Question

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

Answer 1

Answer:

$x=-\dfrac{3}{2}, x=\dfrac{7}{2}$

Step-by-step 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,

$\implies 2x+3=0 \implies x=-\dfrac{3}{2}$

$\implies 2x-7=0 \implies x=\dfrac{7}{2}$

Answer 2

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