Answer :
Answer:
[tex]\boxed{\bf\:k = \dfrac{7}{8} \: } \\ [/tex]
Step-by-step explanation:
Given that, - 5 is a root of the quadratic equation 2x² + 2px - 15 = 0
So, we have
[tex]\sf\: 2 {( - 5)}^{2} + 2p( - 5) - 15 = 0 \\ [/tex]
[tex]\sf\: 50 - 10p - 15 = 0 \\ [/tex]
[tex]\sf\: 35 - 10p = 0 \\ [/tex]
[tex]\sf\: 35 = 10p\\ [/tex]
[tex]\implies\sf\:p = \dfrac{7}{2} \\ [/tex]
Now, Further given that, the quadratic equation p(x² + x) + k = 0 has equal roots.
On comparing with general quadratic equation ax² + bx + c = 0, we get a = p, b = p, c = k
As it is given that, px² + px + k = 0 has equal roots.
[tex]\implies\sf\:Discriminant, D = 0\\[/tex]
[tex]\implies\sf\: {b}^{2} - 4ac = 0 \\ [/tex]
On substituting the values of a, b and c, we get
[tex]\sf\: {p}^{2} - 4pk = 0 \\ [/tex]
[tex]\sf\: p(p - 4k )= 0 \\ [/tex]
[tex]\sf\: p - 4k = 0 \\ [/tex]
[tex]\sf\: k = \dfrac{p}{4} \\ [/tex]
On substituting the value of p, we get
[tex]\sf\: k = \dfrac{7}{2 \times 4} \\ [/tex]
[tex]\implies\sf\:k = \dfrac{7}{8} \\ [/tex]
Hence,
[tex]\implies\sf\:\boxed{\bf\:k = \dfrac{7}{8} \: } \\ [/tex]
Step-by-step explanation:
Find the value of p from the first equation.
[tex]2 {x}^{2} + 2px - 15 = 0 [/tex]
[tex]2 {( - 5)}^{2} + 2p( - 5) - 15 = 0[/tex]
[tex]50 - 10p - 15 = 0[/tex]
[tex]35 - 10p = 0[/tex]
[tex]10p = 35[/tex]
[tex]p = \frac{35}{10} [/tex]
[tex]p = \frac{7}{2} [/tex]
Substitute the value of p into the second equation and solve for k.
[tex]p( {x}^{2} + x) + k = 0 [/tex]
[tex] \frac{7}{2} ( {x}^{2} + x) + k = 0 [/tex]
[tex] \frac{7}{2} {x}^{2} + \frac{7}{2} x + k = 0 [/tex]