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Let us learn how to graphically solve \(2\) quadratic equations.

Example:

**1**. Draw the graph of \(y = x^2 + 2x - 3\) and hence solve \(x^2 - x - 6 = 0\).

**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 given by:

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

\(y\) | \(5\) | \(0\) | \(-3\) | \(-4\) | \(-3\) | \(1\) | \(5\) |

**Step 2**: To solve the equation \(x^2 - x - 6 = 0\), subtract the equation \(x^2 - x - 6 = 0\) from \(y = x^2 + 2x - 3\).

\(y = x^2 + 2x - 3\)

\(0 = x^2 - x - 6\) (\(-\))

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\(y = x + 3\)

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**Step 3**: Draw the graph of the equation \(y = x + 3\).

The table of values for the equation \(y = x + 3\) is given by:

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

\(y\) | \(-1\) | \(0\) | \(1\) | \(2\) | \(3\) | \(4\) | \(5\) | \(6\) |

**Step 4**: Mark the points of intersection of \(y = x^2 + 2x - 3\) and \(y = x + 3\). The point of intersection is \((-3,0)\) and \((2,5)\).

**Step 5**: The \(x\) - coordinates of the points are \(-3\) and \(2\). Therefore, the solution set for the equation \(x^2 - x - 6 = 0\) is \({-3,2}\).

**2**. Draw the graph of \(y = 2x^2 + x - 2\) and hence solve \(2x^2 = 0\).

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

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

\(x\) | \(-2\) | \(0\) | \(2\) |

\(y\) | \(4\) | \(-2\) | \(8\) |

**Step 2**: To solve the equation \(2x^2 = 0\), subtract \(2x^2 = 0\) from \(y = 2x^2 + x - 2\).

\(y = 2x^2 + x - 2\)

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

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\(y = x - 2\)

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

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

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

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

**Step 4**: Mark the points of intersection of \(y = 2x^2 + x - 2\) and \(y = x - 2\). The point of intersection is \((0,-2)\).

**Step 5**: The \(x\) - coordinates of the points is \(0\). Therefore, the solution set for the equation \(2x^2 = 0\) is \(0\).