Question

Solve the quadratic equation by completing the square
4x^2+24x+1=0
I get stuck when I get to the
(X-3)^2= -1/4
Please help? I also know that you need to add 9 to the -1/4...I'm so confused

Answer 1

\( 4x^2 + 24x + 1 = 0\)

Divide both sides by 4:
\( x^2 + 6x + \frac{1}{4} = 0 \)

Subtract \( \frac{1}{4} \) from both sides:
\( x^2 + 6x ~~~~~~~~~~= -\frac{1}{4} \)

Add \( (\frac{b}{2})^2 \) to both sides:
\( x^2 + 6x + 9= -\frac{1}{4} + 9\)

\( (x + 3)^2 = -\frac{1}{4} + \frac{36}{4} \)

\( (x + 3)^2 = \frac{35}{4} \)

Answer 2

How did you get )x - 3)^2? It's (x + 3)^2.

Answer 3

@JennyMae18 Do you follow what I did?

Answer 4

Oh..I'm sorry. I mistyped. It's -24x not plus...

Answer 5

It's the same steps though, right?

Answer 6

\[4x ^{2}+24x+1=0\]
\[x ^{2}+6x=-\frac{ 1 }{ 4 }\]
adding both sides (6/2 )^2 i.e 9
\[x ^{2}+6x+9=\frac{ -1 }{4}+9=\frac{ 35 }{ 4 }\]
\[\left( x+3 \right)^{2}=\left( \frac{ \sqrt{35} }{2 } \right)^{2}\]
\[x+3=\pm \frac{ \sqrt{35} }{ 2 }\]
\[x=-3+\frac{ \sqrt{35} }{ 2 },x=-3-\frac{ \sqrt{35} }{2}\]

Answer 7

yes

Answer 8

\(4x^2 - 24x + 1 = 0\)

Divide both sides by 4:
\( x^2 - 6x + \frac{1}{4} = 0 \)

Subtract \(\frac{1}{4}\) from both sides:
\( x^2 - 6x ~~~~~~~~~~= -\frac{1}{4} \)

Add \( \frac{b}{2})^2\) to both sides:
\( x^2 - 6x + 9= -\frac{1}{4} + 9\)

\( (x - 3)^2 = -\frac{1}{4} + \frac{36}{4} \)

\( (x - 3)^2 = \frac{35}{4} \)

Answer 9

\( (x - 3)^2 = \frac{35}{4} \)

\( x - 3 = \pm \sqrt{ \frac{35}{4} }\)

\( x = 3 \pm \frac{ \sqrt{35} }{2} \)