Question

Find m, p, and k such that 2x^2-8x+15=m(x+p)^2+k

Answer 1

Answer:

m = 2 ; p = - 2 ; k =7

Step-by-step explanation:

Let 2x^2 - 8x + 15 = D

= > 2x^2 - 8x + 15 = D

= > 2[ x^2 - 4x + 15/2 ] = D

= > x^2 - 4x + (15/2) = (D/2)

= > x^2 - 4x = (D/2) - (15/2)

Adding 4 to both sides :

= > x^2 - 4x + 4 = (D/2) - (15/2) + 4

= > x^2 - 2*2x + 2^2 = (D/2) - (15/2) + 4

= > ( x - 2 )^2 = (D/2) + (-15+8)/2

= > ( x - 2 )^2 = (D/2) - (7)/2

= > ( x - 2 )^2 = (D-7)/2

= > 2( x - 2 )^2 = D - 7

= > 2( x - 2 )^2 + 7 = D

Comparing this with m( x + p )^2 + k :

m = 2 ; p = - 2 ; k = 7

Answer 2

Solution -

In the above question , we have to find values. m, p and k such that it satisfies the given relation -

$\sf{ 2x^2 - 8x + 15 = m { (x + p )}^2 + k }$

Do, let us start solving the above relation -

$\sf{ 2x^2 - 8x + 15 = m { (x + p )}^2 + k } \\ \\ \sf{ => 2x^2 - 8x + 15 = m ( x^2 + 2xp + p^2 ) + k } \\ \\ \sf{ => 2x^2 - 8x + 15 = m x^2 + 2mxp + mp^2 + k } \\ \\ \sf{ => 2x^2 - 8x + 15 = ( m ) x^2 + ( 2mp ) x + ( mp^2 + k } \\ \\ \sf{ Comparing \: the \: coefficents \: - } \\ \\ \sf{ => 2x^2 = mx^2} \\ \\ \sf{ => m = 2 \: ....... [A_{1} ] } \\ \\ \sf{ => 2mp = -8 } \\ \\ \sf{ => 4p = -8 } \\ \\ \sf{ => p = -2 \: ............ [A_{2} ] } \\ \\ \sf{ => mp^2 + k = 15 } \\ \\ \sf{ => 8 + k = 15 } \\ \\ \sf{ => k = 7 \: .. ........ [A_{3} ] }$

Thus the required values of m, p and k are 2 , -2 and 7 respectively ....