Q1. λx²₊2x+λ=0 have sum and product of roots equal then λ=?
Q2. If α and β are roots of the equation x²+6x+λ=0 and 3α+2β=-20 then λ=?
Q3. If α and β are the roots of the equation 2x²-3x+4=0 then find the equation whose roots are α² and β²
*Please give detailed solution*
Answers
Answer:
Sum of the roots of the equation 3x
2
+2x+1=0
Sum of the roots α+β=−
3
2
Product of the roots αβ=
3
1
Sum of the roots of unknown equation α+
β
1
+β+
α
1
=(α+β)+(
α
1
+
β
1
)
=(α+β)+(
αβ
α+β
)
=−
3
2
+
3
1
−
3
2
=−
3
2
−2⇒
3
−8
Product of the roots (α+
β
1
)(β+
α
1
)
=αβ+α.
α
1
+β.
β
1
+
αβ
1
Substituting αβ=
3
1
=
3
1
+2+
3
1
1
=
3
16
The general form of quadratic equation is x
2
−(sum of the roots)x+Product of the roots=0
Substituting the respective values, we get
x
2
−(
3
−8
)x+
3
16
=0
Multiplying by 3 on both sides, we get
3x
2
+8x+16=0
Answer:
Step-by-step explanation:
Sure, here are the detailed solutions to the three questions:
Q1. λx²+2x+λ=0 have sum and product of roots equal then λ=?
Let α and β be the roots of the equation λx²+2x+λ=0. Then we have:
α + β = -b/a = -2/λ
αβ = c/a = λ
We need to find λ such that the sum and product of the roots are equal. That is:
α + β = αβ
Substituting the expressions for α and β, we get:
-2/λ = λ
=> λ² = -2
Since λ is a real number, this equation has no solution. Therefore, there is no value of λ for which the sum and product of the roots of the equation λx²+2x+λ=0 are equal.
Q2. If α and β are roots of the equation x²+6x+λ=0 and 3α+2β=-20 then λ=?
Let α and β be the roots of the equation x²+6x+λ=0. Then we have:
α + β = -b/a = -6/1 = -6
αβ = c/a = λ/1 = λ
We are given that 3α+2β=-20. Substituting the expressions for α and β, we get:
3α + 2β = 3(α + β) - α - 6β = 3(-6) - α - 6β = -18 - α - 6β = -20
Simplifying, we get:
α + 6β = 2
Substituting the expression for α in terms of β from the equation αβ = λ, we get:
β² + 6β + λ = 0
Since α and β are the roots of the equation x²+6x+λ=0, we know that the discriminant of this equation is non-negative. That is:
b² - 4ac >= 0
Substituting the values of a, b, and c, we get:
6² - 4(1)(λ) >= 0
=> 36 - 4λ >= 0
=> λ <= 9
Substituting λ = 9 in the equation β² + 6β + λ = 0, we get:
β² + 6β + 9 = 0
=> (β + 3)² = 0
Therefore, β = -3 and α = -6 - β = -3. Substituting these values in the equation x²+6x+λ=0, we get:
(-3)² + 6(-3) + λ = 0
=> λ = 0
Therefore, the value of λ that satisfies the given conditions is λ = 0.
Q3. If α and β are the roots of the equation 2x²-3x+4=0 then find the equation whose roots are α² and β²
Let α and β be the roots of the equation 2x²-3x+4=0. Then we have:
α + β = -b/a = 3/2
αβ = c/a = 2
We need to find the equation whose roots are α² and β². Let's use the fact that if r is a root of the equation f(x) = 0, then r² is a root of the equation f(sqrt(x)) = 0.
Substituting sqrt(α²) for x, we get:
2(sqrt(α²))