Question

p of x is equal to x cube plus x square plus root five x plus root five then find p of minus root five​

Answer 1

Answer:

Since −5 is a root of the equation 2x  

2

+px−15=0. Therefore,

2(−5)  

2

−5p−15=0⇒p=7

Putting p=7 in p(x  

2

+x)+k=0 we get

7x  

2

+7x+k=0

Here ,a=7,b=7;c=k

This equation will have equal roots, if

Discriminant =b  

2

−4ac=0 ⇒49−4×7×k=0⇒k=  

28

49

​

⇒k=  

4

7

​

Step-by-step explanation:

Answer 2

Answer:

Answer:  The answer is 47.

Step-by-step explanation:  We are given that

p=\dfrac{3-\sqrt5}{3+\sqrt5},~~q=\dfrac{3+\sqrt5}{3-\sqrt5}.p=3+53−5,  q=3−53+5.

We are to find the value of p² + q².

We have

\begin{gathered}p^2+q^2\\\\=\left(\dfrac{3-\sqrt5}{3+\sqrt5}\right)^2+\left(\dfrac{3+\sqrt5}{3-\sqrt5}\right)^2\\\\\\=\dfrac{(3-\sqrt5)^4+(3+\sqrt5)^4}{(9-5)^2}\\\\\\=\dfrac{(14-6\sqrt5)^2+(14+6\sqrt5)^2}{16}\\\\\\=\dfrac{196-168\sqrt5+180+196+168\sqrt5+180}{16}\\\\\\=\dfrac{752}{16}\\\\=47.\end{gathered}p2+q2=(3+53−5)2+(3−53+5)2=(9−5)2(3−5)4+(3+5)4=16(14−65)2+(14+65)2=16196−1685+180+196+1685+180=16752=47.

Thus, the required value is 47.