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Let us discuss how to determine the nature of the solutions for the given quadratic equations.

Example:

**1**. Find the nature of the solution of the equation \(y = x^2 - 2x - 3\).

**Solution**:

**Step 1**: Draw the graph of the equation \(y = x^2 - 2x - 3\).

The table of values for the equation \(y = x^2 - 2x - 3\) is:

\(x\) | \(-2\) | \(-1\) | \(0\) | \(1\) | \(2\) | \(3\) | \(4\) |

\(x^2\) | \(4\) | \(1\) | \(0\) | \(1\) | \(4\) | \(9\) | \(16\) |

\(2x\) | \(-4\) | \(-2\) | \(0\) | \(2\) | \(4\) | \(6\) | \(8\) |

\(3\) | \(3\) | \(3\) | \(3\) | \(3\) | \(3\) | \(3\) | \(3\) |

\(y\) | \(5\) | \(0\) | \(-3\) | \(-4\) | \(-3\) | \(0\) | \(5\) |

**Step 2**: Plot the points in the graph using a suitable scale.

**Step 3**: Join the points by a smooth curve.

**Step 4**: In the graph, observe that the curve intersects the \(X\) - axis at \(2\) points \((-1,0)\) and \((3,0)\). Therefore, the roots of the equation are \(-1\) and \(3\).

Since there are two points of intersection with the \(X\) - axis, the given equation has real and unequal roots.

**2**. Find the nature of the solutions of the equation \(y = x^2 - 10x + 25\).

**Solution**:

**Step 1**: Draw the graph of the equation \(y = x^2 - 10x + 25\).

The table of values for the equation \(y = x^2 - 10x + 25\) is given by:

\(x\) | \(3\) | \(4\) | \(5\) | \(6\) | \(7\) |

\(x^2\) | \(9\) | \(16\) | \(25\) | \(36\) | \(49\) |

\(10x\) | \(30\) | \(40\) | \(50\) | \(60\) | \(70\) |

\(25\) | \(25\) | \(25\) | \(25\) | \(25\) | \(25\) |

\(y\) | \(4\) | \(1\) | \(0\) | \(1\) | \(4\) |

**Step 2**: Plot the points in the graph.

**Step 3**: Join the points by a smooth curve.

**Step 4**: Here, the curve meets the \(X\) - axis at only one point. Therefore, the point of intersection of the parabola with \(X\) - axis for the given equation is \((5,0)\).

Since the point of intersection is only one point with \(X\) - axis, the given quadratic equation has real and equal roots.