic equation.

Solution:

Let’s express the given equation in the standard form:

 \(y^{2} - 5y + 10 = 0\)

In this equation, \(y\) is the only variable and the maximum index of the variable is 2.

∴ This is a quadratic equation.

Solution:

Multiplying both sides by \(y\),

 \(\displaystyle y^{2} \times y + \frac {1}{\cancel {y}} \times \cancel {y} = 2 \times y\)

∴ \(\displaystyle y^{3} + 1 = 2y\)

∴ \(\displaystyle y^{3} - 2y + 1 = 0\)

In this equation, \(y\) is the only variable and the maximum index of the variable is 3.

∴ This is not a quadratic equation.

Solution:

Multiplying both sides by \(x\),

 \(\displaystyle x \times x + \frac {1}{\cancel {x}} \times \cancel {x} = - 2 \times x\)

∴ \(\displaystyle x^{2} + 1 = - 2x\)

∴ \(\displaystyle x^{2} + 2x + 1 = 0\)

In this equation, \(x\) is the only variable and the maximum index of the variable is 2.

∴ This is a quadratic equation.

Solution:

 \((m + 2)(m - 5) = 0\)

∴ \(m(m - 5) + 2(m - 5) = 0\)

∴ \(m^{2} - 5m + 2m - 10 = 0\)

∴ \(m^{2} - 3m - 10 = 0\)

In this equation, \(m\) is the only variable and the maximum index of the variable is 2.

∴ This is a quadratic equation.

Solution:

 \(m^{3} + 3m^{2} - 2 = 3m^{3}\)

∴ \(m^{3} + 3m^{2} - 2 - 3m^{3} = 0\)

∴ \(-2m^{3} + 3m^{2} - 2 = 0\)

In this equation, \(m\) is the only variable and the maximum index of the variable is 3.

∴ This is not a quadratic equation.

Solution:

 \(m^{3} + 3m^{2} - 2 = 3m^{3}\)

∴ \(m^{3} + 3m^{2} - 2 - 3m^{3} = 0\)

∴ \(-2m^{3} + 3m^{2} - 2 = 0\)

In this equation, \(m\) is the only variable and the maximum index of the variable is 3.

∴ This is not a quadratic equation.

Solution:

 \(2y = 10 - y^{2}\)
∴ \(2y - 10 + y^{2} = 0\)
i.e. \(y^{2} + 2y - 10 = 0\)

∴ \(a = 1, b = 2, c = - 10\)

  \((x - 1)^{2} = 2x + 3\)
∴ \(x^{2} - 2x + 1 = 2x + 3\)
∴ \(x^{2} - 2x + 1 - 2x - 3 = 0\)
∴ \(x^{2} - 4x - 2 = 0\)

∴ \(a = 1, b = - 4, c = - 2\)

Solution:

  \(x^{2} + 5x = -(3 - x)\)
∴ \(x^{2} + 5x = - 3 + x\)
∴ \(x^{2} + 5x + 3 - x = 0\)
∴ \(x^{2} + 4x + 3 = 0\)

∴ \(a = 1, b = 4, c = 3\)

Solution:

 \(3m^{2} - 2m^{2} = - 9\)
∴ \(m^{2} = - 9\)
∴ \(m^{2} + 9 = 0\)
i.e. \(m^{2} + 0m + 9 = 0\)

∴ \(a = 1, b = 0, c = 9\)

Solution:

 \(p(3 + 6p) = - 5\)
∴ \(3p + 6p^{2} = - 5\)
i.e. \(6p^{2} + 3p + 5 = 0\)

∴ \(a = 6, b = 3, c = 5\)

Solution:

 \(x^{2} - 9 = 13\)
∴ \(x^{2} - 9 - 13 = 0\)
∴ \(x^{2} - 22 = 0\)
∴ \(x^{2} + 0x - 22 = 0\)

∴ \(a = 1, b = 0, c = - 22\)

Solution:

\(x = 1\)
Putting \(x = 1\) in the polynomial \(x^{2} + 4x - 5 = 0\)
 \(x^{2} + 4x - 5\)
= \(1^{2} + 4 \times 1 - 5\)
= \(1 + 4 - 5\)
= \(0\)
∴ \(x = 1\) is the root of the equation \(x^{2} + 4x - 5 = 0\)

\(x = - 1\)
Putting \(x = - 1\) in the polynomial \(x^{2} + 4x - 5 = 0\)
 \(x^{2} + 4x - 5\)
= \((- 1)^{2} + 4 \times (- 1) - 5\)
= \(1 - 4 - 5\)
= \(- 8 \ne 0\)
∴ \(x = - 1\) is not the root of the equation \(x^{2} + 4x - 5 = 0\)

\(m = 2\)
Putting \(m = 2\) in the polynomial \(2m^{2} - 5m\)
 \(2m^{2} - 5m\)
= \(2 \times 2^{2} - 5 \times 2\)
= \(8 - 10\)
= \(- 2 \ne 0\)
∴ \(m = 2\) is not the root of the equation \(2m^{2} - 5m = 0\)

\(\displaystyle m = \frac {5}{2}\)

Putting \(\displaystyle m = \frac {5}{2}\) in the polynomial \(2m^{2} - 5m\)

 \(\displaystyle 2m^{2} - 5m\)

= \(\displaystyle 2 \times \left(\frac {5}{2}\right)^{2} - 5 \times \frac {5}{2}\)

= \(\displaystyle \cancelto {1}{2} \times \frac {25}{\cancelto {2}{4}} - \frac {25}{2}\)

= \(\displaystyle \frac {25}{2} - \frac {25}{2}\)

= \(0\)

∴ \(\displaystyle m = \frac {5}{2}\) is the root of the equation \(2m^{2} - 5m = 0\)

Solution:

\(x = 3\) is a root of equation \(kx^2 - 10x + 3 = 0\)
∴ \(k(3)^2 - 10(3) + 3 = 0\)
∴ \(9k - 30 + 3 = 0\)
∴ \(9k - 27 = 0\)
∴ \(9k = 27\)

∴ \(\displaystyle k = \frac {\cancelto {3}{27}}{\cancelto {1}{9}}\)

∴ \(k = 3\)

Solution:

\(\bbox[white, 5pt, border: 2px solid red]{\displaystyle \frac {- 7}{5} \ }\) is a root of quadratic equation \(5m^2 + 2m + k = 0\).

∴ Put \(\displaystyle m = \bbox[white, 5pt, border: 2px solid red]{\frac {- 7}{5}} \ \) in the given equation,

\(\displaystyle 5 \times \bbox[white, 5pt, border: 2px solid red]{\frac {- 7}{5}}^{2} + 2 \times \bbox[white, 5pt, border: 2px solid red]{\ \frac {- 7}{5}\ } + k = 0\)

\(\displaystyle \cancelto {1}{5} \times \frac {49}{\cancelto {5}{25}} - \frac {14}{5} + k = 0\)

\(\displaystyle \bbox[white, 5pt, border: 2px solid red]{\ \frac {49}{5}\ } - \bbox[white, 5pt, border: 2px solid red]{\ \frac {14}{5}\ } + k = 0\)

\(\displaystyle \bbox[white, 5pt, border: 2px solid red]{\ \frac {\cancelto{7}{35}}{\cancelto {1}{5}}\ } + k = 0\)

\(\displaystyle k = \bbox[white, 5pt, border: 2px solid red]{\ - 7\ } \)