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Four times the square of a number is 21 more than eight times the number.

What is the number?
Worth fifty points!

2 Answers

2 votes

Number be n

  • 4n²=21+8n
  • 4n²-8n-21=0
  • 4n²-14n+6n-21=0
  • 2n(2n-7)+3(2n-7)=0
  • (2n+3(2n-7)=0

n=-3/2 and 7/2

User Zatamine
by
7.3k points
6 votes

Answer:


x=-(3)/(2), x=(7)/(2)

Explanation:

Let the unknown number = x


\implies 4x^2=21+8x


\implies 4x^2-8x-21=0

Factor the equation

Find factors of -84 that sum to -8: -14 and 6

Rewrite -8x as the sum of these 2 numbers:


\implies 4x^2-14x+6x-21=0

Factorize the first two terms and the last two terms separately:


\implies 2x(2x-7)+3(2x-7)=0

Factor out the common term
(2x-7):


\implies (2x+3)(2x-7)=0

Therefore,


\implies 2x+3=0 \implies x=-(3)/(2)


\implies 2x-7=0 \implies x=(7)/(2)

User Kent Nguyen
by
8.3k points

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