Question

solve both questions ..

Answer 1

HELLO DEAR,

------------------------(1)---------------------------

given that:-

(x=3)

now put this value in the Equation

we get,

x²-x+k=0

=> 3²-3+k=0

=> 9-3+k=0

=> 6+k=0

=> k= - 6

--------------------and----------------------

given that
Equation are EQUAL roots

hence,
D=0

x²-k(2x+k+x)+p=0

=> x²-k²-3kx +p

=> x²-3kx +(p-k²)=0

a=1

b=-3k

c=(p-k²)

$d = \sqrt{ {b}^{2} - 4ac} \\ \\ d = \sqrt{ {( - 3k)}^{2} - 4 \times 1 \times (p - {k}^{2} )} = 0 \\ \\ = 9 {k}^{2} - 4p + 4 {k}^{2} = 0 \\ \\ = 9 {( - 6)}^{2} - 4p + 4 {( - 6)}^{2} = 0 \\ \\ = 324 - 4p + 144 = 0 \\ \\ = 468 = 4p \\ \\ p = \frac{468}{4} \\ \\ p = 117$

-------------(2)--------------

let the no. are x,(x+1),(x+2)

according to question

x²+(x+1)(x+2)=46

=> x²+x²+2x+x+2=46

=> 2x²+3x-44=0

=> 2x²+11x-8x-44=0

=> x(2x-11) -4(2x-11)=0

=> (x-4)(2x-11)=0

=> x=4

----------------or--------------------

x=11/2

hence,the other no. will be,

[ 4 , 5 , 6 ]

I HOPE ITS HELP YOU DEAR,
THANKS

Answer 2

Answer:

(1)

2

+(1)−2=0

so(x−1)isazeroofthispolynomial∣becausex−1=0sox=1∣

similarly(x+2)isalsoazero

let(x+1)beaand(x+2)beβ

so

x−1

1

−

x+2

1

=

(x−1)(x+2)

x+2−(x−1)

=

(x−1)(x+2)

3

(2)--------------

let the no. are x,(x+1),(x+2)

according to question

x²+(x+1)(x+2)=46

=> x²+x²+2x+x+2=46

=> 2x²+3x-44=0

=> 2x²+11x-8x-44=0

=> x(2x-11) -4(2x-11)=0

=> (x-4)(2x-11)=0

=> x=4

----------------or--------------------

x=11/2

hence,the other no. will be,

[ 4 , 5 , 6 ]

hope it help u