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Solving Quadratic Equations

Solving Quadratic Equations 1.4 - Solution

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a
Let's start by solving the given equation. We can isolate xx by taking the square root on both sides.
x2=dx^2=d
x=±dx = \pm \sqrt{d}
As we can see, to find the value for xx we just need to calculate the square root of dd and make sure that -d\text{-} \sqrt{d} and d\sqrt{d} have separate real value solutions. Thus, dd can be any positive number. We can see an example below.
±4=±2\pm \sqrt{{\color{#0000FF}{4}}} =\pm 2
22=2(2)=42^2 = 2(2) = {\color{#0000FF}{4}} (-2)2=-2(-2)=4(\text{-}2)^2 = \text{-}2(\text{-}2) = {\color{#0000FF}{4}}

With this in mind, we can write the required equation. x2=4\begin{gathered} x^2 = 4 \end{gathered} Note that this is just an example. Since any d>0d>0 will satisfy the requirements of the exercise, there are infinitely many solutions.

b

In Part A we found that x=±d.x=\pm \sqrt{d}. If we want our equation to have just one real solution, then dd should be 0.0. This is because 00 is the only real number that doesn't have an opposite to allow for two solutions. x2=0\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.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. x2=-4\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.