If x = - 4 is a root of the equation $x^{2} + 2 x + 4 p = 0$ , find the values of k for which the equation $x^{2} + p x (1 + 3 k) + 7 (3 + 2 k) = 0$ has equal roots.

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Solution

The given quadratic equation is :
$x^{2} + 2 x + 4 p = 0$
Since -4 is a root of the above equation, then it must satisfy it.
Now, (-4)^2 + 2(-4) + 4p = 0
$\Rightarrow 16 - 8 + 4 p = 0$
$\Rightarrow 4 p = - 8$
$\Rightarrow p = - 2$
Now, the other quadratic equation is :
$x^{2} + p x (1 + 3 k) + 7 (3 + 2 k) = 0$
$x^{2} - x (2 + 6 k) + (21 + 14 k) = 0$
Now, a = 1, b = - 2 (2 + 6k); c = 21 + 14k
$N o w, D = b^{2} - 4 a c = 4 + 36 k^{2} + 24 k - 84 - 56 k = 36 k 2 - 32 k - 80$
For equal roots :
$D = 0 \Rightarrow 36 k^{2} - 32 k - 80 = 0 \Rightarrow 4 [9 k^{2} - 8 k - 20] = 0 \Rightarrow 9 k^{2} - 8 k - 20 = 0 \Rightarrow 9 k^{2} - 18 k + 10 k - 20 = 0 \Rightarrow 9 k (k - 2) + 10 (k - 2) = 0 \Rightarrow (9 k + 10) (k - 2) = 0 \Rightarrow 9 k + 10 = 0 o r k - 2 = 0 \Rightarrow k = - \frac{10}{9} o r k = 2$

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