Solving Quadratic Equations

a

Let's start by solving the given equation. We can isolate

x

by taking the square root on both sides.

x^2=d

SqrtEqn

\sqrt{\text{LHS}}=\sqrt{\text{RHS}}

x = \pm \sqrt{d}

As we can see, to find the value for

x

we just need to calculate the square root of

d

and make sure that

\text{-} \sqrt{d}

and

\sqrt{d}

have separate real value solutions. Thus,

d

can be any positive number. We can see an example below.

$\pm \sqrt{{\color{#0000FF}{4}}} =\pm 2$
$2^2 = 2(2) = {\color{#0000FF}{4}}$	$(\text{-}2)^2 = \text{-}2(\text{-}2) = {\color{#0000FF}{4}}$

With this in mind, we can write the required equation. $\begin{gathered} x^2 = 4 \end{gathered}$ Note that this is just an example. Since any $d>0$ will satisfy the requirements of the exercise, there are infinitely many solutions.

b

In Part A we found that $x=\pm \sqrt{d}.$ If we want our equation to have just one real solution, then $d$ should be $0.$ This is because $0$ is the only real number that doesn't have an opposite to allow for two solutions. $\begin{gathered} x^2 = 0 \end{gathered}$

c

If we want the equation to have no real solutions, we can use any negative number for $d.$ This is because there is not a real number such that when multiplied by itself results in a negative number. With this in mind, we can write the required equation. $\begin{gathered} x^2 = \text{-}4 \end{gathered}$ Notice that this is just an example. Since any negative number will satisfy the requirements of the exercise, there are infinitely many solutions.

Solving Quadratic Equations 1.4 - Solution