Question

Twice the square of the sum of a number and 3 is 98. Find the number.

Answer 1

r={-10, 4}

PREMISES

2(r+3)^2=98

ASSUMPTIONS

Let r=the number

CALCULATIONS

2(r+3)^2=98 (Expand the polynomial and simplify)

2[(r+3)(r+3)]=98

2[r^2+(3r+3r)+9]=98

2(r^2+6r+9)=98

2(r^2+6r+9)/2=98/2

r^2+6r+9=49

r^2+6r+9–49=49–49

r^2+6r+(9–49)=0

r^2+6r+-40=0

r^2+6r-40=0

(r-4)(r+10)=0 (Factor the expanded polynomial)

r-4=0 and r=4 and r+10=0 and r=-10

r=

-10, 4

PROOF

If r={-10, 4}, then the inverse of the mathematical proposition 2(r+3)^2=98 brings

y/2=(r+3)^2, given that (r+3)=-7, 7

98/2=(+/- 7)^2

98/2={7^2, (-7)^2}

49={49, 49} and

49=49 establishes the roots (zeros) r={-10, 4} of the proposition 2(r+3)^2=98

Answer 2

Answer:

46

Step-by-step explanation:

let the no. be x

acc. to question, 2(x+3)=98

x+3 =98/2=49

x=49-3

x=46