EXERCISE 4.1
Quadratic Equations • 2 Questions
Question 1
Hint available
Check whether the following are : (i) (x + 1)2 = 2(x – 3) (ii) x2 – 2x = (–2) (3 – x) (iii) (x – 2)(x + 1) = (x – 1)(x + 3) (iv) (x – 3)(2x +1) = x(x + 5) (v) (2x – 1)(x – 3) = (x + 5)(x – 1) (vi) x2 + 3x + 1 = (x – 2)2 (vii) (x + 2)3 = 2x (x2 – 1) (viii) x3 – 4x2 – x + 1 = (x – 2)3
Key Idea
Expand both sides of each equation, bring all terms to one side, simplify and factor (or use the quadratic formula) to obtain a polynomial equation. If the resulting polynomial reduces to 0 for all values of \(x\) it is an identity; if it reduces to a non‑zero constant it is a contradiction; otherwise it is a conditional equation whose solutions are obtained by solving the resulting polynomial.
Step-by-Step Solution
i) \((x+1)^2 = 2(x-3)\)
Expand: \(x^2+2x+1 = 2x-6\)
Bring all terms to left: \(x^2+2x+1-2x+6 = 0 \Rightarrow x^2+7=0\)
\(x^2 = -7\) has no real solution ⇒ contradiction.
ii) \(x^2-2x = (-2)(3-x)\)
RHS = \(-6+2x\)
Equation: \(x^2-2x = -6+2x\)
\(x^2-4x+6 = 0\)
Discriminant \(D = 16-24 = -8 < 0\) ⇒ no real solution ⇒ contradiction.
iii) \((x-2)(x+1) = (x-1)(x+3)\)
Expand: \(x^2 - x - 2 = x^2 + 2x - 3\)
Simplify: \(-x-2 = 2x-3\)
\(-3x +1 = 0 \Rightarrow x = \frac{1}{3}\)
Hence a conditional equation with solution \(x=\frac13\).
iv) \((x-3)(2x+1) = x(x+5)\)
Expand: \(2x^2 -5x -3 = x^2 +5x\)
\(x^2 -10x -3 = 0\)
Using quadratic formula: \(x = \frac{10 \pm \sqrt{100+12}}{2}=5 \pm 2\sqrt{7}\)
Conditional equation with two real solutions.
v) \((2x-1)(x-3) = (x+5)(x-1)\)
Expand: \(2x^2 -7x +3 = x^2 +4x -5\)
\(x^2 -11x +8 = 0\)
Solutions: \(x = \frac{11 \pm \sqrt{89}}{2}\)
Conditional equation with two real solutions.
vi) \(x^2 +3x +1 = (x-2)^2\)
RHS = \(x^2 -4x +4\)
Cancel \(x^2\): \(3x +1 = -4x +4\)
\(7x = 3 \Rightarrow x = \frac{3}{7}\)
Conditional equation with single solution.
vii) \((x+2)^3 = 2x(x^2-1)\)
Expand: \(x^3 +6x^2 +12x +8 = 2x^3 -2x\)
Bring to left: \(-x^3 +6x^2 +14x +8 = 0\)
Multiply by \(-1\): \(x^3 -6x^2 -14x -8 = 0\)
The cubic does not factor over rational numbers; it has one real root (approximately \(x \approx 7.9\)) and two complex conjugate roots. Hence it is a conditional equation (real solution \(x \approx 7.9\)).
viii) \(x^3 -4x^2 -x +1 = (x-2)^3\)
RHS = \(x^3 -6x^2 +12x -8\)
Cancel \(x^3\): \(-4x^2 -x +1 = -6x^2 +12x -8\)
Bring all to left: \(2x^2 -13x +9 = 0\)
Solutions: \(x = \frac{13 \pm \sqrt{97}}{2}\)
Conditional equation with two real solutions.
Expand: \(x^2+2x+1 = 2x-6\)
Bring all terms to left: \(x^2+2x+1-2x+6 = 0 \Rightarrow x^2+7=0\)
\(x^2 = -7\) has no real solution ⇒ contradiction.
ii) \(x^2-2x = (-2)(3-x)\)
RHS = \(-6+2x\)
Equation: \(x^2-2x = -6+2x\)
\(x^2-4x+6 = 0\)
Discriminant \(D = 16-24 = -8 < 0\) ⇒ no real solution ⇒ contradiction.
iii) \((x-2)(x+1) = (x-1)(x+3)\)
Expand: \(x^2 - x - 2 = x^2 + 2x - 3\)
Simplify: \(-x-2 = 2x-3\)
\(-3x +1 = 0 \Rightarrow x = \frac{1}{3}\)
Hence a conditional equation with solution \(x=\frac13\).
iv) \((x-3)(2x+1) = x(x+5)\)
Expand: \(2x^2 -5x -3 = x^2 +5x\)
\(x^2 -10x -3 = 0\)
Using quadratic formula: \(x = \frac{10 \pm \sqrt{100+12}}{2}=5 \pm 2\sqrt{7}\)
Conditional equation with two real solutions.
v) \((2x-1)(x-3) = (x+5)(x-1)\)
Expand: \(2x^2 -7x +3 = x^2 +4x -5\)
\(x^2 -11x +8 = 0\)
Solutions: \(x = \frac{11 \pm \sqrt{89}}{2}\)
Conditional equation with two real solutions.
vi) \(x^2 +3x +1 = (x-2)^2\)
RHS = \(x^2 -4x +4\)
Cancel \(x^2\): \(3x +1 = -4x +4\)
\(7x = 3 \Rightarrow x = \frac{3}{7}\)
Conditional equation with single solution.
vii) \((x+2)^3 = 2x(x^2-1)\)
Expand: \(x^3 +6x^2 +12x +8 = 2x^3 -2x\)
Bring to left: \(-x^3 +6x^2 +14x +8 = 0\)
Multiply by \(-1\): \(x^3 -6x^2 -14x -8 = 0\)
The cubic does not factor over rational numbers; it has one real root (approximately \(x \approx 7.9\)) and two complex conjugate roots. Hence it is a conditional equation (real solution \(x \approx 7.9\)).
viii) \(x^3 -4x^2 -x +1 = (x-2)^3\)
RHS = \(x^3 -6x^2 +12x -8\)
Cancel \(x^3\): \(-4x^2 -x +1 = -6x^2 +12x -8\)
Bring all to left: \(2x^2 -13x +9 = 0\)
Solutions: \(x = \frac{13 \pm \sqrt{97}}{2}\)
Conditional equation with two real solutions.
Question 2
Hint available
Represent the following situations in the form of : (i) The area of a rectangular plot is 528 m2. The length of the plot (in metres) is one more than twice its breadth. We need to find the length and breadth of the plot. 42 (ii) The product of two consecutive positive integers is 306. We need to find the integers. (iii) Rohan’s mother is 26 years older than him. The product of their ages (in years) 3 years from now will be 360. We would like to find Rohan’s present age. (iv) A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it would have taken 3 hours more to cover the same distance. We need to find the speed of the train.
Key Idea
Translate each word problem into a quadratic equation of the form $ax^2+bx+c=0$ by introducing a suitable variable, using the given relationships, and then solve the quadratic by factorisation or the quadratic formula. The discriminant $D=b^2-4ac$ determines the nature of the roots; a perfect square discriminant gives integer solutions, which is typical for CBSE problems.
Step-by-Step Solution
### (i) Rectangular plot
1. Let breadth = $x$ metres.
2. Length = $2x+1$ metres (one more than twice the breadth).
3. Area = length \times breadth = $x(2x+1)=528$.
4. Form the quadratic: $$2x^2 + x - 528 = 0.$$
5. Discriminant: $D = 1^2 - 4\times2\times(-528) = 1+4224 = 4225 = 65^2$.
6. Using the quadratic formula: $$x = \frac{-1 \pm 65}{2\times2} = \frac{-1 \pm 65}{4}.$$
7. Positive root: $x = \frac{64}{4}=16$.
8. Hence breadth = $16\,\text{m}$, length = $2(16)+1 = 33\,\text{m}$.
### (ii) Consecutive integers
1. Let the smaller integer be $n$; the next integer is $n+1$.
2. Product condition: $n(n+1)=306$.
3. Quadratic form: $$n^2 + n - 306 = 0.$$
4. Discriminant: $D = 1^2 - 4\times1\times(-306) = 1+1224 = 1225 = 35^2$.
5. Roots: $$n = \frac{-1 \pm 35}{2}.$$
6. Positive integer root: $n = \frac{34}{2}=17$.
7. Therefore the integers are $17$ and $18$.
### (iii) Rohan and his mother
1. Let Rohan’s present age be $r$ years; mother’s age = $r+26$.
2. After 3 years: ages become $r+3$ and $r+29$.
3. Product condition: $$(r+3)(r+29)=360.$$
4. Expand: $$r^2 + 32r + 87 = 360 \;\Rightarrow\; r^2 + 32r - 273 = 0.$$
5. Discriminant: $D = 32^2 - 4\times1\times(-273) = 1024 + 1092 = 2116 = 46^2$.
6. Roots: $$r = \frac{-32 \pm 46}{2}.$$
7. Positive root: $r = \frac{14}{2}=7$.
8. Hence Rohan is $7$ years old (mother is $33$ years old).
### (iv) Train speed
1. Let the uniform speed be $v$ km/h.
2. Time taken at speed $v$: $t = \frac{480}{v}$ hours.
3. If speed were $v-8$, time would be $t' = \frac{480}{v-8}$ and it is given that $t' = t + 3$.
4. Equation: $$\frac{480}{v-8} = \frac{480}{v} + 3.$$
5. Multiply by $v(v-8)$: $$480v = 480(v-8) + 3v(v-8).$$
6. Simplify: $$480v = 480v - 3840 + 3v^2 - 24v.$$
7. Cancel $480v$ and rearrange: $$3v^2 - 24v - 3840 = 0 \;\Rightarrow\; v^2 - 8v - 1280 = 0.$$
8. Discriminant: $D = (-8)^2 - 4\times1\times(-1280) = 64 + 5120 = 5184 = 72^2$.
9. Roots: $$v = \frac{8 \pm 72}{2}.$$
10. Positive feasible speed: $v = \frac{80}{2}=40$ km/h.
Thus the required values are:
- (i) Length = $33\,\text{m}$, Breadth = $16\,\text{m}$
- (ii) Integers = $17$ and $18$
- (iii) Rohan’s present age = $7$ years
- (iv) Speed of the train = $40$ km/h
1. Let breadth = $x$ metres.
2. Length = $2x+1$ metres (one more than twice the breadth).
3. Area = length \times breadth = $x(2x+1)=528$.
4. Form the quadratic: $$2x^2 + x - 528 = 0.$$
5. Discriminant: $D = 1^2 - 4\times2\times(-528) = 1+4224 = 4225 = 65^2$.
6. Using the quadratic formula: $$x = \frac{-1 \pm 65}{2\times2} = \frac{-1 \pm 65}{4}.$$
7. Positive root: $x = \frac{64}{4}=16$.
8. Hence breadth = $16\,\text{m}$, length = $2(16)+1 = 33\,\text{m}$.
### (ii) Consecutive integers
1. Let the smaller integer be $n$; the next integer is $n+1$.
2. Product condition: $n(n+1)=306$.
3. Quadratic form: $$n^2 + n - 306 = 0.$$
4. Discriminant: $D = 1^2 - 4\times1\times(-306) = 1+1224 = 1225 = 35^2$.
5. Roots: $$n = \frac{-1 \pm 35}{2}.$$
6. Positive integer root: $n = \frac{34}{2}=17$.
7. Therefore the integers are $17$ and $18$.
### (iii) Rohan and his mother
1. Let Rohan’s present age be $r$ years; mother’s age = $r+26$.
2. After 3 years: ages become $r+3$ and $r+29$.
3. Product condition: $$(r+3)(r+29)=360.$$
4. Expand: $$r^2 + 32r + 87 = 360 \;\Rightarrow\; r^2 + 32r - 273 = 0.$$
5. Discriminant: $D = 32^2 - 4\times1\times(-273) = 1024 + 1092 = 2116 = 46^2$.
6. Roots: $$r = \frac{-32 \pm 46}{2}.$$
7. Positive root: $r = \frac{14}{2}=7$.
8. Hence Rohan is $7$ years old (mother is $33$ years old).
### (iv) Train speed
1. Let the uniform speed be $v$ km/h.
2. Time taken at speed $v$: $t = \frac{480}{v}$ hours.
3. If speed were $v-8$, time would be $t' = \frac{480}{v-8}$ and it is given that $t' = t + 3$.
4. Equation: $$\frac{480}{v-8} = \frac{480}{v} + 3.$$
5. Multiply by $v(v-8)$: $$480v = 480(v-8) + 3v(v-8).$$
6. Simplify: $$480v = 480v - 3840 + 3v^2 - 24v.$$
7. Cancel $480v$ and rearrange: $$3v^2 - 24v - 3840 = 0 \;\Rightarrow\; v^2 - 8v - 1280 = 0.$$
8. Discriminant: $D = (-8)^2 - 4\times1\times(-1280) = 64 + 5120 = 5184 = 72^2$.
9. Roots: $$v = \frac{8 \pm 72}{2}.$$
10. Positive feasible speed: $v = \frac{80}{2}=40$ km/h.
Thus the required values are:
- (i) Length = $33\,\text{m}$, Breadth = $16\,\text{m}$
- (ii) Integers = $17$ and $18$
- (iii) Rohan’s present age = $7$ years
- (iv) Speed of the train = $40$ km/h