QUADRATIC EQUATIONS-Exercise 4.2

Q1. Find the roots of the following quadratic equations by factorisation:

1. \(x^2 – 3x – 10 = 0\)
   
   

   Solution:

   We are given the quadratic equation $$x^{2}-3x-10=0$$ To factorise by splitting the middle term, find two numbers whose product equals \(a\cdot c = 1\times(-10)=-10\) and whose sum equals \(b=-3\)

   The pair \(-5\) and \(2\) works because \((-5)\times 2=-10\) and \((-5)+2=-3\).

   Use these to split the middle term: $$\begin{aligned} x^{2}-3x-10 \\ = x^{2}-5x+2x-10 \end{aligned}$$

   Group and factor: $$ \begin{aligned} &x^{2}-5x+2x-10 \\&=\ x(x-5)+2(x-5) \\ &=\ (x-5)(x+2) \end{aligned} $$

   Set each factor to zero to find the roots: $$\begin{aligned}x-5&=0 \\\Rightarrow x&=5 \\\text{or}\\ x+2&=0 \\\Rightarrow x=&-2\end{aligned}$$

   Hence the roots of the quadratic are \(x=5\) and \(x=-2\).

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2. \(2x^{2}+x-6=0\)
   
   

   Solution:

   Consider the quadratic equation $$2x^{2}+x-6=0$$ To factorise by splitting the middle term, compute \(a\cdot c = 2\times(-6)=-12\) and find two integers whose product is \(-12\) and whose sum is \(b=1\).

   The integers \(4\) and \(-3\) satisfy these conditions because \(4\times(-3)=-12\) and \(4+(-3)=1\).

   Split the middle term using these numbers: $$ \begin{aligned} 2x^{2}+x-6 \\&= 2x^{2}+4x-3x-6 \end{aligned}$$

   Group and factor: $$ \begin{aligned} &2x^{2}+4x-3x-6 \\&=\ 2x(x+2)-3(x+2) \\&=\ (x+2)(2x-3) \end{aligned} $$

   Set each factor equal to zero to find the roots: $$ \begin{aligned} x+2&=0 \\\Rightarrow x&=-2 \\\text{or}\\ 2x-3&=0 \\\Rightarrow x&=\dfrac{3}{2} \end{aligned} $$

   Therefore the roots of the quadratic are \(x=-2\) and \(x=\dfrac{3}{2}\).

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3. \(\sqrt{2}\,x^{2}+7x+5\sqrt{2}=0\)
   
   

   Solution:

   We are given the quadratic equation $$\sqrt{2}\,x^{2}+7x+5\sqrt{2}=0$$ To factorise, compute \(a\cdot c=\sqrt{2}\times 5\sqrt{2}=10\). We need two numbers whose product is \(10\) and whose sum is \(7\); the numbers \(5\) and \(2\) satisfy this.

   Split the middle term using \(5\) and \(2\): $$ \begin{aligned} &\sqrt{2}\,x^{2}+7x+5\sqrt{2}\\ &=\ \sqrt{2}\,x^{2}+5x+2x+5\sqrt{2} \end{aligned} $$ Group and factor: $$ \begin{aligned} &\sqrt{2}\,x^{2}+5x+2x+5\sqrt{2}\\ &=\ (\sqrt{2}\,x^{2}+5x)+(2x+5\sqrt{2}) \end{aligned} $$

   Factor each group: $$ \begin{aligned} &\sqrt{2}\,x^{2}+5x \\ &=\ x(\sqrt{2}\,x+5)\\ \text{and}\\ &2x+5\sqrt{2}\\ &=\ \sqrt{2}\bigl(x+\sqrt{2}\bigr) \end{aligned} $$ Observing the common binomial, we can write the full factorisation as $$ \begin{aligned} &\sqrt{2}\,x^{2}+7x+5\sqrt{2}\\ &=(\sqrt{2}\,x+5)(x+\sqrt{2}) \end{aligned} $$

   Set each factor equal to zero to find the roots: $$ \begin{aligned} \sqrt{2}\,x+5&=0 \\ \Rightarrow x&=-\dfrac{5}{\sqrt{2}}\\ &=-\dfrac{5\sqrt{2}}{2}\\ x+\sqrt{2}&=0 \\ \Rightarrow x&=-\sqrt{2} \end{aligned} $$

   Therefore the roots of the quadratic are $$x=-\sqrt{2}\quad\text{and}\quad x=-\dfrac{5\sqrt{2}}{2}.$$

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4. \(2x^{2}-x+\dfrac{1}{8}=0\)
   
   

   Solution

   Consider the quadratic equation $$2x^{2}-x+\dfrac{1}{8}=0$$ To eliminate the fraction, multiply the entire equation by \(8\), giving $$16x^{2}-8x+1=0.$$ We now factorise by splitting the middle term.

   The product of the coefficient of \(x^{2}\) and the constant term is \(16 \times 1 = 16\). We seek two numbers whose product is \(16\) and whose sum is \(8\); the pair \(4\) and \(4\) satisfies this condition.

   Using these numbers to split the middle term, we write $$ \begin{aligned} &16x^{2}-8x+1 \\ &=\ 16x^{2}-4x-4x+1 \end{aligned} $$

   Grouping the terms and factoring, we get $$ \begin{aligned} &16x^{2}-4x-4x+1 \\ &= 4x(4x-1)-1(4x-1) \\ &= (4x-1)(4x-1) \end{aligned} $$ Thus the quadratic becomes a perfect square.

   Equating the factor to zero, $$ \begin{aligned} 4x-1&=0 \\\Rightarrow x&=\dfrac{1}{4} \end{aligned} $$ Since both factors are identical, the root is repeated.

   Therefore, the equation has a repeated root $$x=\dfrac{1}{4},\ \dfrac{1}{4}$$

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5. \(\;100x^{2}-20x+1=0\)
   
   

   Solution:

   \(\;100x^{2}-20x+1=0\)

   First, observe that the given quadratic expression can be compared with the identity \(\,a^{2}-2ab+b^{2}=(a-b)^{2}\,\).

   [1]

   Rewrite the middle term so that the expression fits this identity. Here, \[100x^{2}=(10x)^{2}\] the term \[-20x\] can be written as \[-2\cdot 10x\cdot 1\] and the constant term is \[1^{2}\] Thus, the equation becomes \[(10x)^{2}-2\cdot 10x\cdot 1+1^{2}=0\] which is of the form
   \[a^{2}-2ab+b^{2}=0\] with \[a=10x\] and \[b=1\]

   [2]

   Using the identity, factorise the left-hand side as \[(10x-1)^{2}=0\] This means \[(10x-1)(10x-1)=0\] so the only solution comes from \[10x-1=0\]

   [3]

   Solving \[10x-1=0\] gives \[10x=1\] hence \[x=\dfrac{1}{10}\] Since the factor \((10x-1)\) is repeated, the quadratic has equal (repeated) roots, both equal to \[\dfrac{1}{10},\ \dfrac{1}{10}\]

Q2. Solve the problems given in Example 1.

1. \(x^2 - 45x - 324 = 0\)
   
   

   Solution:

   Given: \(x^2 - 45x - 324 = 0\)

   To solve by factorisation, split the middle term \(-45x\) into two terms whose coefficients have a product equal to the constant term \(-324\) and sum equal to \(-45\)

   The factors of \(324\) that multiply to \(324\) and add to \(45\) are \(36\) and \(9\), so the required factors for the split are \(-36\) and \(-9\) since \[(-36) \times (-9) = 324\] and \[(-36) + (-9) = -45\]

   Rewrite the equation as \[x^2 - 36x - 9x - 324 = 0\] Now regroup the terms: \[ \begin{aligned} &x^2 - 36 - 9x - 324 \\ \Rightarrow & x(x - 36) - 9(x - 36)= 0\\ \Rightarrow & (x-36)(x-9)=0 \end{aligned} \]

   Set each factor to zero: \[ \begin{aligned} x - 36 &= 0 \\ \Rightarrow x &= 36 \end{aligned} \] or \[\begin{aligned} x - 9 &= 0 \\\Rightarrow x &= 9 \end{aligned} \]

   Thus, the roots are \(x = 36\) and \(x = 9\).

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2. \(x^{2} - 55x + 750 = 0\)
   
   

   Solution:

   Given: \(x^{2} - 55x + 750 = 0\)

   To factorise, we need two numbers whose product is the constant term \(750\) and whose sum is the coefficient of the middle term, \( -55 \).

   The pair of factors of \(750\) that add up to \(55\) are \(25\) and \(30\). Since the middle term is negative, these factors will both be negative: \(-25\) and \(-30\), as \[(-25) + (-30) = -55\] and \[(-25) \cdot (-30) = 750\]

   Rewrite the quadratic as \[ \begin{aligned} &x^{2} - 25x - 30x + 750 = 0\\ \Rightarrow &x(x -25) - 30(x - 25) = 0\\ \Rightarrow &(x - 25)(x - 30) = 0 \end{aligned} \]

   Setting each factor equal to zero, \[ \begin{aligned} x - 25 &= 0 \\ \Rightarrow x &= 25 \end{aligned} \] and \[ \begin{aligned} x - 30 &= 0\\ \Rightarrow x &= 30 \end{aligned} \]

   Therefore, the roots of the quadratic equation are \(x = 25\) and \(x = 30\).

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Q3. Find two numbers whose sum is 27 and product is 182.

Solution:

Given: Find two numbers whose sum is 27 and product is 182.

Let the first number be \(x\) and the second number be \(27 - x\). Their product gives the quadratic equation: \[ \begin{aligned} x(27 - x) &= 182\\ 27x - x^2 &= 182\\ x^2 - 27x + 182 &= 0 \end{aligned} \]

To factorise, find two numbers whose product is 182 and sum is 27

The factors 13 and 14 satisfy this condition since \[13 \times 14 = 182\] and \[13 + 14 = 27\] Rewrite as \[x^2 - 13x - 14x + 182 = 0\]

Regroup terms: \[ \begin{aligned} x(x - 13) - 14(x - 13) &= 0\\ (x - 13)(x -14) &= 0 \end{aligned} \]

Thus, \[ \begin{aligned} x - 13 &= 0 \\ \Rightarrow x &= 13 \end{aligned} \] or \[ \begin{aligned} x - 14 &= 0 \\ \Rightarrow x &= 14 \end{aligned} \] The two numbers are 13 and 14

Verification: \(13 + 14 = 27\) and \(13 \times 14 = 182\)

Q4. Find two consecutive positive integers, sum of whose squares is 365.

Solution:

Given: Find two consecutive positive integers whose squares sum to 365.

Let the first positive integer be \(x\).
Then the second consecutive positive integer is \(x + 1\).
Their squares sum gives: \[ \begin{aligned} x^2 + (x + 1)^2 &= 365\\ \Rightarrow x^2 + (x^2 + 2x + 1) &= 365\\ \Rightarrow 2x^2 + 2x + 1 &= 365\\ \Rightarrow 2x^2 + 2x - 364 &= 0 \end{aligned} \]

Dividing throughout by 2 gives the simpler quadratic: \[x^2 + x - 182 = 0\]

To factorise, find two numbers whose product is \(-182\) and sum is \(+1\). The factors 14 and \(-13\) satisfy this since \[14 \times (-13) = -182\] and \[14 + (-13) = 1\] Rewrite as \[ \begin{aligned} x^2 + 14x - 13x - 182 &= 0\\ \Rightarrow x(x + 14) - 13(x + 14) &= 0\\ \Rightarrow (x + 14)(x - 13) &= 0 \end{aligned} \]

Thus, \[ \begin{aligned} x + 14 &= 0 \\ \Rightarrow x &= -14 \end{aligned} \] or \[ \begin{aligned} x - 13 &= 0 \\ \Rightarrow x &= 13 \end{aligned} \]

Since the numbers must be positive integers, discard \(x = -14\). Therefore, \(x = 13\), so the first number is 13 and the second is \(13 + 1 = 14\)

Verification: \(13^2 + 14^2 = 169 + 196 = 365\)

Q5. The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.

Solution:

Given: Altitude of a right triangle is 7 cm less than its base. Hypotenuse is 13 cm. Find the other two sides.

Let the base be \(x\) cm.
Then the altitude is \(x - 7\) cm.
By Pythagoras theorem, \[ \begin{aligned} x^2 + (x - 7)^2 &= 13^2\\ \Rightarrow x^2 + (x^2 - 14x + 49) &= 169\\ \Rightarrow 2x^2 - 14x + 49 &= 169\\ \Rightarrow 2x^2 - 14x - 120 &= 0 \end{aligned} \]

Dividing by 2 yields \[x^2 - 7x - 60 = 0\]

To factorise, find two numbers whose product is \(-60\) and sum is \(-7\). The factors \(-12\) and \(+5\) satisfy this since \[(-12) \times 5 = -60\] and \[-12 + 5 = -7\] Rewrite as \[ \begin{aligned} x^2 - 12x + 5x - 60 &= 0\\ x(x - 12) + 5(x - 12) &= 0\\ (x - 12)(x + 5) = &0 \end{aligned} \]

Thus, \[ \begin{aligned} x - 12 &= 0 \\ \Rightarrow x &= 12 \end{aligned} \] or \[ \begin{aligned} x + 5 &= 0 \\ \Rightarrow x &= -5 \end{aligned} \]

Since length cannot be negative, \(x = 12\) cm (base). Altitude is \(12 - 7 = 5\) cm.

Verification: \(5^2 + 12^2 = 25 + 144 = 169 = 13^2\).

Q6. A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was ` 90, find the number of articles produced and the cost of each article.

Solution:

Given: Cost of each pottery article is 3 more than twice the number of articles produced. Total cost is ₹90. Find number of articles and cost per article.

Let number of articles produced be \(x\).
Then cost per article is \(2x + 3\) rupees.
Total cost equation: \[ \begin{aligned} x(2x + 3) &= 90\\ 2x^2 + 3x - 90 &= 0\\ \end{aligned} \]

To factorise, consider factors of \(2 \times (-90) = -180\) whose sum is \(+3\). The factors \(15\) and \(-12\) work since \[15 + (-12) = 3\] and \[15 \times (-12) = -180\]

Rewrite: \[ \begin{aligned} 2x^2 + 15x - 12x - 90 &= 0\\ \Rightarrow x(2x + 15) - 6(2x + 15) &= 0\\ \Rightarrow (2x + 15)(x - 6) &= 0 \end{aligned} \]

Thus: \[ \begin{aligned} x - 6 &= 0 \\ \Rightarrow x &= 6 \end{aligned}\] or \[ \begin{aligned} 2x + 15 &= 0 \\ \Rightarrow x &= -\dfrac{15}{2} \end{aligned} \] Since number of articles must be positive integer, \(x = 6\).

Cost per article: \(2(6) + 3 = 15\) rupees. Verification: \(6 \times 15 = 90\).

Thus, 6 articles produced at ₹15 each.