How do you solve by completing the square: $x^{2} + 3 x - 4 = 0$ $x^2+3x-4=0$ ?

Question

How do you solve by completing the square: $x^{2} + 3 x - 4 = 0$ $x^2+3x-4=0$ ?

1 Answer

Impact of this question

11758 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-04-03T17:18:28

x=1 , -4
Take half of the coefficient of x, that would be $\frac{3}{2}$ $3/2$ in this case. Square this number and add it and subtract it as shown below.

$x^{2}$ $x^2$ +3x + $\frac{9}{4}$ $9/4$ - $\frac{9}{4}$ $9/4$ -4=0

${(x + \frac{3}{2})}^{2}$ $(x+3/2)^2$ = $\frac{25}{4}$ $25/4$ . Take square root of both sides to have x+ $\frac{3}{2}$ $3/2$ = $\frac{5}{2}$ $5/2$ or $- \frac{5}{2}$ $-5/2$

x+ $\frac{3}{2}$ $3/2$ = $\frac{5}{2}$ $5/2$ would give x= $\frac{5}{2}$ $5/2$ - $\frac{3}{2}$ $3/2$ = 1 and

X+ $\frac{3}{2}$ $3/2$ = $- \frac{5}{2}$ $-5/2$ would give x= $- \frac{5}{2}$ $-5/2$ $- \frac{3}{2}$ $-3/2$ = -4

Science

Math

Humanities

... and beyond

How do you solve by completing the square: $x^{2} + 3 x - 4 = 0$ $x^2+3x-4=0$ ?

1 Answer

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you solve by completing the square: x^2+3x-4=0x2+3x−4=0x^2+3x-4=0?

1 Answer

Related questions

Impact of this question

How do you solve by completing the square: $x^{2} + 3 x - 4 = 0$ $x^2+3x-4=0$ ?