EXERCISE 4.3
Quadratic Equations • 5 Questions
Question 1
Hint available
Find the nature of the roots of the following . If the real roots exist, find them: (i) 2x2 – 3x + 5 = 0 (ii) 3x2 – 4 3 x + 4 = 0 (iii) 2x2 – 6x + 3 = 0
Key Idea
For a quadratic equation $ax^2+bx+c=0$, the discriminant $D=b^2-4ac$ determines the nature of its roots:
• $D>0$ – two distinct real roots,
• $D=0$ – equal real roots,
• $D<0$ – no real roots (complex conjugate roots). Real roots, when they exist, are given by $x=\frac{-b\pm\sqrt{D}}{2a}$.
• $D>0$ – two distinct real roots,
• $D=0$ – equal real roots,
• $D<0$ – no real roots (complex conjugate roots). Real roots, when they exist, are given by $x=\frac{-b\pm\sqrt{D}}{2a}$.
Step-by-Step Solution
1. Equation (i): $2x^2-3x+5=0$
• Identify $a=2$, $b=-3$, $c=5$.
• Compute discriminant: $$D = b^2-4ac = (-3)^2-4\times2\times5 = 9-40 = -31.$$
• Since $D<0$, the equation has no real roots; the roots are complex conjugates.
2. Equation (ii): $3x^2-4\sqrt{3}\,x+4=0$
• Identify $a=3$, $b=-4\sqrt{3}$, $c=4$.
• Compute discriminant: $$D = b^2-4ac = (-4\sqrt{3})^2-4\times3\times4 = 16\times3-48 = 48-48 = 0.$$
• Since $D=0$, the equation has equal real roots.
• Root: $$x = \frac{-b}{2a}=\frac{4\sqrt{3}}{2\times3}=\frac{2\sqrt{3}}{3}.$$
• Hence the double root is $x=\frac{2\sqrt{3}}{3}$.
3. Equation (iii): $2x^2-6x+3=0$
• Identify $a=2$, $b=-6$, $c=3$.
• Compute discriminant: $$D = b^2-4ac = (-6)^2-4\times2\times3 = 36-24 = 12.$$
• Since $D>0$, the equation has two distinct real roots.
• Roots: $$x = \frac{-b\pm\sqrt{D}}{2a}=\frac{6\pm\sqrt{12}}{4}=\frac{6\pm2\sqrt{3}}{4}=\frac{3\pm\sqrt{3}}{2}.$$
• Therefore the roots are $x_1=\frac{3+\sqrt{3}}{2}$ and $x_2=\frac{3-\sqrt{3}}{2}$.
• Identify $a=2$, $b=-3$, $c=5$.
• Compute discriminant: $$D = b^2-4ac = (-3)^2-4\times2\times5 = 9-40 = -31.$$
• Since $D<0$, the equation has no real roots; the roots are complex conjugates.
2. Equation (ii): $3x^2-4\sqrt{3}\,x+4=0$
• Identify $a=3$, $b=-4\sqrt{3}$, $c=4$.
• Compute discriminant: $$D = b^2-4ac = (-4\sqrt{3})^2-4\times3\times4 = 16\times3-48 = 48-48 = 0.$$
• Since $D=0$, the equation has equal real roots.
• Root: $$x = \frac{-b}{2a}=\frac{4\sqrt{3}}{2\times3}=\frac{2\sqrt{3}}{3}.$$
• Hence the double root is $x=\frac{2\sqrt{3}}{3}$.
3. Equation (iii): $2x^2-6x+3=0$
• Identify $a=2$, $b=-6$, $c=3$.
• Compute discriminant: $$D = b^2-4ac = (-6)^2-4\times2\times3 = 36-24 = 12.$$
• Since $D>0$, the equation has two distinct real roots.
• Roots: $$x = \frac{-b\pm\sqrt{D}}{2a}=\frac{6\pm\sqrt{12}}{4}=\frac{6\pm2\sqrt{3}}{4}=\frac{3\pm\sqrt{3}}{2}.$$
• Therefore the roots are $x_1=\frac{3+\sqrt{3}}{2}$ and $x_2=\frac{3-\sqrt{3}}{2}$.
Question 2
Hint available
Find the values of k for each of the following , so that they have two equal roots. (i) 2x2 + kx + 3 = 0 (ii) kx (x – 2) + 6 = 0
Key Idea
For a quadratic equation $ax^2+bx+c=0$, the roots are equal (i.e., a repeated root) iff its discriminant $D=b^2-4ac$ is zero.
Step-by-Step Solution
### (i) $\;2x^{2}+kx+3=0$
1. Identify the coefficients: $a=2$, $b=k$, $c=3$.
2. Write the discriminant: \[D = b^{2}-4ac = k^{2}-4\cdot2\cdot3 = k^{2}-24.\]
3. Set $D=0$ for equal roots: \[k^{2}-24=0 \;\Rightarrow\; k^{2}=24.\]
4. Solve for $k$: \[k = \pm\sqrt{24}=\pm 2\sqrt{6}.\]
### (ii) $\;kx(x-2)+6=0$
1. Expand the expression: \[kx(x-2)+6 = kx^{2}-2kx+6 = 0.\]
2. Identify the coefficients: $a=k$, $b=-2k$, $c=6$.
3. Write the discriminant: \[D = b^{2}-4ac = (-2k)^{2}-4\cdot k \cdot 6 = 4k^{2}-24k.\]
4. Set $D=0$: \[4k^{2}-24k = 0 \;\Rightarrow\; 4k(k-6)=0.\]
5. Solve for $k$: \[k=0 \quad\text{or}\quad k=6.\]
6. Since $k=0$ would eliminate the $x^{2}$ term, the equation would no longer be quadratic (it would reduce to $6=0$, which is impossible). Hence discard $k=0$.
7. Therefore, the required value is $k=6$. For $k=6$, the equation becomes $6x^{2}-12x+6=0 \Rightarrow (x-1)^{2}=0$, confirming equal roots.
1. Identify the coefficients: $a=2$, $b=k$, $c=3$.
2. Write the discriminant: \[D = b^{2}-4ac = k^{2}-4\cdot2\cdot3 = k^{2}-24.\]
3. Set $D=0$ for equal roots: \[k^{2}-24=0 \;\Rightarrow\; k^{2}=24.\]
4. Solve for $k$: \[k = \pm\sqrt{24}=\pm 2\sqrt{6}.\]
### (ii) $\;kx(x-2)+6=0$
1. Expand the expression: \[kx(x-2)+6 = kx^{2}-2kx+6 = 0.\]
2. Identify the coefficients: $a=k$, $b=-2k$, $c=6$.
3. Write the discriminant: \[D = b^{2}-4ac = (-2k)^{2}-4\cdot k \cdot 6 = 4k^{2}-24k.\]
4. Set $D=0$: \[4k^{2}-24k = 0 \;\Rightarrow\; 4k(k-6)=0.\]
5. Solve for $k$: \[k=0 \quad\text{or}\quad k=6.\]
6. Since $k=0$ would eliminate the $x^{2}$ term, the equation would no longer be quadratic (it would reduce to $6=0$, which is impossible). Hence discard $k=0$.
7. Therefore, the required value is $k=6$. For $k=6$, the equation becomes $6x^{2}-12x+6=0 \Rightarrow (x-1)^{2}=0$, confirming equal roots.
Question 3
Hint available
Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m2? If so, find its length and breadth.
Key Idea
Translate the geometric condition into an algebraic equation. Let the breadth be \(x\) metres; then the length is \(2x\) metres. Using the area formula \(\text{Area}=\text{length}\times\text{breadth}\) gives a quadratic equation in \(x\). Solve the quadratic and select the positive root because dimensions cannot be negative.
Step-by-Step Solution
1. Let the breadth be \(x\) metres.
Then the length = \(2x\) metres (given).\
2. Write the area condition:
\[\text{Area}=\text{length}\times\text{breadth}=2x\times x = 2x^{2}=800\]\
3. Form the quadratic equation:
\[2x^{2}=800 \;\Rightarrow\; x^{2}=400\]
This can be written as \[x^{2}-400=0\], which is a quadratic in \(x\).\
4. Solve the quadratic:
\[x^{2}=400 \;\Rightarrow\; x=\pm\sqrt{400}=\pm20\]
Since a length cannot be negative, we take \(x=20\) metres.\
5. Find the length:
Length = \(2x = 2\times20 = 40\) metres.\
6. Conclusion:
The required rectangular mango grove can be designed with breadth \(20\) m and length \(40\) m.
Then the length = \(2x\) metres (given).\
2. Write the area condition:
\[\text{Area}=\text{length}\times\text{breadth}=2x\times x = 2x^{2}=800\]\
3. Form the quadratic equation:
\[2x^{2}=800 \;\Rightarrow\; x^{2}=400\]
This can be written as \[x^{2}-400=0\], which is a quadratic in \(x\).\
4. Solve the quadratic:
\[x^{2}=400 \;\Rightarrow\; x=\pm\sqrt{400}=\pm20\]
Since a length cannot be negative, we take \(x=20\) metres.\
5. Find the length:
Length = \(2x = 2\times20 = 40\) metres.\
6. Conclusion:
The required rectangular mango grove can be designed with breadth \(20\) m and length \(40\) m.
Question 4
Hint available
Is the following situation possible? If so, determine their present ages. The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48.
Key Idea
Form a system of equations using the given sum and product conditions, substitute one variable from the linear equation into the quadratic equation, and solve the resulting quadratic by finding its discriminant. If the discriminant is negative, the quadratic has no real roots, indicating the situation is impossible.
Step-by-Step Solution
1. Let the present ages of the two friends be $x$ years and $y$ years.\
2. From the given sum: $$x + y = 20 \[4pt] \Rightarrow y = 20 - x.$$\
3. Four years ago their ages were $x-4$ and $y-4$. The product of those ages was 48: $$ (x-4)(y-4) = 48. $$\
4. Substitute $y = 20 - x$ into the product equation: $$ (x-4)\big[(20 - x)-4\big] = 48 \Rightarrow (x-4)(16 - x) = 48. $$\
5. Expand the left‑hand side: $$ (x-4)(16 - x) = 16x - x^{2} - 64 + 4x = -x^{2} + 20x - 64. $$\
6. Set the expanded expression equal to 48 and bring all terms to one side: $$ -x^{2} + 20x - 64 = 48 \Rightarrow -x^{2} + 20x - 112 = 0. $$\
7. Multiply by $-1$ to obtain the standard quadratic form: $$ x^{2} - 20x + 112 = 0. $$\
8. Compute the discriminant $D$: $$ D = b^{2} - 4ac = (-20)^{2} - 4\cdot1\cdot112 = 400 - 448 = -48. $$\
9. Since $D < 0$, the quadratic equation has no real solutions. Hence there are no real (and therefore no possible integer) ages $x$ and $y$ that satisfy both conditions.\
10. Conclusion: The described situation is not possible.
2. From the given sum: $$x + y = 20 \[4pt] \Rightarrow y = 20 - x.$$\
3. Four years ago their ages were $x-4$ and $y-4$. The product of those ages was 48: $$ (x-4)(y-4) = 48. $$\
4. Substitute $y = 20 - x$ into the product equation: $$ (x-4)\big[(20 - x)-4\big] = 48 \Rightarrow (x-4)(16 - x) = 48. $$\
5. Expand the left‑hand side: $$ (x-4)(16 - x) = 16x - x^{2} - 64 + 4x = -x^{2} + 20x - 64. $$\
6. Set the expanded expression equal to 48 and bring all terms to one side: $$ -x^{2} + 20x - 64 = 48 \Rightarrow -x^{2} + 20x - 112 = 0. $$\
7. Multiply by $-1$ to obtain the standard quadratic form: $$ x^{2} - 20x + 112 = 0. $$\
8. Compute the discriminant $D$: $$ D = b^{2} - 4ac = (-20)^{2} - 4\cdot1\cdot112 = 400 - 448 = -48. $$\
9. Since $D < 0$, the quadratic equation has no real solutions. Hence there are no real (and therefore no possible integer) ages $x$ and $y$ that satisfy both conditions.\
10. Conclusion: The described situation is not possible.
Question 5
Hint available
Is it possible to design a rectangular park of perimeter 80 m and area 400 m2? If so, find its length and breadth.
Key Idea
Use the relations between perimeter and area of a rectangle. Express one side in terms of the other using the perimeter condition, substitute into the area condition, and obtain a quadratic equation. Solve the quadratic using the discriminant method taught in NCERT.
Step-by-Step Solution
1. Let the length be \(l\) metres and the breadth be \(b\) metres.
2. Perimeter condition:
$$2(l+b)=80 \;\Rightarrow\; l+b=40 \quad\text{(Equation 1)}$$
3. Area condition:
$$l\times b = 400 \quad\text{(Equation 2)}$$
4. Express \(b\) from Equation 1:
$$b = 40 - l$$
5. Substitute \(b\) in Equation 2:
$$l(40 - l) = 400$$
6. Form the quadratic equation:
$$40l - l^{2} = 400 \;\Rightarrow\; l^{2} - 40l + 400 = 0$$
7. Identify coefficients:
$$a = 1,\; b = -40,\; c = 400$$
8. Compute the discriminant \(D\):
$$D = b^{2} - 4ac = (-40)^{2} - 4\times1\times400 = 1600 - 1600 = 0$$
9. Since \(D = 0\), the quadratic has equal real roots:
$$l = \frac{-b}{2a} = \frac{40}{2} = 20\;\text{metres}$$
10. Find \(b\) using Equation 1:
$$b = 40 - l = 40 - 20 = 20\;\text{metres}$$
11. Conclusion: Both length and breadth are 20 m, i.e., the rectangle is a square. Hence, such a park can be designed.
Answer: Length = 20 m, Breadth = 20 m.
2. Perimeter condition:
$$2(l+b)=80 \;\Rightarrow\; l+b=40 \quad\text{(Equation 1)}$$
3. Area condition:
$$l\times b = 400 \quad\text{(Equation 2)}$$
4. Express \(b\) from Equation 1:
$$b = 40 - l$$
5. Substitute \(b\) in Equation 2:
$$l(40 - l) = 400$$
6. Form the quadratic equation:
$$40l - l^{2} = 400 \;\Rightarrow\; l^{2} - 40l + 400 = 0$$
7. Identify coefficients:
$$a = 1,\; b = -40,\; c = 400$$
8. Compute the discriminant \(D\):
$$D = b^{2} - 4ac = (-40)^{2} - 4\times1\times400 = 1600 - 1600 = 0$$
9. Since \(D = 0\), the quadratic has equal real roots:
$$l = \frac{-b}{2a} = \frac{40}{2} = 20\;\text{metres}$$
10. Find \(b\) using Equation 1:
$$b = 40 - l = 40 - 20 = 20\;\text{metres}$$
11. Conclusion: Both length and breadth are 20 m, i.e., the rectangle is a square. Hence, such a park can be designed.
Answer: Length = 20 m, Breadth = 20 m.