Question

If alpha and beta are zeroes of a quadratic polynomial 2x^2+5x+k, find the value of k succh that (alpha + beta)^2-apha×beta=15.

Answer 1

Solution :-

2x² + 5x + k

α, β are the zeroes of the quadratic polynomial

Comparing 2x² + 5x + k with ax² + bx + c we get,

• a = 2
• b = 5
• c = k

Sum of zeroes = α + β = - b/a = - 5/2

Product of zeroes = αβ = c/a = k/2

Given :-

⇒ (α + β)² - αβ = 15

Substituting the values

⇒ ( - 5/2 )² - k/2 = 15

⇒ ( - 5 )² / 2² - k/2 = 15

⇒ 25/4 - k/2 = 15

⇒ 25/4 - 15 = k/2

⇒ ( 25 - 60 )/4 = k/2

⇒ - 35/4 = k/2

⇒ - 35/4 * 2 = k

⇒ - 35/2 = k

⇒ k = - 35/2

Therefore the value of k is - 35/2.

Answer 2

$\huge{\underline{\underline{\red{\mathfrak{AnSwEr :}}}}}$

Equation is 2x² + 5x + k

Where,

• a = 2
• b = 5
• c = k

As we are given α and β are zeroes. So,

Use formula for sum of zeros

$\large{\boxed{\sf{Sum \: = \: \dfrac{-b}{a}}}} \\ \\ \implies {\sf{\alpha \: + \: \beta \: = \: \dfrac{-5}{2}}}$

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Now use formula for product of zeros :

$\large{\boxed{\sf{Product \: = \: \dfrac{c}{a}}}} \\ \\ \implies {\sf{\alpha \beta \: = \: \dfrac{k}{2}}}$

$\rule{200}{2}$

And it is also given that :

$\sf{(\alpha \: + \: \beta)^2 \: - \: \alpha \beta \: = \: 15} \\ \\ \implies {\sf{15 \: = \: \bigg(\dfrac{-5}{2} \bigg) ^2 \: - \: \dfrac{k}{2}}} \\ \\ \implies {\sf{15 \: = \: \dfrac{25}{4} \: - \: \dfrac{k}{2}}} \\ \\ \implies {\sf{15 \: = \: \dfrac{25 \: - \: 2k}{4}}} \\ \\ \implies {\sf{15 \: \times \: 4 \: = \: 25 \: - \: 2k}} \\ \\ \implies {\sf{60 \: = \: 25 \: - \: 2k}} \\ \\ \implies {\sf{60 \: - \: 25 \: = \: -2k}} \\ \\ \implies {\sf{35 \: = \: -2k}} \\ \\ \implies {\sf{k \: = \: \dfrac{-35}{2}}} \\ \\ \underline{\sf{\therefore \: Value \: of \: k \: is \: \: \dfrac{-35}{2}}}$