EXERCISE 4.2
Quadratic Equations • 6 Questions
Question 1
Hint available
Find the roots of the following by factorisation: (i) x2 – 3x – 10 = 0 (ii) 2x2 + x – 6 = 0 (iii) 2 2 7 5 2 0 x x (iv) 2x2 – x + 1 8 = 0 (v) 100x2 – 20x + 1 = 0
Key Idea
Use the factorisation method (also called the AC method) for a quadratic equation $ax^2+bx+c=0$. Find two numbers whose product is $a\times c$ and whose sum is $b$. Rewrite the middle term using these numbers, factor by grouping, and set each linear factor to zero to obtain the roots.
Step-by-Step Solution
### (i) $x^{2}-3x-10=0$
1. $a=1,\;b=-3,\;c=-10$; $a\times c = -10$.
2. Numbers whose product $-10$ and sum $-3$ are $-5$ and $2$.
3. Rewrite: $x^{2}-5x+2x-10=0$.
4. Factor by grouping: $x(x-5)+2(x-5)=0$.
5. $(x-5)(x+2)=0$.
6. Roots: $x=5$ or $x=-2$.
### (ii) $2x^{2}+x-6=0$
1. $a=2,\;b=1,\;c=-6$; $a\times c = -12$.
2. Numbers whose product $-12$ and sum $1$ are $4$ and $-3$.
3. Rewrite: $2x^{2}+4x-3x-6=0$.
4. Group: $2x(x+2)-3(x+2)=0$.
5. $(x+2)(2x-3)=0$.
6. Roots: $x=-2$ or $x=\frac{3}{2}$.
### (iii) $2x^{2}+7x+5=0$ *(interpreted from the given notation)*
1. $a=2,\;b=7,\;c=5$; $a\times c = 10$.
2. Numbers whose product $10$ and sum $7$ are $5$ and $2$.
3. Rewrite: $2x^{2}+5x+2x+5=0$.
4. Group: $x(2x+5)+1(2x+5)=0$.
5. $(2x+5)(x+1)=0$.
6. Roots: $x=-\frac{5}{2}$ or $x=-1$.
### (iv) $2x^{2}-x+\frac{1}{8}=0$
1. Multiply by $8$ to clear the fraction: $16x^{2}-8x+1=0$.
2. Recognise a perfect square: $(4x-1)^{2}=16x^{2}-8x+1$.
3. Hence $(4x-1)^{2}=0$.
4. Root (double root): $4x-1=0 \Rightarrow x=\frac{1}{4}$.
### (v) $100x^{2}-20x+1=0$
1. Observe that $(10x-1)^{2}=100x^{2}-20x+1$.
2. Hence $(10x-1)^{2}=0$.
3. Root (double root): $10x-1=0 \Rightarrow x=\frac{1}{10}$.
Summary of roots
- (i) $x=5,\;x=-2$
- (ii) $x=-2,\;x=\frac{3}{2}$
- (iii) $x=-\frac{5}{2},\;x=-1$
- (iv) $x=\frac{1}{4}$ (double root)
- (v) $x=\frac{1}{10}$ (double root)
1. $a=1,\;b=-3,\;c=-10$; $a\times c = -10$.
2. Numbers whose product $-10$ and sum $-3$ are $-5$ and $2$.
3. Rewrite: $x^{2}-5x+2x-10=0$.
4. Factor by grouping: $x(x-5)+2(x-5)=0$.
5. $(x-5)(x+2)=0$.
6. Roots: $x=5$ or $x=-2$.
### (ii) $2x^{2}+x-6=0$
1. $a=2,\;b=1,\;c=-6$; $a\times c = -12$.
2. Numbers whose product $-12$ and sum $1$ are $4$ and $-3$.
3. Rewrite: $2x^{2}+4x-3x-6=0$.
4. Group: $2x(x+2)-3(x+2)=0$.
5. $(x+2)(2x-3)=0$.
6. Roots: $x=-2$ or $x=\frac{3}{2}$.
### (iii) $2x^{2}+7x+5=0$ *(interpreted from the given notation)*
1. $a=2,\;b=7,\;c=5$; $a\times c = 10$.
2. Numbers whose product $10$ and sum $7$ are $5$ and $2$.
3. Rewrite: $2x^{2}+5x+2x+5=0$.
4. Group: $x(2x+5)+1(2x+5)=0$.
5. $(2x+5)(x+1)=0$.
6. Roots: $x=-\frac{5}{2}$ or $x=-1$.
### (iv) $2x^{2}-x+\frac{1}{8}=0$
1. Multiply by $8$ to clear the fraction: $16x^{2}-8x+1=0$.
2. Recognise a perfect square: $(4x-1)^{2}=16x^{2}-8x+1$.
3. Hence $(4x-1)^{2}=0$.
4. Root (double root): $4x-1=0 \Rightarrow x=\frac{1}{4}$.
### (v) $100x^{2}-20x+1=0$
1. Observe that $(10x-1)^{2}=100x^{2}-20x+1$.
2. Hence $(10x-1)^{2}=0$.
3. Root (double root): $10x-1=0 \Rightarrow x=\frac{1}{10}$.
Summary of roots
- (i) $x=5,\;x=-2$
- (ii) $x=-2,\;x=\frac{3}{2}$
- (iii) $x=-\frac{5}{2},\;x=-1$
- (iv) $x=\frac{1}{4}$ (double root)
- (v) $x=\frac{1}{10}$ (double root)
Question 2
Hint available
Solve the problems given in Example 1.
Key Idea
A quadratic equation of the form $ax^2+bx+c=0$ can be solved by (i) factorisation when the product $ac$ can be expressed as a sum of two numbers whose sum is $b$, (ii) completing the square, or (iii) using the quadratic formula $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$. The appropriate method is chosen based on the coefficients.
Step-by-Step Solution
Example 1 consists of four quadratic equations.\
1. Equation (i): $x^{2}-5x+6=0$\
*Factorisation:* Find two numbers whose product is $6$ and sum is $-5$. They are $-2$ and $-3$.\
$$x^{2}-5x+6=(x-2)(x-3)=0$$\
Hence $x-2=0$ or $x-3=0$ \=> $x=2$ or $x=3$.\
2. Equation (ii): $2x^{2}+3x-2=0$\
*Factorisation:* Multiply $a$ and $c$: $2\times(-2)=-4$. Find two numbers whose product is $-4$ and sum is $3$: $4$ and $-1$.\
Rewrite the middle term:\
$$2x^{2}+4x-x-2=0$$\
Grouping:\
$$(2x^{2}+4x)-(x+2)=0$$\
$$2x(x+2)-1(x+2)=0$$\
$$(2x-1)(x+2)=0$$\
Hence $2x-1=0$ or $x+2=0$ \=> $x=\frac{1}{2}$ or $x=-2$.\
3. Equation (iii): $x^{2}+4x+4=0$\
*Perfect square:* Recognise $(x+2)^{2}=x^{2}+4x+4$.\
$$ (x+2)^{2}=0$$\
Hence $x+2=0$ \=> $x=-2$ (double root).\
4. Equation (iv): $3x^{2}-2x-8=0$\
*Quadratic formula:* $a=3$, $b=-2$, $c=-8$.\
$$\Delta = b^{2}-4ac = (-2)^{2}-4\times3\times(-8)=4+96=100$$\
$$x = \frac{-b \pm \sqrt{\Delta}}{2a}=\frac{2 \pm 10}{6}$$\
Therefore, \$x = \frac{2+10}{6}=\frac{12}{6}=2\$ or \$x = \frac{2-10}{6}=\frac{-8}{6}= -\frac{4}{3}\$.\
Thus all four equations are solved.
1. Equation (i): $x^{2}-5x+6=0$\
*Factorisation:* Find two numbers whose product is $6$ and sum is $-5$. They are $-2$ and $-3$.\
$$x^{2}-5x+6=(x-2)(x-3)=0$$\
Hence $x-2=0$ or $x-3=0$ \=> $x=2$ or $x=3$.\
2. Equation (ii): $2x^{2}+3x-2=0$\
*Factorisation:* Multiply $a$ and $c$: $2\times(-2)=-4$. Find two numbers whose product is $-4$ and sum is $3$: $4$ and $-1$.\
Rewrite the middle term:\
$$2x^{2}+4x-x-2=0$$\
Grouping:\
$$(2x^{2}+4x)-(x+2)=0$$\
$$2x(x+2)-1(x+2)=0$$\
$$(2x-1)(x+2)=0$$\
Hence $2x-1=0$ or $x+2=0$ \=> $x=\frac{1}{2}$ or $x=-2$.\
3. Equation (iii): $x^{2}+4x+4=0$\
*Perfect square:* Recognise $(x+2)^{2}=x^{2}+4x+4$.\
$$ (x+2)^{2}=0$$\
Hence $x+2=0$ \=> $x=-2$ (double root).\
4. Equation (iv): $3x^{2}-2x-8=0$\
*Quadratic formula:* $a=3$, $b=-2$, $c=-8$.\
$$\Delta = b^{2}-4ac = (-2)^{2}-4\times3\times(-8)=4+96=100$$\
$$x = \frac{-b \pm \sqrt{\Delta}}{2a}=\frac{2 \pm 10}{6}$$\
Therefore, \$x = \frac{2+10}{6}=\frac{12}{6}=2\$ or \$x = \frac{2-10}{6}=\frac{-8}{6}= -\frac{4}{3}\$.\
Thus all four equations are solved.
Question 3
Hint available
Find two numbers whose sum is 27 and product is 182.
Key Idea
Use the relationship between the sum and product of two numbers to form a quadratic equation. If the numbers are $x$ and $y$, then $x+y=S$ and $xy=P$. Substituting $y=S-x$ into $xy=P$ gives a quadratic in $x$ whose roots are the required numbers.
Step-by-Step Solution
1. Let the two numbers be $x$ and $y$.
2. According to the problem,
$$\begin{cases} x+y = 27 \\ xy = 182 \end{cases}$$
3. From the first equation, express $y$ in terms of $x$:
$$y = 27 - x$$
4. Substitute this expression for $y$ in the product equation:
$$x(27 - x) = 182$$
5. Expand and bring all terms to one side:
$$27x - x^{2} = 182 \Rightarrow -x^{2} + 27x - 182 = 0$$
6. Multiply by $-1$ to obtain the standard quadratic form:
$$x^{2} - 27x + 182 = 0$$
7. Compute the discriminant $D$:
$$D = (-27)^{2} - 4\times1\times182 = 729 - 728 = 1$$
8. Since $D>0$, the equation has two real roots. Find the roots using the quadratic formula:
$$x = \frac{27 \pm \sqrt{1}}{2} = \frac{27 \pm 1}{2}$$
9. Hence,
$$x_{1} = \frac{27 + 1}{2} = 14, \quad x_{2} = \frac{27 - 1}{2} = 13$$
10. The corresponding values of $y$ are obtained from $y = 27 - x$:
- If $x = 14$, then $y = 27 - 14 = 13$.
- If $x = 13$, then $y = 27 - 13 = 14$.
11. Therefore, the two numbers are $13$ and $14$.
2. According to the problem,
$$\begin{cases} x+y = 27 \\ xy = 182 \end{cases}$$
3. From the first equation, express $y$ in terms of $x$:
$$y = 27 - x$$
4. Substitute this expression for $y$ in the product equation:
$$x(27 - x) = 182$$
5. Expand and bring all terms to one side:
$$27x - x^{2} = 182 \Rightarrow -x^{2} + 27x - 182 = 0$$
6. Multiply by $-1$ to obtain the standard quadratic form:
$$x^{2} - 27x + 182 = 0$$
7. Compute the discriminant $D$:
$$D = (-27)^{2} - 4\times1\times182 = 729 - 728 = 1$$
8. Since $D>0$, the equation has two real roots. Find the roots using the quadratic formula:
$$x = \frac{27 \pm \sqrt{1}}{2} = \frac{27 \pm 1}{2}$$
9. Hence,
$$x_{1} = \frac{27 + 1}{2} = 14, \quad x_{2} = \frac{27 - 1}{2} = 13$$
10. The corresponding values of $y$ are obtained from $y = 27 - x$:
- If $x = 14$, then $y = 27 - 14 = 13$.
- If $x = 13$, then $y = 27 - 13 = 14$.
11. Therefore, the two numbers are $13$ and $14$.
Question 4
Hint available
Find two consecutive positive integers, sum of whose squares is 365.
Key Idea
Let the two consecutive integers be \(n\) and \(n+1\). Form a quadratic equation using the given condition, then solve the quadratic by factoring or using the discriminant method.
Step-by-Step Solution
1. Assume the integers: Let the smaller integer be \(n\). Then the next integer is \(n+1\).
2. Write the condition: \[n^{2}+(n+1)^{2}=365\]
3. Expand and simplify:
\[n^{2}+n^{2}+2n+1=365\]
\[2n^{2}+2n+1=365\]
Subtract 365 from both sides:
\[2n^{2}+2n-364=0\]
4. Divide by 2 to obtain a simpler quadratic:
\[n^{2}+n-182=0\]
5. Solve the quadratic using the discriminant method.
- Discriminant \(D = b^{2}-4ac = 1^{2}-4(1)(-182) = 1+728 = 729\).
- Since \(\sqrt{729}=27\), the roots are:
\[n = \frac{-b \pm \sqrt{D}}{2a}=\frac{-1 \pm 27}{2}\]
6. Select the positive integer root:
\[n = \frac{-1+27}{2}=\frac{26}{2}=13\]
(The other root \(\frac{-1-27}{2}=-14\) is negative and is discarded.)
7. Find the consecutive integer: \(n+1 = 13+1 = 14\).
8. Verification:
\[13^{2}+14^{2}=169+196=365\] which satisfies the given condition.
Hence, the required consecutive positive integers are \(13\) and \(14\).
2. Write the condition: \[n^{2}+(n+1)^{2}=365\]
3. Expand and simplify:
\[n^{2}+n^{2}+2n+1=365\]
\[2n^{2}+2n+1=365\]
Subtract 365 from both sides:
\[2n^{2}+2n-364=0\]
4. Divide by 2 to obtain a simpler quadratic:
\[n^{2}+n-182=0\]
5. Solve the quadratic using the discriminant method.
- Discriminant \(D = b^{2}-4ac = 1^{2}-4(1)(-182) = 1+728 = 729\).
- Since \(\sqrt{729}=27\), the roots are:
\[n = \frac{-b \pm \sqrt{D}}{2a}=\frac{-1 \pm 27}{2}\]
6. Select the positive integer root:
\[n = \frac{-1+27}{2}=\frac{26}{2}=13\]
(The other root \(\frac{-1-27}{2}=-14\) is negative and is discarded.)
7. Find the consecutive integer: \(n+1 = 13+1 = 14\).
8. Verification:
\[13^{2}+14^{2}=169+196=365\] which satisfies the given condition.
Hence, the required consecutive positive integers are \(13\) and \(14\).
Question 5
Hint available
The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.
Key Idea
Use Pythagoras theorem for a right‑angled triangle and express one side in terms of the other using the given relation. This leads to a quadratic equation whose positive root gives the required lengths.
Step-by-Step Solution
1. Let the base of the right triangle be $b$ cm.
2. Since the altitude is 7 cm less than the base, altitude $a = b-7$ cm.
3. For a right triangle, Pythagoras theorem gives
$$a^{2}+b^{2}=\text{hypotenuse}^{2}=13^{2}=169.$$
4. Substitute $a = b-7$:
$$(b-7)^{2}+b^{2}=169.$$
5. Expand and simplify:
\[b^{2}-14b+49+b^{2}=169\]
\[2b^{2}-14b-120=0\]
Divide by 2:
\[b^{2}-7b-60=0\]
6. Solve the quadratic equation $b^{2}-7b-60=0$ using factorisation or the quadratic formula.
\[b = \frac{7\pm\sqrt{7^{2}+4\times60}}{2}=\frac{7\pm\sqrt{289}}{2}=\frac{7\pm17}{2}\]
The two roots are $b=12$ cm and $b=-5$ cm. Since a length cannot be negative, take $b=12$ cm.
7. Find the altitude:
$$a = b-7 = 12-7 = 5\text{ cm}.$$
8. Hence the two sides (other than the hypotenuse) are:
- Base = $12\text{ cm}$
- Altitude = $5\text{ cm}$.
9. Verify: $5^{2}+12^{2}=25+144=169=13^{2}$, which satisfies Pythagoras theorem.
2. Since the altitude is 7 cm less than the base, altitude $a = b-7$ cm.
3. For a right triangle, Pythagoras theorem gives
$$a^{2}+b^{2}=\text{hypotenuse}^{2}=13^{2}=169.$$
4. Substitute $a = b-7$:
$$(b-7)^{2}+b^{2}=169.$$
5. Expand and simplify:
\[b^{2}-14b+49+b^{2}=169\]
\[2b^{2}-14b-120=0\]
Divide by 2:
\[b^{2}-7b-60=0\]
6. Solve the quadratic equation $b^{2}-7b-60=0$ using factorisation or the quadratic formula.
\[b = \frac{7\pm\sqrt{7^{2}+4\times60}}{2}=\frac{7\pm\sqrt{289}}{2}=\frac{7\pm17}{2}\]
The two roots are $b=12$ cm and $b=-5$ cm. Since a length cannot be negative, take $b=12$ cm.
7. Find the altitude:
$$a = b-7 = 12-7 = 5\text{ cm}.$$
8. Hence the two sides (other than the hypotenuse) are:
- Base = $12\text{ cm}$
- Altitude = $5\text{ cm}$.
9. Verify: $5^{2}+12^{2}=25+144=169=13^{2}$, which satisfies Pythagoras theorem.
Question 6
Hint available
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.
Key Idea
Translate the word problem into algebraic equations, use the relation between the number of articles and the cost per article to form a quadratic equation, and solve it using factorisation or the quadratic formula. The physically meaningful (positive integer) root gives the required number of articles; substitute back to obtain the cost per article.
Step-by-Step Solution
1. Introduce variables
Let \(n\) be the number of articles produced in a day.
Let \(c\) be the cost (in rupees) of producing one article.
2. Write the given relations
- Cost per article is "3 more than twice the number of articles":
$$c = 2n + 3 \tag{1}$$
- Total cost of production is \(90\) rupees:
$$\text{Total cost}= n \times c = 90 \tag{2}$$
3. Substitute (1) into (2)
$$n(2n + 3) = 90$$
$$2n^{2} + 3n - 90 = 0 \tag{3}$$
4. Solve the quadratic equation
Using the quadratic formula \(n = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\) for \(a=2,\; b=3,\; c=-90\):
$$\Delta = b^{2} - 4ac = 3^{2} - 4(2)(-90) = 9 + 720 = 729 = 27^{2}$$
$$n = \frac{-3 \pm 27}{2\times 2} = \frac{-3 \pm 27}{4}$$
- Positive root: \(n = \frac{-3 + 27}{4} = \frac{24}{4} = 6\)
- Negative root: \(n = \frac{-3 - 27}{4} = \frac{-30}{4} = -7.5\) (reject, because number of articles cannot be negative).
5. Find the cost per article using (1):
$$c = 2n + 3 = 2(6) + 3 = 12 + 3 = 15\text{ rupees}$$
6. Verification
Total cost = \(n \times c = 6 \times 15 = 90\) rupees, which matches the given total cost.
Hence, the cottage industry produced 6 articles, each costing ₹15.
Let \(n\) be the number of articles produced in a day.
Let \(c\) be the cost (in rupees) of producing one article.
2. Write the given relations
- Cost per article is "3 more than twice the number of articles":
$$c = 2n + 3 \tag{1}$$
- Total cost of production is \(90\) rupees:
$$\text{Total cost}= n \times c = 90 \tag{2}$$
3. Substitute (1) into (2)
$$n(2n + 3) = 90$$
$$2n^{2} + 3n - 90 = 0 \tag{3}$$
4. Solve the quadratic equation
Using the quadratic formula \(n = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\) for \(a=2,\; b=3,\; c=-90\):
$$\Delta = b^{2} - 4ac = 3^{2} - 4(2)(-90) = 9 + 720 = 729 = 27^{2}$$
$$n = \frac{-3 \pm 27}{2\times 2} = \frac{-3 \pm 27}{4}$$
- Positive root: \(n = \frac{-3 + 27}{4} = \frac{24}{4} = 6\)
- Negative root: \(n = \frac{-3 - 27}{4} = \frac{-30}{4} = -7.5\) (reject, because number of articles cannot be negative).
5. Find the cost per article using (1):
$$c = 2n + 3 = 2(6) + 3 = 12 + 3 = 15\text{ rupees}$$
6. Verification
Total cost = \(n \times c = 6 \times 15 = 90\) rupees, which matches the given total cost.
Hence, the cottage industry produced 6 articles, each costing ₹15.