2.2 Solving quadratic equations (Page 2/2)

Maths grade 10 rought draft Page 2 / 2

You can combine these in many ways and so the best way to develop your intuition for the best thing to do is practice problems. A combined set of operations could be, for example,

\begin{matrix} \frac{1}{\sqrt{a x^{2} + b x}} & = & c \\ {(\frac{1}{a x^{2} + b x})}^{- 1} & = & {(c)}^{- 1} & (invert both sides) \\ \frac{\sqrt{a x^{2} + b x}}{1} & = & \frac{1}{c} \\ \sqrt{a x^{2} + b x} & = & \frac{1}{c} \\ {(\sqrt{a x^{2} + b x})}^{2} & = & {(\frac{1}{c})}^{2} & (square both sides) \\ a x^{2} + b x & = & \frac{1}{c^{2}} \end{matrix}

Solve for $x$ : $\sqrt{x + 2} = x$

Square both sides of the equation
Both sides of the equation should be squared to remove the square root sign.

$x + 2 = x^{2}$
Write equation in the form $a x^{2} + b x + c = 0$
$\begin{matrix} x + 2 & = & x^{2} & (subtract x^{2} from both sides) \\ x + 2 - x^{2} & = & 0 & (divide both sides by - 1) \\ - x - 2 + x^{2} & = & 0 \\ x^{2} - x + 2 & = & 0 \end{matrix}$
Factorise the quadratic
$x^{2} - x + 2$

The factors of $x^{2} - x + 2$ are $(x - 2) (x + 1)$ .
Write the equation with the factors
$(x - 2) (x + 1) = 0$
Determine the two solutions
We have

$x + 1 = 0$

or

$x - 2 = 0$

Therefore, $x = - 1$ or $x = 2$ .
Check whether solutions are valid
Substitute $x = - 1$ into the original equation $\sqrt{x + 2} = x$ :

$\begin{matrix} LHS & = & \sqrt{(- 1) + 2} \\ = & \sqrt{1} \\ = & 1 \\ but \\ RHS & = & (- 1) \end{matrix}$

Therefore LHS $â‰$ RHS. The sides of an equation must always balance, a potential solution that does not balance the equation is not valid. In this case the equation does not balance.

Therefore $x â‰ - 1$ .

Now substitute $x = 2$ into original equation $\sqrt{x + 2} = x$ :

$\begin{matrix} LHS & = & \sqrt{2 + 2} \\ = & \sqrt{4} \\ = & 2 \\ and \\ RHS & = & 2 \end{matrix}$

Therefore LHS = RHS

Therefore $x = 2$ is the only valid solution
Write the final answer
$\sqrt{x + 2} = x$ for $x = 2$ only.

Solve the equation: $x^{2} + 3 x - 4 = 0$ .

Check if the equation is in the form $a x^{2} + b x + c = 0$
The equation is in the required form, with $a = 1$ .
Factorise the quadratic
You need the factors of 1 and 4 so that the middle term is $+ 3$ So the factors are:

$(x - 1) (x + 4)$
Solve the quadratic equation
$x^{2} + 3 x - 4 = (x - 1) (x + 4) = 0$

Therefore $x = 1$ or $x = - 4$ .
Check the solutions
$1^{2} + 3 (1) - 4 = 0$

${(- 4)}^{2} + 3 (- 4) - 4 = 0$

Both solutions are valid.
Write the final solution
Therefore the solutions are $x = 1$ or $x = - 4$ .

Find the roots of the quadratic equation $0 = - 2 x^{2} + 4 x - 2$ .

Determine whether the equation is in the form $a x^{2} + b x + c = 0$ , with no common factors.
There is a common factor: -2. Therefore, divide both sides of the equation by -2.

$\begin{matrix} - 2 x^{2} + 4 x - 2 & = & 0 \\ x^{2} - 2 x + 1 & = & 0 \end{matrix}$
Factorise $x^{2} - 2 x + 1$
The middle term is negative. Therefore, the factors are $(x - 1) (x - 1)$

If we multiply out $(x - 1) (x - 1)$ , we get $x^{2} - 2 x + 1$ .
Solve the quadratic equation
$x^{2} - 2 x + 1 = (x - 1) (x - 1) = 0$

In this case, the quadratic is a perfect square, so there is only one solution for $x$ : $x = 1$ .
Check the solution
$- 2 {(1)}^{2} + 4 (1) - 2 = 0$ .
Write the final solution
The root of $0 = - 2 x^{2} + 4 x - 2$ is $x = 1$ .

Solving quadratic equations

Solve for $x$ : $(3 x + 2) (3 x - 4) = 0$
Solve for $x$ : $(5 x - 9) (x + 6) = 0$
Solve for $x$ : $(2 x + 3) (2 x - 3) = 0$
Solve for $x$ : $(2 x + 1) (2 x - 9) = 0$
Solve for $x$ : $(2 x - 3) (2 x - 3) = 0$
Solve for $x$ : $20 x + 25 x^{2} = 0$
Solve for $x$ : $4 x^{2} - 17 x - 77 = 0$
Solve for $x$ : $2 x^{2} - 5 x - 12 = 0$
Solve for $x$ : $- 75 x^{2} + 290 x - 240 = 0$
Solve for $x$ : $2 x = \frac{1}{3} x^{2} - 3 x + 14 \frac{2}{3}$
Solve for $x$ : $x^{2} - 4 x = - 4$
Solve for $x$ : $- x^{2} + 4 x - 6 = 4 x^{2} - 5 x + 3$
Solve for $x$ : $x^{2} = 3 x$
Solve for $x$ : $3 x^{2} + 10 x - 25 = 0$
Solve for $x$ : $x^{2} - x + 3$
Solve for $x$ : $x^{2} - 4 x + 4 = 0$
Solve for $x$ : $x^{2} - 6 x = 7$
Solve for $x$ : $14 x^{2} + 5 x = 6$
Solve for $x$ : $2 x^{2} - 2 x = 12$
Solve for $x$ : $3 x^{2} + 2 y - 6 = x^{2} - x + 2$

<< Chapter < Page Page > Chapter >>

Read also:

Get Jobilize Job Search Mobile App in your pocket Now!

100% Free Mobile Applications
Receive real-time job alerts and never miss the right job again

Source: OpenStax, Maths grade 10 rought draft. OpenStax CNX. Sep 29, 2011 Download for free at http://cnx.org/content/col11363/1.1

Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Maths grade 10 rought draft' conversation and receive update notifications?

Ask

©flickr: Abraham	Multicellular Organisms Test By Monty Hartfield Start Test
	8 AP 08 Appendicular Skeleton MCQ By OpenStax Start Quiz
©flickr: iClassical	Music Apperciation By Courntey Hub Start Test
	16 Dr. Amberg Pharm quiz 2 By Brooke Delaney Start Exam
	13 Sociology 13 Aging and the Elderly MCQ By OpenStax Start Quiz
	Principles of microeconomics for ap® courses By OpenStax Read Online Course
	14 Sociology 14 Marriage and Family MCQ By OpenStax Start Quiz
	10 Psychology MCQ 2011 Final Exam By John Gabrieli Start Exam
	NCE Ch 08 Appraisal By Anh Dao Start Quiz
©flickr:	Core Java Unit Test-1(CAJ003) By Abishek Devaraj Start Quiz