🟢 EXERCISE 4.1
Q1. Check whether the following are quadratic equations:
(i) x2+1=0x^2 + 1 = 0x2+1=0
Solution:
Highest power of x is 2
∴ It is a quadratic equation
(ii) 2x2−5x+1=02x^2 – 5x + 1 = 02×2−5x+1=0
Solution:
Standard form ax2+bx+c=0ax^2 + bx + c = 0ax2+bx+c=0
∴ It is a quadratic equation
(iii) x3−2x+1=0x^3 – 2x + 1 = 0x3−2x+1=0
Solution:
Highest power of x is 3
∴ Not a quadratic equation
(iv) x2−3x=−4x^2 – 3x = -4x2−3x=−4
Solution:
Rewriting:x2−3x+4=0x^2 – 3x + 4 = 0x2−3x+4=0
∴ It is a quadratic equation
Q2. Represent the following situations mathematically:
(i) The area of a rectangular plot is 528 sq m. The length is one more than twice the breadth.
Solution:
Let breadth = xxx m
Length = 2x+12x + 12x+1
Area = length × breadthx(2x+1)=528x(2x + 1) = 528x(2x+1)=528 2x2+x−528=02x^2 + x – 528 = 02×2+x−528=0
🟢 EXERCISE 4.2
(Solve by Factorisation)
Q1. Find the roots of: x2−3x−10=0x^2 – 3x – 10 = 0x2−3x−10=0
Solution:
Split middle term:x2−5x+2x−10=0x^2 – 5x + 2x – 10 = 0x2−5x+2x−10=0 (x−5)(x+2)=0(x – 5)(x + 2) = 0(x−5)(x+2)=0 x=5, −2x = 5,\ -2x=5, −2
Q2. Solve: 2×2+7x+3=02x^2 + 7x + 3 = 02×2+7x+3=0
Solution:
Split middle term:2x2+6x+x+3=02x^2 + 6x + x + 3 = 02×2+6x+x+3=0 (2x+1)(x+3)=0(2x + 1)(x + 3) = 0(2x+1)(x+3)=0 x=−12, −3x = -\frac{1}{2},\ -3x=−21, −3
🟢 EXERCISE 4.3
(Solve by Quadratic Formula)
Q1. Solve: x2−4x+1=0x^2 – 4x + 1 = 0x2−4x+1=0
Solution:
Given:
- a=1a = 1a=1, b=−4b = -4b=−4, c=1c = 1c=1
x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}x=2a−b±b2−4ac x=4±16−42x = \frac{4 \pm \sqrt{16 – 4}}{2}x=24±16−4 x=4±122x = \frac{4 \pm \sqrt{12}}{2}x=24±12 x=2±3x = 2 \pm \sqrt{3}x=2±3
Q2. Find roots of: 3×2−5x−2=03x^2 – 5x – 2 = 03×2−5x−2=0
Solution:a=3, b=−5, c=−2a = 3,\ b = -5,\ c = -2a=3, b=−5, c=−2 x=5±25+246x = \frac{5 \pm \sqrt{25 + 24}}{6}x=65±25+24 x=5±76x = \frac{5 \pm 7}{6}x=65±7 x=2, −13x = 2,\ -\frac{1}{3}x=2, −31
🟢 EXERCISE 4.4
(Nature of Roots)
Q1. Find the nature of roots of 2×2−4x+3=02x^2 – 4x + 3 = 02×2−4x+3=0
Solution:D=b2−4acD = b^2 – 4acD=b2−4ac D=(−4)2−4(2)(3)D = (-4)^2 – 4(2)(3)D=(−4)2−4(2)(3) D=16−24=−8D = 16 – 24 = -8D=16−24=−8
Since D<0D < 0D<0,
No real roots
🟢 EXERCISE 4.5
(Word Problems)
Q1. Find two consecutive positive integers whose product is 306.
Solution:
Let numbers be xxx and x+1x + 1x+1x(x+1)=306x(x + 1) = 306x(x+1)=306 x2+x−306=0x^2 + x – 306 = 0x2+x−306=0
Using quadratic formula:x=−1±1+12242x = \frac{-1 \pm \sqrt{1 + 1224}}{2}x=2−1±1+1224 x=−1±352x = \frac{-1 \pm 35}{2}x=2−1±35 x=17 (positive)x = 17\ (\text{positive})x=17 (positive)
Numbers are 17 and 18
⭐ IMPORTANT NCERT EXAM TIPS
✔ Always write Given / Required
✔ Show discriminant calculation clearly
✔ Use quadratic formula when unsure
✔ Final answer must be boxed
✔ Choose only valid values in word problems