Question

4n² +7n=0 what do I do if there is no part c?

Answer 1

Answer:

• Find the coefficients using ax^2 + bx + c = 0

a = 4

b = 7

c = 0

• Make the coefficient equal 1. But how do we do that? Division.

4n² + 7n = 0

4/4n² + 7n/4 = 0/4

• Simplify

= n² + 7/4n = 0

a = 1

b = 7/4

c = 0

• Complete the square using b/2².

b = 7/4

(b/2)² = (7/4 ÷ 2)²

Use fraction rule.

7/4 ÷ 2² = 7/4² / 2²

square the numbers

7/4² / 2² = 49/16 ÷ 4

49/16 ÷ 4 = 49/16 × 1/4

49/16 × 1/4 = 49/64

Add to both sides of the equation using 49/64

b = 7/4

b/2 = 7/4 ÷ 2

simplify

b/2 = 7 ÷ (4 × 2)

simplify(arithmetic)

b/2 = 7/8

n² + 7/4n + 49/64 = 49/64

(n + 7/8)² = 49/64

• Solve for the unknown x

$(n+\frac{7}{8})^2=\frac{49}{64}\\\sqrt{(n+\frac{7}{8})^2}=\sqrt{\frac{49}{64}}$

Next, cancel out the square on the left side.

$n+\frac{7}{8}=+/-\sqrt{\frac{49}{64}}$

Subtract 7/8 from both of the sides.

$n+\frac{7}{8}-\frac{7}{8}=-\frac{7}{8}+/-\sqrt{\frac{49}{64}}$

Simplify the left side.

$n=-\frac{7}{8}+/-\sqrt{\frac{49}{64}}\\n=-\frac{7}{8}+/-\frac{\sqrt{49}}{\sqrt{64}}\\n=-\frac{7}{8}+/-\frac{7}{8}\\n_1=0\\n_2=-\frac{7}{4}$

That wraps it up for this equation. Hope it helped!