Question

Can the quadratic polynomial x ^2+kx+k have equal zeroes for some integer K > 1?​

Answer 1

Answer:

$\large\boxed{\sf{k=4}}$

Explanation:

Given that a quadratic equation such that,

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

We have to find the conditon for equal zeroes.

We know that, zeroes are equal only when Descriminant is equal to zero.

Also, Descriminant is given by formula,

$\large\bold{D={b}^{2}-4ac}$

Here, in the given equation,

• a = 1
• b = k
• c = k

Now, we have to equate D = 0

Substituting the values, we get,

$= > {b}^{2} - 4ac = 0 \\ \\ = > {k}^{2} - 4 \times 1 \times k = 0 \\ \\ = > {k}^{2} - 4k = 0 \\ \\ = > k(k - 4) = 0 \\ \\ = > k = 0 \: \: \: and \: \: \: \: \: k = 4$

Here, only k = 4 satisfied the value k > 1.

Hence, required value of k = 4.

Answer 2

$\huge\boxed{\underline{\underline{\mathcal{\red{A}\green{N}\pink{S}\orange{W}\green{E}\blue{R}}}}}$

$\blue{for\:an\:equation\:to\:have\:equal\:zeros\:it\:must\:satisfy\:the\: equation}$

$\large\bold{{b}^{2} - 4ac = 0}$

$\red{here\: a=1}$

$\green{b=k}$

$\pink{c=k}$

$\large{substituting\: the \:values}$

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

$k(k-4)=0$

$\blue{thus \:k\: =\: 0\: or \:k\: =\:4}$

$\bold{\green{thus\: the\: equation\: can\: have\: k \:greater\: than \:1}}$

$\huge{\pink{HOPE\:IT\:HELPS}}$

Mark as th brainliest^_^