Question

If a and ß are the zeros of the polynomial f(x) = 2x² + 5x + k satisfying the relation
a²+ß² +aß =
$\frac{21}{4}$
, then find the value of k for this to be possible.​

Answer 1

${a ^{2} + \beta } ^{2} + 2 \alpha \beta = - 5 \div 2 \\ \alpha \beta + 21 \div 4 = - 5 \div 2 \\ \alpha \beta = - 31 \div 4 \\ k \div 2 = - 31 \div 4 \\ k = - 31 \div 2$

Answer 2

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 ,

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

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

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

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

Sub. values ,

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

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

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

$:\implies \rm 25-2k=21$

$:\implies \rm 25-21 = 2k$

$:\implies \rm 4=2k$

$:\implies \rm 2=k$

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

Value of k is 2 .