Quadratic Equations

UK Secondary (7-11)

Solve quadratics by algebraic manipulation

Lesson

Solving basic or simple quadratics involves using algebraic manipulation. Let's look at a few examples of this process.

Solve the quadratic equation $x^2-9=0$`x`2−9=0.

**Think**: We want to end up with an expression like $x=\dots$`x`=…, and to get there we will need to isolate $x$`x`.

**Do**: Considering the order of operations, we will first move the constant term to the other side, then undo the square:

$x^2-9$x2−9 |
$=$= | $0$0 |

$x^2$x2 |
$=$= | $9$9 |

$x$x |
$=$= | $\pm3$±3 |

Since there are two numbers that, when squared, give $9$9, the solutions to this equation are $x=3$`x`=3 or $x=-3$`x`=−3.

Find solutions to $\left(2x+7\right)^2=64$(2`x`+7)2=64.

**Think**: Identify the order of operations necessary to isolate the $x$`x` variable. In this case we will deal with the square root first, then the addition and finally the multiplication by $2$2.

**Do**:

$\left(2x+7\right)^2$(2x+7)2 |
$=$= | $64$64 |

$2x+7$2x+7 |
$=$= | $\pm\sqrt{64}$±√64 |

We use inverse operations to remove the square from the $LHS$`L``H``S`, the opposite of a square is a square root. Next we have

$2x+7$2x+7 |
$=$= | $\pm8$±8 | (from here on, this is just like solving linear equations) |

Where we have evaluate the square root. There are two solutions now, let's continue to solve both for $x$`x`:

$2x$2x |
$=$= | $8-7$8−7 | or | $2x$2x |
$=$= | $-8-7$−8−7 |

$2x$2x |
$=$= | $1$1 | or | $2x$2x |
$=$= | $-15$−15 |

$x$x |
$=$= | $\frac{1}{2}$12 | or | $x$x |
$=$= | $\frac{-15}{2}$−152 |

Find the solutions to $x^2-17=0$`x`2−17=0.

**Think**: Identify the order of operations necessary to isolate the $x$`x` variable. In this case we will deal with the subtraction first, then the square.

**Do:**

$x^2-17$x2−17 |
$=$= | $0$0 |

$x^2$x2 |
$=$= | $17$17 |

Here we use inverse operations to remove the $-17$−17 from the $LHS$`L``H``S`, the opposite of a $-17$−17 is a $+$+ $17$17. Then we have

$x$x |
$=$= | $\sqrt{17}$√17 |

OR | ||

$x$x |
$=$= | $-\sqrt{17}$−√17 |

Where we use inverse operations to remove the square from the $LHS$`L``H``S`, the opposite of a square is a square root.

Remember that all square roots can have a positive or negative answer. This is normally written $\pm$±.

For example, what is $4^2$42 and what is $\left(-4\right)^2$(−4)2? They are *both* equal to $16$16! Hence $\sqrt{16}=\pm4$√16=±4.

A special note about exact solutions

For nearly all of our work with solutions to functions and equations it is standard practice to leave our final expression in exact form. In this case, $\sqrt{17}$√17 is as far as we will go, we cannot simplify the root any further.

In questions involving applications of quadratics we may be asked to evaluate the square root at the very end and then approximate to a specific number of decimal places. For example, perhaps we want the time taken to travel a certain distance, or an estimate of the area of a block of land.

Solve for $p$`p`:

$5\left(p^2-3\right)=705$5(`p`2−3)=705

Solve $\left(x-4\right)^2=10$(`x`−4)2=10 for $x$`x`.

- Write all solutions on the same line, separated by commas.