Exercises 1

This page allows you to practice some exercises on Solving Equations. If you notice that you have difficulties, we advise you to go over the corresponding sections in the book.

These questions correspond to the following sections: - Section 3.1 - Solving Equations - Section 3.2 - Equations and Their Parameters - Section 3.3 - Quadratic Equations

Question 1

Solve each of the following equations. (a) $4x + 2(x-4) - 3 = 2(3x -5) - 1$

(b) $(3x - 1)^2 + (4x + 1)^2 = (5x - 1)(5x + 1) + 1$

(c) $\frac{x-3}{x+3} = \frac{x-4}{x+4}$

(d) $\frac{3}{x-3} - \frac{2}{x+3} = \frac{9}{x^2 - 9}$

(e) $\frac{2 - x/(1-x)}{1+x} = \frac{6}{2x + 1}$

(f) $\frac{1}{2}\left(\frac{x}{2} - \frac{3}{4}\right) - \frac{1}{4}\left(1 - \frac{x}{3}\right) - \frac{1}{3} (1-x) = -\frac{1}{3}$

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(a) Any $x$ is a solution.

(b) $x = −1$.

(c) $x = 0$.

(d) $x = −6$.

(e) $x = 4$.

(f) $x = 15/16.$

Question 2

Solve the following equations for the indicated variables. (a) $\frac{1}{2}py^{-1/2} - w = 0 \quad \text{for } y$

(b) $\sqrt{1 + z} + \frac{az}{\sqrt{1+z}} = 0 \quad \text{for } z$

(c) $(3 + a^2)^{x} = 1 \quad \text{for } x$

(d) $\sqrt{pq} - 3q = 5 \quad \text{for } p$

(e) $\frac{\frac{1}{2}K^{-1/2}L^{1/4}}{\frac{1}{4}L^{-3/4}K^{1/2}} = \frac{r}{w} \quad \text{for } L$

(f) $\frac{1}{2}pK^{-1/4}\left(\frac{1}{2} \frac{r}{w} \right)^{1/4} = r \quad \text{for } K$

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(a) $y = \frac{p^2}{4w^2}$.

(b) $z = -\frac{1}{1+a}$.

(c) $x = 0$.

(d) $p = (3q + 5)^2/q.$

(e) $L = rK /2w$.

(f) $K = \frac{1}{32}p^{4}r^{-3}w^{-1} = p^{4} / (32r^{3}w)$.

Question 3

Solve the following quadratic equations, if they have solutions. (a) $x(x+1) = 2x(x-1)$

(b) $x^2 - 4x + 4 = 0$

(c) $-\frac{1}{4}x^2 + \frac{1}{2}x + \frac{1}{2} = 0$

(d) $x(x-5) - 3 = 0$

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

(f) $y^2 - 3y + 2 = 0$

(g) $9y^2 + 42y + 44 = 0$

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(a) $x = 0$ and $x = 3$.

(b) $x = 2$. Note that $x^2 − 4x + 4 = (x − 2)^2$.

(c) $−\frac{1}{4}x^2 + \frac{1}{2}x + \frac{1}{2} = −\frac{1}{4} \left[x − (1 + \sqrt{3})\right]\left[x − (1 − \sqrt{3}) \right] = 0$ for $x = 1 ± \sqrt{3}$.

(d) $x^2 − 5x − 3 = \left[x − \frac{1}{2}(5 + \sqrt{37}) \right] \left[x − \frac{1}{2}(5 − \sqrt{37}) \right] = 0$ for $x = \frac{1}{2}(5 \pm \sqrt{37})$.

(e) $z = −\sqrt{3}, z = \sqrt{2}$.

(f) $y = 1, y = 2$.

(g) $y = \frac{1}{3}( −7 \pm \sqrt{5})$.

Question 4

In a right-angled triangle, the hypotenuse is $34$ cm. One of the short sides is $14$ cm longer than the other. Find the lengths of the two short sides.

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The length $x$ of the shortest side satisfies $x^2 + (x + 14)^2 = 34^2$ , or $2x^2 + 28x = 1156 − 196 = 960$, or $x^2 + 14x - 480 = 0$. The lengths are $16$ cm and $30$ cm.

Question 5

(a) $x^3 - 4x = 0$

(b) $x^4 - 5x^2 + 4 = 0$

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(a) $x = −2, x = 0, x = 2.$ Note that $x(x^2 − 4) = 0$ or equivalenty, $x(x + 2)(x − 2) = 0$. (b) $x = −2, x = −1, x = 1, x = 2.$ The easiest way to solve this is by setting e.g. $x^2 = u$.