Question

1. Can the quadratic polynomial x2

+ kx + k have equal zeroes for some odd integer k > 1? Justify.​

Answer 1

Answer:

For a Quadratic Equation to have equal roots, it must satisfy the condition:

b² - 4ac = 0

Given equation is x² + kx + k = 0

a = 1, b = k, x = k

So Substituting in the equation we get,

=> k² - 4 ( 1 ) ( k ) = 0

=> k² - 4k = 0

=> k ( k - 4 ) = 0

=> k = 0 , k = 4

But in the question, it is given that k is greater than 1.

Hence the value of k is 4 if the equation has common roots.

Hence if the value of k = 4, then the equation ( x² + kx + k ) will have equal roots.

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Answer 2

Step-by-step explanation:

${x}^{2} + kx + k = 0$

we have,

$x = \frac{ - k + \sqrt{ {k }^{2} - 4k} }{2}$

for equal roots,

${k}^{2} - 4k = 0$

$k(k - 4) = 0$

so,

$k = 0 \: \: or \: 4$

so , there is no exist odd even integer