# micellaneious equations

Miscellaneous Equations

This method of factoring used in the previous section can also be applied to certain nonquadratic equations. Specifically, if an equation can be expresses in factored form, with zero on one side, then solutions may often be obtained by setting each factor equal to zero, as illustrated in the next two examples.

Example 1 :

Solve

$x^3+2x^2-x-2=0$

Solution :

The left side may be factored by grouping, as follows :

$x^3+2x^2-x-2=0$

$x^2(x+2)-(x+2)=0$

$(x^2-1)(x+2)=0$

$(x+1)(x-1)(x+2)=0$

Setting each factor equal to 0 gives us

x + 1 = 0 , x – 1 = 0, x + 2 = 0,

or equivalently,

x = -1, x = 1, x = -2.

That these three numbers are solutions may be checked by substitution in the original equation.

That these three numbers are solutions may be checked by substitution in the original equation.

Example 2 :

Solve

$x^{\frac{3}{2}}=x^{\frac{1}{2}}$

Solution :

We may proceed as follows :

$x^{\frac{3}{2}}-x^{\frac{1}{2}}=0$

$x^{\frac{1}{2}}(x-1)=0$

Hence,

$x^{\frac{1}{2}}=0\text{ or}x-1=0$

From which we obtain the solution x = 0 and x = 1.

A common error, when an equation such as $x^{\frac{3}{2}}=x^{\frac{1}{2}}$ is given, is to divide both sides by $x^{\frac{1}{2}}$, obtaining x =1. Observe that this lead to the loss of the solution x = 0.

Certain equation that are not linear or quadratic can be transformed into one of those forms by a suitable manipulation. If equations involve radicals or fractional exponents, the method of raising both sides to a positive integral power is often used. When this is done, the solutions of the new equation always contain the solutions of the original equation. For example, the solutions of

$2x-3=\sqrt{x+6}$

Are also solutions of

$(2x-3)^2=(\sqrt{x+6})^2$

In some cases the new equation has more solutions than the original equation. To illustrate, if we start with the equation x = 3 and square both sides, we obtain $x^2=9$. Note that the given equation has only one solution, 3, whereas the new equation has two solutions, 3 and -3. Any solution of the new equation which is not a solution of the original equation is called an extraneous solution. Since extraneous solutions may arise, it is absolutely essential to check all solutions obtained after raising both sides of an equation to some power.

Example 3 :

Solve

$\sqrt[3]{x^2-1}=2$

Solution :

If we cube both sides, then the solutions of the given equation are included among the solutions of the following :

$(\sqrt[3]{x^2-1})^3=2^3$

$x^2-1=8$

$x^2=9$

Hence, the only possible solutions of

$\sqrt[3]{x^2-1}=2$

Are 3 or -3.

We next check each of these numbers by substitution in $\sqrt[3]{x^2-1}=2$, substituting 3 for x in the equation we obtain $\sqrt[3]{3^2-1}=2$, or $\sqrt[3]{8}=2$, which is a true statement. Thus, 3 is a solution. Similarly, -3 is a solution. Hence, the solutions of the given equation are 3 and -3.

Example 4 :

Solve

$3+\sqrt{3x+1}=x$

Solution :

We begin by isolating the radical on one side as follows :

$\sqrt{3x+1}=x-3$

Next we square both sides and simplify, obtaining

$(\sqrt{3x+1})^2=(x-3)^2$

$3x+1=x^2-6x+9$

$x^2-9x+8=0$

$9x-1)(x-8)=0$

Since the last equation has solution 1 and 8, it follows that 1 and 8 are the only possible solutions of the original equation.

We now check each of these by substitution in $3+\sqrt{3x+1}=x$. letting x = 1 given us

$3+\sqrt{4}=1\text{ or }5=1$

Which is false. Consequently. 1 is not a solution. Letting x = 8 in the given equation we obtain

$3+\sqrt{25}=8\text{ or }3+5=8$

Which is true. Hence, the equation $3+\sqrt{3x+1}=x$ has only one solution, x = 8.

For certain equation involving radicals it is necessary to use the process of raising sides to powers several times, as illustrated in the next example.

Example 5 :

Solve

$\sqrt{2x-3}-\sqrt{x+7}+2=0$

Solution :

Let us begin by writing

$\sqrt{2x-3}=\sqrt{x+7}-2=0$

Squaring both sides we obtain

$2x-3=(x+7)-4\sqrt{x+7}+4$

Which simplifies to

$x-14=-4\sqrt{x+7}$

Squaring both sides of the last equation and simplifying given us

$x^2-28x+196=16(x+7)$

$x^2-28x+196=16x+112$

$x^2-44x+84=0$

$(x-42)(x-2)=0$

Hence, the only possible solutions of the given equation are 42 and 2.

We next check each of these by substitution in the original equation. Substituting x = 42 gives us

$\sqrt{84-3}-\sqrt{42+7}+2=0$

Or

9 – 7 + 2 = 0

Which is false. Hence, 42 is not a solution. If we substitute x = 2 we obtain

$\sqrt{4-3}-\sqrt{2+7}+2=0$

Or

1 – 3 + 2 = 0

Which is true. Hence, the given equation has one solution, x = 2.

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