Question

find the positive value of k for which the equation X square + kx + 64 equal to zero and x square - 8 x + K is equal to zero will have real roots​

Answer 1

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Answer 2

Given, $x^{2}+kx=64=0$ and $x^{2}-8x+k=0$will have real roots.

So, for real roots

$b^{2}-4ac\geq 0\\\therefore k^{2}-4\times 64\geq 0 ......(1) and (-8)^{2}-4\times k \geq 0 .....(2)$

From inequality (1), $-16\leq k\leq 16 ....(3)$

From inequality (2),$k\leq 32 .....(4)$

Solving inequalities (3) and (4),$-16 \leq k\leq 16$

But, given k should be positive, $\therefore k > 0$

So, the positive values of k for which the equations will have real roots are: $k\leq 16$

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