Extra Practice Questions
Quadratic Equations (R.D. Sharma Level)
1. Solve for x: 4x2 – 4a2x + (a4 – b4) = 0
We use the quadratic formula: x = [-b ± √(b²-4ac)] / 2a.
Here, the variable is x, and a,b,c are coefficients: A=4, B=-4a², C=(a⁴-b⁴).
D = B²-4AC = (-4a²)² – 4(4)(a⁴-b⁴) = 16a⁴ – 16(a⁴-b⁴) = 16a⁴ – 16a⁴ + 16b⁴ = 16b⁴.
x = [ -(-4a²) ± √(16b⁴) ] / (2 × 4) = [ 4a² ± 4b² ] / 8 = [ a² ± b² ] / 2.
Answer: The roots are (a²+b²)/2 and (a²-b²)/2.
2. Find the value of k for which the quadratic equation (k-12)x2 + 2(k-12)x + 2 = 0 has equal roots.
For equal roots, Discriminant D = 0.
Here, a = (k-12), b = 2(k-12), c = 2.
D = b² – 4ac = [2(k-12)]² – 4(k-12)(2) = 0
4(k-12)² – 8(k-12) = 0.
Divide by 4: (k-12)² – 2(k-12) = 0.
Factor out (k-12): (k-12) [ (k-12) – 2 ] = 0
(k-12)(k-14) = 0.
This gives k=12 or k=14. If k=12, the x² term becomes zero, so it is not a quadratic equation. Thus, we must have k=14.
Answer: k = 14.
3. A two-digit number is such that the product of its digits is 18. When 63 is subtracted from the number, the digits interchange their places. Find the number.
Let the digits be x (tens) and y (units). The number is 10x + y.
Given: xy = 18 => y = 18/x.
Also given: (10x + y) – 63 = 10y + x.
9x – 9y = 63 => x – y = 7.
Substitute y: x – (18/x) = 7.
Multiply by x: x² – 18 = 7x => x² – 7x – 18 = 0.
Factorize: (x-9)(x+2) = 0. Since x is a digit, x=9 (x cannot be -2).
y = 18/9 = 2.
Answer: The number is 92.
4. Solve for x: 1x+4 – 1x-7 = 1130, x ≠ -4, 7.
Take LCM on the left side: (x-7) – (x+4)(x+4)(x-7) = 1130
-11x2-3x-28 = 1130
Divide by 11: -1x2-3x-28 = 130
Cross-multiply: -30 = x² – 3x – 28.
x² – 3x + 2 = 0.
Factorize: (x-1)(x-2) = 0.
Answer: The roots are 1 and 2.
5. The hypotenuse of a right-angled triangle is 6 metres more than twice the shortest side. If the third side is 2 metres less than the hypotenuse, find the sides of the triangle.
Let the shortest side be x.
Hypotenuse = 2x + 6.
Third side = (2x + 6) – 2 = 2x + 4.
By Pythagoras theorem: x² + (2x+4)² = (2x+6)².
x² + 4x² + 16x + 16 = 4x² + 24x + 36.
x² – 8x – 20 = 0.
Factorize: (x-10)(x+2) = 0. Since side length must be positive, x=10.
Shortest side = 10 m.
Hypotenuse = 2(10)+6 = 26 m.
Third side = 2(10)+4 = 24 m.
Answer: The sides are 10 m, 24 m, and 26 m.
6. Find the roots of the equation x2 – (√2+1)x + √2 = 0 by factorization.
Expand the middle term:
x2 – √2x – x + √2 = 0
Factor by grouping:
x(x – √2) – 1(x – √2) = 0
(x – 1)(x – √2) = 0
Answer: The roots are 1 and √2.
7. A motor boat whose speed is 18 km/h in still water takes 1 hour more to go 24 km upstream than to return downstream to the same spot. Find the speed of the stream.
Let the speed of the stream be x km/h.
Upstream speed = 18 – x. Downstream speed = 18 + x.
Time upstream (t₁) = 24 / (18-x). Time downstream (t₂) = 24 / (18+x).
Given: t₁ – t₂ = 1.
2418-x – 2418+x = 1
24[ (18+x) – (18-x)(18-x)(18+x) ] = 1
24[ 2x324-x2 ] = 1
48x = 324 – x² => x² + 48x – 324 = 0.
Factorize: (x+54)(x-6)=0. Since speed cannot be negative, x=6.
Answer: The speed of the stream is 6 km/h.
8. If the roots of the equation (a–b)x2 + (b–c)x + (c–a) = 0 are equal, prove that 2a = b + c.
For equal roots, Discriminant D=0.
D = (b–c)2 – 4(a–b)(c–a) = 0
(b2 – 2bc + c2) – 4(ac – a2 – bc + ab) = 0
b2 – 2bc + c2 – 4ac + 4a2 + 4bc – 4ab = 0
4a2 + b2 + c2 – 4ab + 2bc – 4ac = 0
This can be rearranged as: (2a)2 + (-b)2 + (-c)2 + 2(2a)(-b) + 2(-b)(-c) + 2(2a)(-c) = 0.
This is the expansion of (2a – b – c)2 = 0.
Therefore, 2a – b – c = 0, which means 2a = b + c. (Proved)
9. The sum of the areas of two squares is 468 m². If the difference of their perimeters is 24 m, find the sides of the two squares.
Let the sides be x and y. Given x² + y² = 468.
Difference of perimeters: 4x – 4y = 24 => x – y = 6 => x = y + 6.
Substitute x into the first equation:
(y+6)² + y² = 468
y² + 12y + 36 + y² = 468
2y² + 12y – 432 = 0
y² + 6y – 216 = 0.
Factorize: (y+18)(y-12) = 0. Since side length must be positive, y=12.
x = y + 6 = 12 + 6 = 18.
Answer: The sides of the squares are 18 m and 12 m.
10. A dealer sells an article for ₹24 and gains as much percent as the cost price of the article. Find the cost price of the article.
Let the cost price (CP) be ₹x.
Gain percent = x%.
Gain = (x/100) × CP = x²/100.
Selling Price (SP) = CP + Gain.
24 = x + x²/100.
Multiply by 100: 2400 = 100x + x².
x² + 100x – 2400 = 0.
Factorize: x² + 120x – 20x – 2400 = 0.
x(x+120) – 20(x+120) = 0 => (x-20)(x+120) = 0.
Since price cannot be negative, x=20.
Answer: The cost price of the article is ₹20.
