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

Question

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

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Iris 1 month 2021-08-14T06:19:19+00:00 2 Answers 0 views 0

Answers ( )

    0
    2021-08-14T06:20:56+00:00

    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

    0
    2021-08-14T06:20:59+00:00

    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

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