Question

If alpha and beta are the zeroes of the polynomial f(x) = 2x2 + 5x + k satisfying the relation alpha square+ Beta square + alpha beta =21/4 then find the value of k for this to be possible.​

Answer 1

➤ Answer

• The value of k = 2.

➤ Given

• α and β are the zeros of the quadratic polynomial 2x² + 5x + k satisfying the relation ${ \alpha }^{2} + { \beta }^{2} + \alpha \beta = \cfrac{21}{4}$

➤ To Find

• The value of k.

➤ Step By Step Explanation

Given that α and β are the zeros of the polynomial 2x² + 5k + k satisfying the relation ${ \alpha }^{2} + { \beta }^{2} + \alpha \beta = \cfrac{21}{4}$. We need to find the value of k.

So let's do it !!

➠ Formula Used

$\dag\underline {\boxed{\purple{\alpha + \beta = \cfrac{ - b}{a}}}} \\ \\ \dag\underline{ \boxed{\pink{\alpha \beta = \cfrac{c}{a}}}}$

➠ Identity Used

$\red{{( \alpha + \beta )}^{2} = { \alpha }^{2} + { \beta }^{2} + 2 \alpha \beta}$

➠ Solution

$\longmapsto\tt { \alpha }^{2} + { \beta }^{2} + \alpha \beta = \cfrac{21}{4} \\ \\\longmapsto\tt {( \alpha + \beta )}^{2} - \alpha \beta = \cfrac{21}{4} \\ \\ \longmapsto\tt {\bigg(\cfrac{ - 5}{2} \bigg)}^{2} - \cfrac{k}{2} = \cfrac{21}{4} \\ \\\longmapsto\tt \cfrac{25}{4} - \cfrac{k}{2} = \cfrac{21}{4} \\ \\ \longmapsto\tt \cfrac{25}{4} - \cfrac{21}{4} = \cfrac{k}{2} \\ \\ \longmapsto\tt \cfrac{4}{4} = \cfrac{k}{2} \\ \\ \cfrac{4 \times 2}{4} = k \\ \\ \longmapsto\tt \cancel\cfrac{8}{4} = k \\ \\ \longmapsto\bf{\green{ 2 = k}}$

➵ Therefore, the value of k = 2.

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