Very Short Answer on Quadratic Equations Class 11 OP Malhotra Exe-10G ISC Maths Solutions Ch-10. In this article you would learn to solve all types questions on Quadratic Equations. Step by step solutions of latest textbook has been given as latest syllabus. Visit official Website CISCE for detail information about ISC Board Class-11
Quadratic Equation Class 11 OP Malhotra Very Short Answer ISC Math Solutions Ch-10
| Board | ICSE |
| Publications | S Chand |
| Subject | Maths |
| Class | 11th |
| Chapter-10 | Quadratic Equations |
| Writer | OP Malhotra |
| Exe-10(G) | Very Short Answer Type Questions. |
Very Short Answer on Quadratic Equations
OP Malhotra ISC Class 11 Maths Solutions
Que-1: Find the value of α + β and αβ if α, β are the roots of the equation
5x2 − 10x − 7 = 0.
Sol: For ax2 + bx + c = 0
α + β = −b/a
αβ = c/a
Here a = 5, b = −10, c = −7
α + β = −(−10)/5 = 2
αβ = −7/5
Que-2: Find the arithmetic mean of the roots of the equation
x2 − 14x + 48 = 0.
Sol: Arithmetic Mean = (α + β)/2
α + β = −b/a = 14
Arithmetic Mean = 14/2 = 7
Que-3: 3p2 = 5p + 2 and 3q2 = 5q + 2 where p ≠ q, then pq = ….
Sol: 3p2 − 5p − 2 = 0
Thus p and q are roots of:
3x2 − 5x − 2 = 0
Product of roots = c/a
pq = −2/3
Que-4: If α + β = −11 or αβ = 7, form the equation whose roots are α and β.
Sol: Required quadratic equation:
x² − (α + β)x + αβ = 0
= x² − (−11)x + 7 = 0
Equation: x² + 11x + 7 = 0
Que-5: Prove that both the roots of the equation x2 − x − 3 = 0 are irrational.
Sol: Discriminant D = b² − 4ac
= (−1)² − 4(1)(−3)
= 1 + 12 = 13
Since √13 is irrational, therefore both roots are irrational.
Que-6: For what values of m will the equation x2 − 2mx + 7m − 12 = 0 have (i) equal roots (ii) reciprocal roots.
Sol: (i) For equal roots: D = 0
(−2m)² − 4(1)(7m − 12) = 0
4m² − 28m + 48 = 0
m² − 7m + 12 = 0
(m − 3)(m − 4) = 0
m = 3 or m = 4
(ii) Reciprocal roots
For reciprocal roots: c/a = 1
7m − 12 = 1
7m = 13
m = 13/7
Que-7: If one root of 2x2 − 5x + k = 0 be double the other, find the value of k.
Sol: Let roots = α and 2α
α + 2α = 5/2
3α = 5/2
α = 5/6
Product α·2α = k/2
2α² = k/2
2(25/36) = k/2
k = 25/9
Que-8: If x₁ and x₂ are the roots of 3x2 − 2x − 6 = 0, then find the value of x₁2 + x₂2.
Sol: x₁ + x₂ = 2/3
x₁x₂ = −2
Formula: x₁² + x₂² = (x₁ + x₂)² − 2x₁x₂
= (2/3)² − 2(−2)
= 4/9 + 4
= 40/9
Que-9: If the roots of the equation ax2 + bx + c = 0 be in the ratio 3 : 4, show that 12b2 = 49ac.
Sol: Let roots = 3k and 4k
Sum = −b/a
3k + 4k = 7k = −b/a
k = −b / 7a
Product = c/a
3k · 4k = 12k² = c/a
12(b² / 49a²) = c/a
Multiply by 49a²:
12b² = 49ac
Que-10: If α, β be the roots of the equation x2 − x − 1 = 0, determine the value of (i) α2 + β2 (ii) α3 + β3.
Sol: For equation ax² + bx + c = 0
α + β = −b/a , αβ = c/a
Here: α + β = 1 , αβ = −1
(i) α² + β² = (α + β)² − 2αβ
= 1² − 2(−1) = 1 + 2 = 3
(ii) α³ + β³ = (α + β)³ − 3αβ(α + β)
= 1³ − 3(−1)(1) = 1 + 3 = 4
Que-11: Check whether the roots of the equation x2 + 3x + 9 = 0 are imaginary or irrational.
Sol: Discriminant D = b² − 4ac
= 3² − 4(1)(9)
= 9 − 36 = −27
Since D < 0, the roots are Imaginary.
Que-12: Form the quadratic equation whose roots are the cubes of the roots of the equation x2 + 2x + 3 = 0.
Sol: α + β = −2 , αβ = 3
α³ + β³ = (α+β)³ − 3αβ(α+β)
= (−2)³ − 3(3)(−2)
= −8 + 18 = 10
α³β³ = (αβ)³ = 27
Required equation:
x² − (α³+β³)x + α³β³ = 0
x² − 10x + 27 = 0
True or False
Que-13: If α, β are the roots of ax2 + bx + c = 0 and if α − β = 0 then the roots are real and equal.
Sol: True
Because α − β = 0 ⇒ α = β (equal roots).
Que-14: If 1 − i is a root of the equation x2 + ax + b = 0, then find the value of b.
Sol: Since coefficients are real, conjugate root is 1 + i.
Product of roots = αβ = b
(1 − i)(1 + i) = 1 + 1 = 2
Therefore b = 2
Solve
Que-15: x2 + 3 = 0
Sol: x² = −3
x = ± √−3
x = ± i√3
Roots are imaginary.
Que-16: 3x2 − 4x + 20/3 = 0
Sol: Multiply by 3:
9x2 − 12x + 20 = 0
Discriminant D = b² − 4ac
= (−12)² − 4(9)(20)
= 144 − 720 = −576
D < 0 ⇒ Roots are imaginary.
Que-17: x2 + x + 1 = 0
Sol: D = b² − 4ac
= 1² − 4(1)(1)
= −3
D < 0 ⇒ Roots are imaginary.
Que-18: √3 x2 − √2 x + 3√3 = 0
Sol: D = b² − 4ac
= (√2)² − 4(√3)(3√3)
= 2 − 36 = −34
D < 0 ⇒ Roots are imaginary.
Que-19: If x is real, prove that the quadratic expression
(i) (x − 2)(x + 3) + 7 is always positive.
(ii) 4x − 3x2 − 2 is always negative.
Sol: (i) x² + x − 6 + 7
= x² + x + 1
D = 1 − 4 = −3 < 0
Leading coefficient > 0
∴ Expression is always positive.
(ii) −3x² + 4x − 2
D = 16 − 24 = −8 < 0
Leading coefficient negative
∴ Expression is always negative.
Que-20: If the roots of the equation qx2 + 2px + 2q = 0 are real and unequal, prove that the roots of the equation
(p + q)x2 + 2qx + (p − q) = 0 are imaginary.
Sol: For first equation:
D = (2p)² − 4(q)(2q)
= 4(p² − 2q²) > 0
⇒ p² > 2q²
For equation:
(p+q)x² + 2qx + (p−q) = 0
D = (2q)² − 4(p+q)(p−q)
= 4(q² − p² + q²)
= 4(2q² − p²)
Since p² > 2q² ⇒ D < 0
∴ Roots are imaginary.
–: End Quadratic Equations Class 11 OP Malhotra Exe-10G ISC Maths Ch-10 Solutions :–
Return to :- OP Malhotra ISC Class-11 S Chand Publication Maths Solutions
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