Question

Find K , if the sum of the zeroes of the polynomial x2 – (K +6)x + 2 (2 K -1) is half

their product​

Answer 1

Answer:

if a & ß are the zeroes of the quadratic polynomial f(x) = x2 – 3x - 2, then find a quadratic polynomial

Answer 2

Aɴsᴡᴇʀ

we have,

• quadratic polynomial = x² - (k + 6)x + 2(2k - 1)
• sum of zeroes = ½ product of zeroes

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to find

• value of k ?

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firstly compare the polynomial with ax² + bx + c = 0

so, here x² - (k + 6)x + 2(2k - 1)

• coefficient of x², (a) = 1
• coefficient of x, (b) = - (k + 6)
• constant term, (c) = 2(2k - 1)

and we know that,

$\boxed {\bold{sum \: of \: zeroes = \frac{ - coefficient \: of \: x}{coefficient \: of \: {x}^{2} } }}$

$\boxed{ \bold{product \: of \: zeroes = \frac{constant \: term}{coefficient \: of \: {x}^{2} } }}$

now,

$\bold{sum \: of \: zeroes = \frac{ \cancel - \{ \cancel - (k + 6) \} }{1} = k + 6 }$

and

$\bold{product \: of \: zeroes = \frac{2(2k - 1)}{1} =2(2k - 1)}$

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and here it is given that,

sum of zeroes = ½ product of zeroes

$: \implies k + 6 = \frac{1}{ \cancel2} \times \cancel2(2k - 1) \\ \\ : \implies \: k + 6 = 2k - 1 \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ : \implies 2k - k = 6 + 1 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ : \implies \boxed{ \pink{ k = 7} }\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

so, value of k is 7