Question

5 and are the zeroes of the quadratic polynomial p(x) = x2 – (k – 6)x + (2k+1);

find the value of k, if + = ​

Answer 1

p(x) = x² - (k - 6)x + (2k + 1)

Given that 5 is one of the two zeroes of p(x),

∴ p(5) = 0

==> (5)² - (k - 6)*5 + 2k + 1 = 0

==> 25 - 5k + 30 + 2k + 1 = 0

==> -3k + 56 = 0

==> 3k = 56

==> k = 56/3

Answer 2

$\huge\underline\mathfrak\red{Answer}$

The value of k, if + =56/3

STEP BY STEP EXPLANATION:-

^2-(k-6)x+(2k+1). (x=5)

(5)^2-(k-6)(5)+(2k+1)=0

25-(5k-30)+2k+1=0

25-5k+30+2k+1=0

-3k+56=0

-3k=-56

k=56/3✔✔

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