Question

If -5 is a root of the quadratic equation 2x square + px - 15 = 0 and the quadratic equation p(x square + x) + k = 0 has equal roots, find the value of k.

Answer 1

Answer:

k = 63/4

Step-by-step explanation:

In any quadratic equation ax^2 + bx + c,

Sum of roots = -b/a

Product of roots = c/a

Say the second root of the first equation and the cequal root in second equation are m and n respectively.

In the first equation,

(-5)(m) = -15/2 => m = 3/2

-5 + m = -p/2 => -5 + 3/2 = -p/2 => p = 7

In the second equation,

(m)(m) = k/p => 3x3/2x2 = k/7 => k = 63/4

Answer 2

Answer:

$2 {x}^{2} + px - 15 = 0 \\ = > x = \frac{ - p( + - ) \sqrt{ {p}^{2} - 4 \times 2 \times ( - 15)} }{2 \times 2} \\ = > x = \frac{ - p( + - ) \sqrt{ {p}^{2} + 120 } }{4}$

either,

$x = \frac{ - p + \sqrt{ {p}^{2} + 120 } }{4} \\ = > - 5 = \frac{ - p + \sqrt{ {p {}^{2} + 120} } }{4} \\ = > - p + \sqrt{ {p}^{2} + 120} = - 20 \\ = > p - \sqrt{ {p}^{2} + 120 } = 20 \\ = > \sqrt{ {p}^{2} + 120} = p - 20 \\ = > p {}^{2} + 120 = {p}^{2} - 40p + 400 \\ = > 4 0p = 280 \\ = > p = 7$

or,

$x = \frac{ - p - \sqrt{ {p}^{2} + 120 } }{4} \\ = > - 5 = \frac{ - p - \sqrt{ {p}^{2} + 120} }{4} \\ = > - 20 = - p - \sqrt{ {p}^{2} + 120} \\ = > \sqrt{ {p}^{2} + 12 0 } = 20 - p \\ = > {p}^{2} + 120 = 400 - 40p + {p}^{2} \\ = > 40p = 280 \\ = > p = 7$

so in both the cases p=7

now,

$p {x}^{2} + x + k = 0 \\ = > 7( { - 5})^{2} + ( - 5) + k = 0 \\ = > 7 \times 25 - 5 + k = 0 \\ = > 175 - 5 + k = 0 \\ = > 170 + k = 0 \\ = > k = - 170$

k=170

Huff...mah hands are tireddddd......(〒﹏〒)