Algebra Ex 3.14-10th Std Maths-Book Back Question And Answer
Question 1.
Write each of the following expression in terms of α + β and αβ
(i) α3β+β3α
Answer:
(ii) 1α2β+1β2α
Answer:
(iii) (3α – 1) (3β – 1)
Answer:
(3α – 1) (3β – 1) = 9αc – 3α – 3β + 1
= 9αβ – 3(α + β) + 1
(iv) α+3β+β+3α
Answer:
Question 2.
The roots of the equation 2x2 – 7x + 5 = 0 are a and p. Find the value of [without solving the equation]
(i) 1α+1β
Answer:
α and α are the roots of the equation 2x2 – 7x + 5 = 0
α + β = 72 ; αβ = 52
(i) 1α+1β = β+ααβ
= 72 + 52 = 72 × 25 = 75
(ii) αβ+βα
Answer:
= (72)2 – 2 × 52 ÷ 52
= 494 – 5 ÷ 52 = 49−204 ÷ 52
= 294 × 25 = 2910
(iii) α+2β+2+β+2α+2
Answer:
Question 3.
The roots of the equation x2 + 6x – 4 = 0 are a, p. Find the quadratic equation whose roots are
(i) α2 and β2
Answer:
α and β are the roots of x2 + 6x – 4 = 0
α + β = -6; αβ = -4
(i) Sum of the roots = α2 + β2
= (α + β)2 – 2αβ
= 36 – 2 – (4) = 36 + 8
= 44
Product of the roots = α2 + β2
= (αβ)2
= (-4)2
= 16
The Quadratic equation is
x2 – (sum of the roots) x + Product of the roots = 0
x2 – (44)x + 16 = 0
x2 – 44x + 16 = 0
(ii) 2α and 2β
Answer:
Sum of the roots = 2α + 2β
= 2β+2ααβ=2(α+β)αβ
= 2(−6)−4=−12−4=3
Product of the roots = 2α×2β=4αβ
= 4−4 = -1
The Quadratic equation is
x2 – (sum of the roots) x + Product of the roots = 0
x2 – 3x – 1 = 0
(iii) α2β and β2α
Answer:
Sum of the roots = α2β + β2α
= αβ (α + β)
= -4 (-6) = 24
Product of the roots = α2β × β2α
= α2β3 = (αβ)3
= (-4)3 = -64
The Quadratic equation is
x2 – (Sum of the roots) x + Product of the roots = 0
x2 – 24x – 64 = 0
Question 4.
If α, β are the roots of 7x2 + ax + 2 = 0 and if β – α = 137 Find the values of a.
Answer:
α and β are the roots of 7x2 + ax + 2 = 0
α + β = −a7; αβ = 27
Given β – α = – 137 ⇒ α – β = 137
Squaring on both sides
(α – β)2 = (137)2
α2 + β2 = 2αβ = 16949
(- a7)2 -4(27) = 16949 ⇒ a249−87=16949
a249 = 22549 ⇒ a2 = 225×4949
a2 = 225 ⇒ a = ± 225−−−√ = ± 15
The value of a = 15 or – 15
Question 5.
If one root of the equation 2y2, – ay + 64 = 0 is twice the other then find the values of a.
Answer:
Let the roots be α and 2α
Here a = 2, b = – a, c = 64
Sum of the roots = – ba
α + 2α = a2
3α = a2
a = 6α …….(1)
Product of the roots = ca
α × 2α = 642 = 2α2 = 32
α2 = 322 = 16
α = 16−−√ = ± 4
Substitute the value of a in (1)
When α = 4
a = 6(4)
a = 24
The Value of a is 24 or -24
When α = -4
a = 6(-4)
a = -24
Question 6.
If one root of the equation 3x2 + kx + 81 = 0 (having real roots) is the square of the other then find k.
Answer:
Let α and α2 be the root of the equation 3x2 + kx + 81
Here a = 3, b = k, c = 81
Sum of the roots = – ba = – k3
α + α2 = –k3
3α + 3α2 = -k ……..(1)
Product of the roots = ca = 813 = 27
α × α2 = 27
α3 = 27 ⇒ α3 = 33
α = 3
Substitute the value of α = 3 in (1)
3(3) + 3(3)2 = -k
9 + 27 = -k ⇒ 36 = – k
∴ k = -36
The value of k = -36
