I.Solve the following equations graphically.

(i) x^{2} – 4x = 0

(ii) x^{2} + x – 12 = 0

(iii) x^{2} – x – 2 = 0

(iv) x^{2} – 5x + 6 = 0

II.

- Draw the graph of y = x
^{2}and find the value of √3 - Draw the graph of y = 2x
^{2}and find the value of √7 - Draw the graph of y =
^{1}/_{2}y^{2}and find the value of √10

## Quadratic Equations – Exercise 9.10 – Solutions:

**I.Solve the following equations graphically.**

**(i) x ^{2} – 4x = 0**

Solution:

Prepare the table of the values for the equation y = x^{2} – 4x

x | -1 | 0 | 1 | 2 | 3 | 4 | 5 |

y | 5 | 0 | -3 | -4 | -3 | 0 | 5 |

The parabola intersects the x – axis at (0, 0) and (4, 0)

Therefore, the roots of equation are 0 and 4.

** (ii) x ^{2} + x – 12 = 0**

Solution:

Prepare the table of the values for the equation x^{2} + x – 12 = 0

x | -4 | -3 | -2 | -1 | 0 | 1 | 2 | 3 | 4 |

y | 0 | 6 | -10 | -12 | -12 | -10 | -6 | 0 | 8 |

**
**The parabola intersects the x – axis only at (3, 0) and (-4, 0)

Therefore, the roots of equation are 3 and -4.

**(iii) x ^{2} – x – 2 = 0**

Solution:

Prepare the table of the values for the equation x^{2} – x – 2 = 0

x | -2 | -1 | 0 | 1 | 2 | 3 |

y | 4 | 0 | -2 | -2 | 0 | 4 |

The parabola intersects the x – axis at (-1, 0) and (2, 0)

Therefore, the roots of equation are -1 and 2.

** (iv) x ^{2} – 5x + 6 = 0**

Solution:

Prepare the table of the values for the equation y = x^{2} – 5x + 6

x | 0 | 1 | 2 | 3 | 4 | 5 |

y | 6 | 2 | 0 | 0 | 2 | 6 |

The parabola intersects the x – axis at (2, 0) and (3, 0)

Therefore, the roots of equation are 2 and 3.

** **

**II.**

**Draw the graph of y = x**^{2}and find the value of √3

Solution:

y = x^{2}

x | 0 | -1 | 1 | -2 | 2 | √3 |

y | 0 | 1 | 1 | 4 | 4 | 3 |

When x = √3, y = (√3)^{2} = 3

Draw a straight line y = 3 parallel to x-axis

The point on x – axis at which the perpendiculars meets are the values of √3 , x = ±1.7

Therefore, √3 = ±1.7

**Draw the graph of y = 2x**^{2}and find the value of √7

Solution:

y = 2x^{2}

x | 0 | -1 | 1 | -2 | 2 | -3 | 3 | √7 |

y | 0 | 2 | 2 | 8 | 8 | 18 | 18 | 14 |

When x = √7, y = 2(√7)^{2} = 14

Draw a straight line y = 14 parallel to x-axis.

The point on x – axis at which the perpendiculars meets are the values of √7 , x = ±2.6

Therefore, √7 = ±2.6

**Draw the graph of y =**^{1}/_{2}x^{2}and find the value of √10

Solution:

y = ^{1}/_{2}x^{2}

x | 0 | -2 | 2 | -4 | 4 | √10 |

y | 0 | 2 | 2 | 8 | 8 | 5 |

When x = √10, y = ^{1}/_{2}(√10)^{2} = 5

Draw a straight line y = 5 parallel to x-axis.

The point on x – axis at which the perpendiculars meets are the values of √10 , x = ±3.1

Therefore, √10 = ±3.1