# Long Questions

Solve by graphically x^{2} - 3x = 10.

Let, x

^{2}- 3x - 10 = y --- (i)

Comparing equation (i) with ax^{2} + bx + c = y, we get,

`a` = 1, `b` = -3, `c` = -10

`a` is positive. So, the curve is concave upward.

Now, vertex of parabola,

(`x`, `y`) = $(\frac{-b}{2a}, \frac{4ac\text{ - }
b^2}{4a})$ = $(\frac{3}{2}, \frac{\text{-40 - 9}}{4})$

(`x`, `y`) = (1.5, -12.25)

Finding passing points of parabola,

x |
-1 | -2 | 0 | 1 | 1.5 | 2 | 3 | 5 |
---|---|---|---|---|---|---|---|---|

y |
-6 | 0 | -10 | -12 | -12.25 | -12 | -10 | 0 |

Finally, graph of given quadratic equation is,

The parabola cuts the x-axis at (-2, 0) and (5, 0). So, `x` = -2
and `x` = 5.

Solve by graphically x^{2} + 2x - 3 = 0.

Let, x

^{2}+ 2x - 3 = y --- (i)

Comparing equation (i) with ax^{2} + bx + c = y, we get,

`a` = 1, `b` = 2, `c` = -3

`a` is positive. So, the curve is concave upward.

Now, vertex of parabola,

(`x`, `y`) = $(\frac{-b}{2a}, \frac{4ac\text{ - }
b^2}{4a})$ = $(\frac{-2}{2}, \frac{\text{-12 - 4}}{4})$

(`x`, `y`) = (-1, -4)

Finding passing points of parabola,

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

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

Finally, graph of given quadratic equation is,

The parabola cuts the x-axis at (-3, 0) and (1, 0). So, `x` = -3
and `x` = 1.

Solve graphically the quadratic equation x^{2} - 2x - 3 = 0.

Let, x

^{2}- 2x - 3 = y --- (i)

Comparing equation (i) with ax^{2} + bx + c = y, we get,

`a` = 1, `b` = -2, `c` = -3

`a` is positive. So, the curve is concave upward.

Now, vertex of parabola,

(`x`, `y`) = $(\frac{-b}{2a}, \frac{4ac\text{ - }
b^2}{4a})$ = $(\frac{2}{2}, \frac{\text{-12 - 4}}{4})$

(`x`, `y`) = (1, -4)

Finding passing points of parabola,

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

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

Finally, graph of given quadratic equation is,

`x`= -1 and

`x`= 3.

Solve by graphically x^{2} - 3x + 2 = 0.

Let, x

^{2}- 3x + 2 = y --- (i)

Comparing equation (i) with ax^{2} + bx + c = y, we get,

`a` = 1, `b` = -3, `c` = 2

`a` is positive. So, the curve is concave upward.

Now, vertex of parabola,

(`x`, `y`) = $(\frac{-b}{2a}, \frac{4ac\text{ - }
b^2}{4a})$ = $(\frac{3}{2}, \frac{\text{8 - 9}}{4})$

(`x`, `y`) = (1.5, -0.25)

Finding passing points of parabola,

x |
1.5 | 0 | 1 | -1 | 2 | 3 | 4 |
---|---|---|---|---|---|---|---|

y |
-0.25 | 2 | 0 | 6 | 0 | 2 | 6 |

Finally, graph of given quadratic equation is,

The parabola cuts the x-axis at (2, 0) and (3, 0). So, `x` = 2
and `x` = 3.

Solve by graphically: x^{2} = 5x - 6.

Let, x

^{2}- 5x + 6 = y --- (i)

Comparing equation (i) with ax^{2} + bx + c = y, we get,
`a` = 1, `b` = -5, `c` = 6

`a` is positive. So, the curve is concave upward.

Now, vertex of parabola,

(`x`, `y`) = $(\frac{-b}{2a}, \frac{4ac\text{ - }
b^2}{4a})$ = $(\frac{5}{2}, \frac{\text{4(6) - }(-5)^2}{4})$

(`x`, `y`) = $(\frac{5}{2}, \frac{-1}{4})$ = (2.5,
-0.25)

Finding passing points of parabola,

x |
2.5 | 0 | 1 | 2 | 3 | 4 | 5 |
---|---|---|---|---|---|---|---|

y |
-0.25 | 6 | 2 | 0 | 0 | 2 | 6 |

Finally, graph of given quadratic equation is,

The parabola cuts the x-axis at (2, 0) and (3, 0). So, `x` = 2
and `x` = 3.

Solve by graphical method: x^{2} + x - 2 = 0

Let, x

^{2}+ x - 2 = y --- (i)

Comparing equation (i) with ax^{2} + bx + c = y, we get,

`a` = 1, `b` = 1, `c` = -2

`a` is positive. So, the curve is concave upward.

Now, vertex of parabola,

(`x`, `y`) = $(\frac{-b}{2a}, \frac{4ac\text{ - }
b^2}{4a})$ = $(\frac{-1}{2}, \frac{\text{-8 - 1}}{4})$

(`x`, `y`) = $(\frac{-1}{2}, \frac{-9}{4})$ = (-0.5,
-2.25)

Finding passing points of parabola,

x |
-0.5 | 1 | -1 | 2 | -2 | -3 |
---|---|---|---|---|---|---|

y |
-2.25 | 0 | -2 | 4 | 0 | 4 |

Finally, graph of given quadratic equation is,
The parabola cuts the x-axis at (1, 0) and (-2, 0). So, `x` = 1
and `x` = -2.

y = x^{2}; y = 2 - x

Given equations,

y = x^{2} --- (i)

y = 2 - x --- (ii)

From equation (i),

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

y |
0 | 1 | 4 | 9 |

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

y |
2 | 1 | 0 | -1 | -2 |

Finally, graph of given equations is,

From the graph, parabola and straight meets each other at (1, 1) and (-2, 4). So, the solutions are (1, 1) and (-2, 4).x^{2} - 2x - 8 = 0

Let, x

^{2}- 2x - 8 = y --- (i)

Comparing equation (i) with ax^{2} + bx + c = y, we get,

`a` = 1, `b` = -2, `c` = -8

`a` is positive. So, the curve is concave upward.

Now, vertex of parabola,

(`x`, `y`) = $(\frac{-b}{2a}, \frac{4ac\text{ - }
b^2}{4a})$ = $(\frac{2}{2}, \frac{\text{4(-8) - 4}}{4})$

(`x`, `y`) = (1, -9)

Finding passing points of parabola,

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

y |
-9 | -8 | -5 | -8 | 0 |

Finally, graph of given quadratic equation is,

`x`= -2 and

`x`= 4.

x^{2} + 7x + 12 = 0

Let, x

^{2}+ 7x + 12 = y --- (i)

Comparing equation (i) with ax^{2} + bx + c = y, we get,

`a` = 1, `b` = 7, `c` = 12

`a` is positive. So, the curve is concave upward.

Now, vertex of parabola,

(`x`, `y`) = $(\frac{-b}{2a}, \frac{4ac\text{ - }
b^2}{4a})$ = $(\frac{-7}{2}, \frac{\text{4(12) - 49}}{4})$

(`x`, `y`) = $(\frac{-7}{2}, \frac{-1}{4})$ = (-3.5,
-0.25)

Finding passing points of parabola,

x |
-3.5 | -1 | -2 | -3 | -4 | -5 |
---|---|---|---|---|---|---|

y |
-0.25 | 6 | 2 | 0 | 0 | 2 |

Finally, graph of given quadratic equation is,

The parabola cuts the x-axis at (-3, 0) and (-4, 0). So, `x` =
-3 and `x` = -4.