Question

0.7. If a , ß be the heroes of the polynomial 2x² +5x+k
such that (x + 83² - dB = 21, then K= ?
6) - 3 (©) - 2 (d) 2
x+2y + 5 = 0 and-3a-6y +1
one​

Answer 1

Answer:

Question :

If α and β are the zeros of the polynomial p(x) = 2x² + 5x + k satisfying satisfying the relation α²+ β²+αβ = ²¹/₄ , then find the value of k

Solution :

On comparing given polynomial 2x²+5x+k with ax²+bx+c , we get ,

➙ a = 2 , b = 5 , c = k

Sum of zeroes , α + β = - ᵇ/ₐ

⇒ α + β = - ⁵/₂ \pink{\bigstar}★

Product of zeroes , αβ = ᶜ/ₐ

⇒ αβ = ᵏ/₂ \green{\bigstar}★

Given that ,

\rm \alpha ^2 + \beta ^2 + \alpha \beta = \dfrac{21}{4}α

2

+β

2

+αβ=

4

21

\begin{gathered}\bullet\ \; \sf \red{(x+y)^2=x^2+y^2+2xy}\\\\ \bullet\ \; \sf \red{x^2+y^2= (x+y^2)-2xy}\end{gathered}

∙ (x+y)

2

=x

2

+y

2

+2xy

∙ x

2

+y

2

=(x+y

2

)−2xy

:\implies \rm ( \alpha + \beta )^2-2 \alpha \beta + \alpha \beta = \dfrac{21}{4}:⟹(α+β)

2

−2αβ+αβ=

4

21

:\implies \rm ( \alpha + \beta )^2- \alpha \beta = \dfrac{21}{4}:⟹(α+β)

2

−αβ=

4

21

Sub. values ,

:\implies \rm \bigg( - \dfrac{5}{2} \bigg)^2 - \dfrac{k}{2} = \dfrac{21}{4}:⟹(−

2

5

)

2

−

2

k

=

4

21

:\implies \rm \dfrac{25}{4}- \dfrac{k}{2} = \dfrac{21}{4}:⟹

4

25

−

2

k

=

4

21

:\implies \rm \dfrac{25-2k}{4} = \dfrac{21}{4}:⟹

4

25−2k

=

4

21

:\implies \rm 25-2k=21:⟹25−2k=21

:\implies \rm 25-21 = 2k:⟹25−21=2k

:\implies \rm 4=2k:⟹4=2k

:\implies \rm 2=k:⟹2=k

:\implies \rm k=2\ \; \blue{\bigstar}:⟹k=2 ★

Value of k is 2 .

Step-by-step explanation:

sorry no I can't explain