Chapter 4: Quadratic Equations
Key Formulas:
• Standard form: ax²+bx+c=0
• Quadratic formula: x=(-b±√(b²-4ac))/2a
• Discriminant D=b²-4ac: D>0 two distinct roots, D=0 equal roots, D<0 no real roots
• Sum of roots = -b/a, Product of roots = c/a
Exercise 4.1
Q1. Check if quadratic: (i) (x+1)²=2(x-3) (ii) x²-2x=(-2)(3-x) (iii) (x-2)(x+1)=(x-1)(x+3)
Solution(i) x²+2x+1=2x-6 → x²+7=0 → Quadratic (Yes)
(ii) x²-2x=-6+2x → x²-4x+6=0 → Quadratic (Yes)
(iii) x²-x-2=x²+2x-3 → -3x+1=0 → Not quadratic (linear)
Exercise 4.2
Q1. Find roots by factorisation:
(i) x²-3x-10=0 (ii) 2x²+x-6=0 (iii) √2x²+7x+5√2=0 (iv) 2x²-x+1/8=0
Solution
(i) x²-3x-10=0
x²-5x+2x-10=0
(x-5)(x+2)=0
∴ x=5 or x=-2
(ii) 2x²+x-6=0
2x²+4x-3x-6=0
2x(x+2)-3(x+2)=0
(2x-3)(x+2)=0
∴ x=3/2 or x=-2
(iii) √2x²+7x+5√2=0
√2x²+2x+5x+5√2=0
√2x(x+√2)+5(x+√2)=0
(√2x+5)(x+√2)=0
∴ x=-5/√2 or x=-√2
(iv) 2x²-x+1/8=0
Multiply by 8: 16x²-8x+1=0
(4x-1)²=0
∴ x=1/4 (repeated root)
Q4. Sum of squares of two consecutive positive integers is 365. Find them.
Solutionx²+(x+1)²=365
x²+x²+2x+1=365
2x²+2x-364=0
x²+x-182=0
(x+14)(x-13)=0
x=13 (positive)
∴ Integers are 13 and 14
Q5. Altitude of right triangle is 7cm less than base, hypotenuse 13cm. Find sides.
SolutionBase=x, altitude=x-7
x²+(x-7)²=13²
x²+x²-14x+49=169
2x²-14x-120=0
x²-7x-60=0
(x-12)(x+5)=0 → x=12
Base=12cm, Altitude=5cm
∴ Sides are 12cm and 5cm
Q6. Cost per article is 3 more than twice number of articles. Total cost Rs 90. Find number and cost.
Solutionx(2x+3)=90
2x²+3x-90=0
2x²+15x-12x-90=0
x(2x+15)-6(2x+15)=0
(x-6)(2x+15)=0 → x=6
Cost=2(6)+3=15
∴ Articles=6, Cost=Rs 15 each
Exercise 4.3
Q1. Solve by completing the square: (i) 2x²+x-4=0 (ii) 4x²+4√3x+3=0 (iii) 2x²+x+4=0
Solution
(i)
x²+x/2=2
x²+x/2+1/16=2+1/16
(x+1/4)²=33/16
x=(-1±√33)/4
(ii)
x²+√3x+3/4=0
(x+√3/2)²=0
∴ x=-√3/2 (equal roots)
(iii)
D=1-4(2)(4)=1-32=-31<0
∴ No real roots
Q5. Find nature of roots: (i) 2x²-3x+5=0 (ii) 3x²-4√3x+4=0 (iii) 2x²-6x+3=0
Solution(i) D=9-40=-31<0 → No real roots
(ii) D=48-48=0 → Equal roots: x=4√3/6=2/√3
(iii) D=36-24=12>0 → Two distinct: x=(6±2√3)/4=(3±√3)/2
Exercise 4.4
Q2. Find k for equal roots: (i) 2x²+kx+3=0 (ii) kx(x-2)+6=0
Solution(i) D=k²-24=0 → k²=24 → k=±2√6
(ii) kx²-2kx+6=0
D=4k²-24k=0 → 4k(k-6)=0
k≠0 so k=6
Q3. Can a rectangular grove have length twice breadth and area 800m²?
SolutionLet breadth=x, length=2x
2x×x=800 → 2x²=800 → x²=400 → x=20
∴ Yes. Length=40m, Breadth=20m
Q4. Sum of ages of two friends is 20. Four years ago product was 48. Is this possible?
Solution(x-4)(20-x-4)=48
(x-4)(16-x)=48
-x²+20x-64=48
x²-20x+112=0
D=400-448=-48<0
∴ Not possible (no real roots)
Q5. Can a park have perimeter 80m and area 400m²?
Solutionl+b=40, l×b=400
b(40-b)=400
b²-40b+400=0
(b-20)²=0 → b=20, l=20
∴ Yes, it is a square of side 20m