Question

For what value of k the quadratic equation kx2 + 11x - 4 = 0 has real roots

Answer 1

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Here is your answer

a = k
b= 11
c = -4

If the equation has real roots than
${b}^{2} - 4ac = 0 \\ (11) {}^{2} - 4 \times k \times ( - 4) = 0 \\ 121 + 16k = 0 \\ 16k = - 121 \\ k = - \frac{121}{16}$



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Answer 2

ANSWER IS

Here
a = k
b= 11
c = -4

Now for Real roots b^2 - 4ac = 0
Put the value we get

$= {11}^{2} - 4 \times k \times ( - 4) = 0\\ = 121 + 16k = 0 \\ = 16k = - 121 \\ = k = \frac{ - 121}{16}$
Thanks