Mathematics at Aksharam

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Dr. Lalitha Subramanian, About the Author

CBSE X Mathematics

A quadratic expression in one variable has at most three terms: a term with square of the variable, a term with linear power of the variable, and a constant term. Of these, a quadratic expression may miss the linear term or the constant term, but it must have at least the term containing the square of the variable. Examples of quadratic expressions are: $2x^2 - 3x + 4$, $3x^2 + 7$, $5x^2 - 2x$, $2 - 6x^2,$, $8x - x^2$, $10 - 2x^2 + 4x$, etc. A quadratic expression in the standard form is one with the power of the variable in descending order, i.e., first term having the highest power, 2, and subsequent terms terms with decreasing powers of the variable.
A quadratic equation in standard form is of the form $ax^2 + bx + c = 0$, where $a, b, c$ are real numbers and $a \neq 0$
The graph of a quadratic equation $y = ax^2 + bx + c$ is a parabola, which we studied in chapter 2. We also know that the zeros of the graph are the $x$-intercepts, which are also the solution set of the equation $ax^2 + bx + c = 0$. We also learned that a quadratic equation can have a maximum of two solutions. In this chapter, we discuss different methods of solving a quadratic equation.
The solution set of the quadratic equation can be detertmined in any one of the following methods:

1. Factorization method.
2. Taking square root method.
3. Completing the square method.
5. Graphing method.

Factorization method is based on the zero product principle. The zero product principle states that, when the product equals zero, then, any one or more of the factors should be zero. For using this principle, we first factor the quadratic expression on the left side. As the product of these factors equals zero, we set each factor equal to zero and thus find the roots of the equation. We shall use the techniques of factorising a quadratic equations that we had learned in earlier classes.

Example 1:

Solve each of the following equations by factorization: $(a) 2x^2 + 15x + 27 = 0; \quad (b) x^2 + 4x - 12 = 0; \quad (c) y^2 - 16y - 80 = 0; \quad (d) 8x^2 - 18x + 9 = 0$

Solution:

$(a)$ To factorize this equation, recall that we first need to find the two factors of $P = 2 \times 27 = 54$ such that the sum of the factors equals the coefficient of the middle term $15$. Such a pair of factors is $6$ and $9$ as $6 \times 9 = 54$ and $6 + 9 = 15$. Factoring and solving this equation: \begin{eqnarray} 2x^2 + 15x + 27 &=& 0\\ 2x^2 + 6x + 9x + 27 &=& 0\\ 2x(x + 3) + 9(x + 3) &=& 0\\ (x + 3)(2x + 9) &=& 0\\ x + 3 &=& 0 \implies x = - 3\\ 2x + 9 &=& 0 \implies x = -\frac{9}{2}\\ \end{eqnarray}
So, the solution set is $\{-\frac{9}{2}, - 3\}$

$(b)$ Here, product $P = - 12$ and sum $S = + 4$. Such a pair of factors are $+6$ and $-2$.: \begin{eqnarray} x^2 + 4x - 12 &=& 0\\ x^2 + 6x - 2x - 12 &=& 0 \\ x(x + 6) - 2(x + 6) &=& 0\\ ( x + 6)(x - 2) &=& 0\\ x + 6 &=& 0 \implies x= - 6\\ x - 2 &=& 0 \implies x = 2\\ \end{eqnarray}
So, solution set is $\{-6, 2\}$

$(c)$ In this equation, product of factors $P = -80$ and their sum $s = - 16$. Such a pair of factors are $-20$ and $4$: \begin{eqnarray} y^2 - 16y - 80 &=& 0\\ y^2 - 20y + 4y - 80 &=& 0\\ y(y - 20) + 4(y - 20) &=& 0\\ (y - 20)(y + 4) &=& 0\\ y - 20 &=& 0 \implies y = 20\\ y + 4 &=& 0 \implies y = - 4\\ \end{eqnarray}
So, solution set is $\{-4, 20\}$

$(d)$ For this equation, product is $P = 8 \times 9 = 72$ and sum $s = - 18$. Such a pair of factors are $-12$ and $-6$: \begin{eqnarray} 8x^2 - 18x + 9 &=& 0\\ 8x^2 - 12x - 6x + 9 &=& 0\\ 4x(2x - 3) - 3(2x - 3) &=& 0\\ (2x - 3)(4x - 3) &=& 0\\ 2x - 3 &=& 0 \implies x = \frac{3}{2}\\ 4x - 3 &=& 0 \implies x = \frac{3}{4}\\ \end{eqnarray}
So, solution set is $\{\frac{3}{3}. \frac{3}{2}\}$

The method of solving a quadratic equation by square root method is possible when the equation has a square term and a constant term. In this case, we keep the square term on left side and the constant term on right side. Then, take square root on both sides. This will result in two possible values on the right side, one positive and one negative. This is because of the fact that both $(a)^2$ and $(-a)^2$ result in $a^2$. So, when we take square root of $a^2$, the result could be either $a$ or $- a$. The process results in obtaining two different equations which can be solved separately to get the solution sets of the equations.

Example 2:

Solve each of the following equations by taking square root on both sides:
$(a) x^2 - 16 = 0; \quad (b) (y - 3)^2 - 7 = 0; \quad (c) (2x - 3)^2 = 9; \quad (d) (x + 5)^2 = 49; Solution:$(a)$First take the constant term to the right side, and the take square root on both sides: \begin{eqnarray}x^2 &=& 16\\ \sqrt{x^2} &=& \pm \sqrt{16}\\ x &=& \pm 4\\ x &=& 4\\x &=& - 4\\ \end{eqnarray} So, solution set is$\{- 4, 4\}(b)$First we take the constant term to the right side and take square root on both sides: \begin{eqnarray} (y - 3)^2 &=& 7\\ \sqrt{(y - 3)^2} &=& \pm \sqrt{7}\\ y - 3 &=& \pm \sqrt{7}\\ y - 3 &=& +\sqrt{7} \implies y = 3 + \sqrt{7}\\ y - 3 &=& -\sqrt{7} \implies y = 3 - \sqrt{7}\\ \end{eqnarray} So, solution set is$\{3 + \sqrt{7}, 3 - \sqrt{7}\}(c)$We can directly start the process here: \begin{eqnarray} (2x - 3)^2 &=& 9\\ \sqrt{(2x - 3)^2} &=&\pm \sqrt{9}\\ 2x - 3 &=& \pm 3\\ 2x - 3 &=& 3 \implies x = 3\\ 2x - 3 &=& - 3 \implies x = 0\\ \end{eqnarray} So, solution set is$\{0, 3\}(d)$We can directly start the process for this problem also: \begin{eqnarray} (x + 5)^2 &=& 49\\ \sqrt{(x + 5)^2} &=& \pm \sqrt{49}\\ x + 5 &=& \pm 7\\ x + 5 &=& 7 \implies x = 2\\ x + 5 &=& - 7 \implies x = - 12\\ \end{eqnarray} So, solution set is$\{-12, 2\}$. When we solve a quadratic equation by completing the square method, we add suitable constant to both sides of the equation to get the left side of the equation as a perfect square. Then, we use the previous method to solve the equation. Example 3: Solve each of the following equations by completing the square:$(a) x^2 - 6x - 3 = 0; \quad (b) y^2 - 4y + 2 = 0; \quad (c) 2t^2 + 4t + 1 = 0; \quad (d) 2n^2 - 8n - 3 = 0$Solution:$(a)$First step is to take the constant term to the right side. Then, complete the square on the left side by adding appropriate third constant term, on both sides. After this step, the equation can be solved by taking square root on both side, like the previous method: \begin{eqnarray} x^2 - 6x - 3 &=& 0\\ x^2 - 6x &=& 3\\ x^2 - 6x + (3)^2 &=& 3 +9\\ (x - 3)^2 &=& 12\\ \sqrt{(x - 3)^2} &=& \pm \sqrt{12}\\ x - 3 &=& \pm 2 \sqrt{3}\\ x - 3 &=& + 2 \sqrt{3} \implies x = 3 + 2 \sqrt{3}\\ x - 3 &=& - 2 \sqrt{3} \implies x = 3 - 2 \sqrt{3}\\ \end{eqnarray} So, solution is$\{3 + 2 \sqrt{3}, 3 - 2 \sqrt{3}\}(b)$\begin{eqnarray} y^2 - 4y + 2 &=& 0\\ y^2 - 4y &=& - 2 \\ y^2 - 4y + (2)^2 &=& - 2 + (2)^2\\ (y - 2)^2 &=& 2\\ \sqrt{(y - 2)^2} &=& \pm \sqrt{2}\\ - 2 &=& \pm \sqrt{2}\\ y - 2 &=& + \sqrt{2} \implies y = 2 + \sqrt{2}\\ y - 2 &=& -\sqrt{2} \implies y = 2 - \sqrt{2}\\ \end{eqnarray} So, solution set is$\{2 + \sqrt{2}, 2 - \sqrt{2}\}(c)$First step is to take the constant term to the right side. In this problem, coefficient of the leading term$t^2$term is not$1$. So, divide each term on both sides by this coefficient so that the$t^2$term has unity coefficient. After this, proceed as in the previous problems: \begin{eqnarray} 2t^2 + 4t + 1 &=& 0\\ 2t^2 + 4t &=& - 1\\ t^2 + 2t &=& -\frac{1}{2}\\ t^2 + 2t + 1 &=& -\frac{1}{2} + 1\\ (t + 1)^2 &=& \frac{1}{2}\\ \sqrt{(t + 1)^2} &=& \pm \sqrt{\frac{1}{2}}\\ t + 1 &=& \pm \frac{\sqrt{2}}{2}\\ t + 1 &=& \frac{\sqrt{2}}{2} \implies t = - 1 + \frac{\sqrt{2}}{2}\\ t + 1 &=& - \frac{\sqrt{2}}{2} \implies t = - 1 - \frac{\sqrt{2}}{2}\\ \end{eqnarray} So, solution set is$\{ - 1 + \frac{\sqrt{2}}{2}, - 1 - \frac{\sqrt{2}}{2}\}$.$(d)$This equation is to be solved like the previous one: \begin{eqnarray} 2n^2 - 8n - 3 &=& 0\\ 2n^2 - 8n &=& 3\\ n^2 - 4n &=& \frac{3}{2}\\ n^2 - 4n + (2)^2 &=& \frac{3}{2} + 4\\ (n - 2)^2 &=& \frac{11}{2}\\ n - 2 &=& \pm \sqrt{\frac{11}{2}}\\ n - 2 &=& + \frac{\sqrt{11}}{2} \implies n = 2 + \frac{\sqrt{11}}{2}\\ n - 2 &=& - \frac{\sqrt{11}}{2} \implies n = 2 - \frac{\sqrt{11}}{2}\\ \end{eqnarray} So, solution set is$\{2 - \frac{\sqrt{11}}{2}, 2 + \frac{\sqrt{11}}{2}\}$By generalizing the 'completing the square method'. we get the quadratic formula, which can be used directly used to find solutions to a quadratic equation. The standard form of a quadratic equation is$ax^2 + bx + c = 0$. Using the process of completing the square method for this, we get $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ Example 4: Find solution sets of each of the following quadratic equations using the quadratic formula:$(a) 2x^2 - 5x + 3 = 0; \quad (b) 4x^2 + 4\sqrt{3}x + 3 = 0; \quad (c) \frac{1}{x + 4} - \frac{1}{x - 7} = \frac{11}{30}, x \neq 4, 7; \quad (d) 2p^2 - 3p - 2 = 0$. Solution:$(a)$In this equation,$a = 2, b = - 5, c = 3$. Substituting these values in the quadratic formula: \begin{eqnarray} x &=& \frac{5 \pm \sqrt{(-5)^2 - 4(2)(3)}}{2(2)}\\ &=& \frac{5 \pm \sqrt{25 - 24}}{4}\\ &=& \frac{5 \pm 1}{4}\\ x &=& \frac{5 + 1}{4} = \frac{6}{4} = \frac{3}{2}\\ x &=& \frac{5 - 1}{4} = 1\\ \end{eqnarray}. So, solution set is$\{1, \frac{3}{2}\}(b)$Here,$a=4, b=4\sqrt{3}, c=3$: \begin{eqnarray} x &=& \frac{-4 \pm \sqrt{(4\sqrt{3})^2 - 4(4)(3)}}{2(4)}\\ &=& \frac{-4 \pm \sqrt{48 - 48}}{8}\\ &=& \frac{- 4 \pm 0}{8}\\ &=& \frac{- 4}{8}\\ &=& - \frac{1}{2}\\ \end{eqnarray}. So, this equation has only one solution$x = -\frac{1}{2}$. We know that this means the graph of this equation touches the$x$-axis at$\left(\frac{1}{2}, 0\right)$.$(c)$This equation has fractions in each of the three terms. So, first we shall take LCM of the denominators and multiply each term on both sides by the LCM. LCM =$30(x + 4)(x - 7)$. Multiplying each term by this will give us$30(x - 7) - 30(x + 4) = 11(x + 4)(x - 7)$. We need to simplify and rearrange the terms in order to bring it in standard form before we can use the quadratic formula: \begin{eqnarray} 30(x - 7) - 30(x + 4) &=& 11(x + 4)(x - 7)\\ 30x - 210 - 30x - 120 &=& 11(x^2 - 3x - 28)\\ &=& 11x^2 - 33x - 308\\ - 330 &=& 11x^2 - 33x - 308\\ 0 &=& 11x^2 - 33x + 22\\ \end{eqnarray} From the standard form of the equation, we get$a = 11, b = - 33, c = 22$: \begin{eqnarray} x &=& \frac{33 \pm \sqrt{(-33)^2 - 4(11)(22)}}{2(11)}\\ &=& \frac{33 \pm \sqrt{1089 - 968}}{22}\\ &=& \frac{33 \pm \sqrt{121}}{22}\\ &=& \frac{33 \pm 11}{22}\\ x &=& \frac{33 + 11}{22} = 2\\ x &=& \frac{33 - 11}{22} = 1\\ \end{eqnarray} So, solution set is$\{1, 2\}(d)$Here,$a=2, b=-3, c=-2$: \begin{eqnarray} x &=& \frac{3 \pm \sqrt{(-3)^2 - 4(2)(-2)}}{2(2)}\\ &=& \frac{3 \pm \sqrt{9 + 16}}{4}\\ &=& \frac{3 \pm \sqrt{25}}{4}\\ &=& \frac{3 \pm 5}{4}\\ x &=& \frac{3 + 5}{4} = 2\\ x &=& \frac{3 - 5}{4} = \frac{-2}{4} = -\frac{1}{2}\\ \end{eqnarray} So, solution set is$\{-\frac{1}{2}, 2\}$. We can identify the nature of the roots of a quadratic equation using the quadratic formula. The decision maker is the expression inside the radical, in the numerator of the formula:$(b^2 - 4ac)$. For this reason, this radical is called the discriminant . 1. If$(b^2 - 4ac) \gt 0$, the radical gives a real number. So, equation has two distinct real roots. 2. If$(b^2 - 4ac) = 0$, The solution becomes$-\frac{b}{2a}$. So, the equation has only one real root. 3. If$(b^2 - 4ac) \lt 0$, we have a negative number inside the radical, which is not a real number. So, the equation has no real root. Example 5: Without solving each of the following equations, determine the nature of the roots:$(a) t^2 - 5t - 5 = 0; \quad (b) 3x^2 - 4x + 2 = 0; \quad (c) 4x^2 + 12x + 9 = 0$Solution:$(a)$Substituting$a = 1, b = - 5, c = - 5$in the discriminant, we get \begin{eqnarray} b^2 - 4ac &=& (- 5)^2 - 4(1)(- 5)\\ &=& 25 + 20\\ &=& 45\\ \end{eqnarray} As the discriminant is positive, this equation has two distinct real roots.$(b)$Substitutiong$a = 3, b = - 4, c = 2$in the discriminant, we have \begin{eqnarray} b^2 - 4ac &=& (- 4)^2 - 4(3)(2)\\ &=& 16 - 24\\ &=& - 8\\ \end{eqnarray} As the discriminant is negative, this equation has no real roots.$(c)$Substituting$a = 4, b = 12 c = + 9$in the discriminant, we have \begin{eqnarray} b^2 - 4ac &=& (12)^2 - 4(4)(9)\\ &=& 144 - 144\\ &=& 0\\ \end{eqnarray} As the discriminant is zero, this equation has one real root. Example 6: Find the values of the unknown constant$k$in the equation$3x^2 - 6x - k = 0$so that the equation has$(a)$Two distinct real roots;$(b)$One real root;$(c)$No real roots. Solution: The discriminant for this equation is $$(-6)^2 - (4)(3)(-k) = 36 + 12k$$.$(a)$The equation will have two distinct real roots if the discriminant is positive: \begin{eqnarray} 36 + 12k &\gt& 0\\ 12k &\gt& - 36\\ k &\gt& -\frac{36}{12}\\ k &\gt& - 3\\ \end{eqnarray}. So, this equation will have two distinct real roots for$k \gt - 3$.$(b)$The equation will have one real root if the discriminant equals zero: \begin{eqnarray} 36 + 12k &=& 0\\ 12k &=& - 36\\ k &=& - 3\\ \end{eqnarray}. So, the equation will have one real root if$k = - 3$.$(c)$For this equation to have no real roots, the discriminant should be negative: \begin{eqnarray} 36 + 12k &\lt& 0\\ 12k &\lt& - 36\\ k &\lt& - 3\\ \end{eqnarray} So, this equation will have no real roots for any value of$k \lt - 3$. Practice Problems For problems # 1 - 5, solve each given equation by factorization method: Solution set:$\{1, 8\}$Solution set:$\{- 2, \frac{5}{6}\}$Solution set:$\{5, -\frac{3}{4}\}$Solution set:$\{1, \frac{3}{2}\}$. Solution set:$\{\frac{1}{5}, - 1\}$. For problems #6 - 10, solve each equation by taking square root on both sides: Solution set:$ \{- 1, 5\}$Solution set:$\{-\frac{1}{2}, -\frac{9}{2}\}$Solution set:$\{4 + \sqrt{5}, 4 - \sqrt{5}\}$Solution set:$\{1, -\frac{11}{3}\}$Solution set:$\{\frac{- 3 + \sqrt{8}}{2}, \frac{- 3 - \sqrt{8}}{2}\}$For problems # 11-14, solve each equation using 'completing the square' method: Solution set:$\{ - 2, 4\}$No real roots. Solution set:$\{2, 18\}$Solution set:$\{- 3 + \sqrt{11}, - 3 - \sqrt{11}\}$For problems #15 - 19, solve each problem using the quadratic formula: Solution set:$\{- 1, \frac{1}{3}\}$Solution set:$\{\frac{3 + \sqrt{29}}{2}, \frac{3 - \sqrt{29}}{2}\}$Solution set:$\{\frac{3 + 2\sqrt{6}}{5}, \frac{3 - 2\sqrt{6}}{5}\}$Solution set:$\{\frac{- 3 + \sqrt{2}}{2}. \frac{- 3 - \sqrt{2}}{2}\}$. Solution set:$\{\frac{4 + \sqrt{10}}{2}, \frac{4 - \sqrt{10}}{2}\}$Equation has no real roots. This equation has two distinct real roots. This equation has two distinct real roots. This equation will have two distinct real roots for all values of$k \gt - 2$. This equation will have exactly one real root when$k = \pm 1$. This equation will have no real roots for$- 2 \lt k \lt 2$4-2: Problem Solving Involving Quadratic Equations A wide variety of situations in real life can be modeled as quadratic equation. In this section we discuss some of these situations and how to solve real life problems using quadratic equations. Example 1: Ram and Radha together have 45 marbles. Each of them gave away 5 marbles to their sisters respectively. NOw, the product of the marbles they have is 124. How many marbles did each have originally? Solution: Let the number of marbles Ram had originally be$x$. As they had 45 marbles together, number of marbles Radha had originally ls$45 - x$. After they gave away 5 marbles each to their sisters, now, Ram has$x - 5$marbles and Radha has$45 - x - 5 = 40 - x$marbles. According to the given information,$(x - 5)(40 - x) = 124$. We can solve this equation after we first write it in standard form: \begin{eqnarray} (x - 5)(40 - x) &=& 124\\ 40x - x^2 - 200 + 5x &=& 124\\ - x^2 + 45x - 200 &=& 124\\ -x^2 + 45x - 324 &=& 0\\ x^2 - 45x + 320 &=& 0\\ (x - 36)(x - 9) &=& 0\\ x - 36 &=&= 0 \implies x = 36\\ x - 9 &=& 0 \implies x = 9\\ \end{eqnarray} Answer: One of them had 36 marbles and the other had 9 marbles originally. Example 2: Product of two consecutive even integers is 840. Find the two integers. Solution: We note that the difference between two even or two odd consecutive integers is 2. So, if one even integer is$x$, then the next consecutive even integer is$x + 2$. Given that the product of these two is 840. So, we have the equation$x(x + 2) = 840$that models this situation. Rearranging the terms in standard form and solving: \begin{eqnarray} x(x + 2)&=& 840\\ x^2 + 2x &=& 840\\ x^2 + 2x - 840 &=& 0\\ (x - 28)(x - 30) &=& 0\\ x - 28 &=& 0 \implies x = 28\\ x - 30 &=& 0 \implies x = 30\\ \end{eqnarray} Answer: The two even consecutive numbers are 28 and 30. Example 3: A train travels 360 km at a uniform speed.If the speed had been 5 km/h more, it would have taken 1 hour less for the same journey. Find the speed of the train. Solution: For problems involving distance, speed, and time, we use the relation$t = \frac{d}{s}$This problem can be easily modeled by making a table of the two situations: \begin{array}{aaa} distance(km) & speed(Km/h) & time(h)\\ \hline 360 & x & \frac{360}{x}\\ 360 & x + 5 & \frac{360}{x+5}\\ \end{array} According to the given information, with the increased speed, time taken reduces by 1 hour. That is, $$\frac{360}{x+5} = \frac{360}{x} - 1$$. We can multiply each term on both sides by the LCM$x(x+5)$to eliminate the denominator and form a quadratic equation, the solution of which gives us the answer: \begin{eqnarray} 360(x) &=& 360(x + 5) - 1(x)(x + 5)\\ 360x &=& 360x + 1800 - x^2 - 5x\\ x^2 + 5x - 1800 &=& 0\\ (x + 45)(x - 40) &=& 0\\ x + 45 &=& 0 \implies x = - 45\\ x - 40 &=& 0 \implies x = 40\\ \end{eqnarray} We discard the negative value. So, the original speed of the train is$40$km/h. Example 4: A swimming pool 6 m wide and 10 m long is surrounded by a walkway of uniform width. The area of the walkway is equal to the area of the pool. Find the width of the walkway. Solution: This problem is best understood by drawing a diagram to represent the swimming pool and the walkway. The diagram is shown at the bottom. Let the width of the walk way be$x$m. The area of the swimming pool is$6 x 10 = 60 m^2$/ Area of the outer rectangle that includes the pool and the walkway is double this area, as the area of the walkway is given to be equal to the area of the pool. The length and width of the outer rectangle are$6 + 2x$and$10 + 2x$respectively. Setting up the equation and solving: \begin{eqnarray} (6 + 2x)(10 + 2x) &=& 120\\ 60 + 12x + 20x + 4x^2 &=& 120\\ 4x^2 + 32x + 60 - 120 &=& 0\\ 4x^2 + 32x - 60 &=& 0\\ x^2 + 8x - 15 &=& 0\\ x &=& \frac{-8 \pm \sqrt{8^2 - 4(1)(-15)}}{2}\\ &=& \frac{-8 \pm \sqrt{64 + 60}}{2}\\ &=& \frac{-8 \pm \sqrt{124}}{2}\\ &=& \frac{-8 \pm 2\sqrt{31}}{2}\\ &=& -4 \pm \sqrt{31}\\ \end{eqnarray} As$-4 -\sqrt{31}$gives a negative value which is not possible, the width of the walkway is$-4 + \sqrt{31}$m. Example 5: Working alone, Anand takes two more hours to clean the house than Anita does. If they work together, Anand and Anita can clean the house in 2 hours and 24 minutes. How long does Anand take to clean the house if he is working alone Solution: Let$x$be the number of hours Anand takes to clean the house alone. then in one hour, he would clean$\frac{1}{x}$of the house. Given that Anand takes two more hours to clean the house than Anita. So, Anita takes$x - 2$hours to clean the house alone. So,in one hour, Anita would clean$\frac{1}{x-2}$of the house. Both of them together would clean$\frac{1}{x} + \frac{1}{x-2}$of the house in one hour. Given that together, they clean the house in 2 hours and 24 minutes, i.e.,$2\frac{24}{60} = 2\frac{2}{5} = \frac{12}{5}$hours. So, in one hour, they clean$\frac{5}{12}$of the house. This means that $$\frac{1}{x} + \frac{1}{x-2} = \frac{5}{12}$$. Multiplying each term on both sides by the LCM$12x(x-2)$and solving for$x$: \begin{eqnarray} 12(x-2) + 12x &=& 5x(x-2)\\ 12x - 24 + 12x &=& 5x^2 - 10x\\ 5x^2 - 34x + 24 &=& 0\\ x &=& \frac{34 \pm \sqrt{34^2 - 4(5)(24)}}{10}\\ &=& \frac{34 \pm \sqrt{676}}{10}\\ &=& \frac{34 \pm 26}{10}\\ x &=& \frac{34 + 26}{10} \implies x = 6\\ x &=& \frac{34 - 26}{10} \implies x = 0.8\\ \end{eqnarray} If$x = 0.8$,$x-2 = -1.2$, which is negative. So, the correct answer is$x = 6$. That is, Anand takes 6 hours to clean the house alone. Example 6: The numerator of a fraction is 1 less than its denominator. when both numerator and denominator are increased by 2, the fraction is increased by$\frac{1}{12}$. Find the original fraction. Solution: Let the denominator of the original fraction by$x$. Then, its numerator is$x-1$, thus the original fraction is$\frac{x-1}{x}$. When both numerator and the denominator are increased by 2,the fraction becomes$\frac{x-1+2}{x+2} = \frac{x+1}{x+2}$. Given that, $$\frac{x+1}{x+2} = \frac{x-1}{x} + \frac{1}{12}$$. Multiplying each term on both sides by the LCM$12x(x+2)$and solving for$x$, we get: \begin{eqnarray} \frac{x+1}{x+2} &=& \frac{x-1}{x} + \frac{1}{12}\\ 12x(x+1) &=& 12(x+2)(x-1)+x(x+2)\\ 12x^2+12x &=& 12x^2+12x-24+x^2+2x\\ x^2+2x-24 &=& 0\\(x+6)(x-4) &=& 0\\ x+6 &=& 0\implies x=-6\\x-4 &=& 0 \implies x=4\\ \end{eqnarray} Discarding the negative value, we get$x=4$. So, the original fraction is$\frac{3}{4}$. We can verify that$\frac{3}{4}+ \frac{1}{12} = \frac{5}{6}$, which can be obtained by increasing both numerator and denominator by 2. Example 7: Each side of a square is 4 metres. When each side is increased by$x$m, the area of the square is doubled. Find the value of$x$. Solution: Each side of the square is 4m. So, are area is 16sq.m. If each side of the square is increased by$x$m, the length of each side becomes$x+4$m. So, the new area is$(x+4)^2$. This area is given to be twice the area of the original square. Setting up the equation and solving, we get: \begin{eqnarray} (x+4)^2 &=& 32\\ x+4 &=& \pm \sqrt{32}\\ &=& 4\sqrt{2}\\ x &=& -4\pm4\sqrt{2}\\ x &=& -4+4\sqrt{2} \implies x \approx 1.64\\ x &=& -4-4\sqrt{2} \implies x \approx -9.64\\ \end{eqnarray} Discarding the negative value, we have the value of$x = 1.64$m Practice Problems She bought 20 pencils. The integers are 10 and 12 or -10 and -12. Length of the picture is 24 cm and the width is 8 cm. The three integers are 7, 9, and 11. Length is 68 cm and width is 21 cm. Her average speed is 60m km/h Ravi's average speed uphill is 17.8 km/h and his average speed downhill is 52.8 km/h. Then two integers are 15 and 22. The length and width of the rectangle are 8 cm and 6 cm respectively. Solution set:$\{\frac{- 3 + \sqrt{8}}{2}, \frac{- 3 - \sqrt{8}}{2}\}$Solution set:$\{ - 2, 4\}\$

The smaller tap takes 25 hours to fill the tank and the larger tap takes 15 hours.

The walkway is 2m wide.

The original inside diameter of the pipe is approximately 18.94mm.

The dimensions of the field are 50m and 100m.