How do you solve the quadratic $4 x^{2} - 12 x = 16$ $4x^2-12x=16$ using any method?

Question

How do you solve the quadratic $4 x^{2} - 12 x = 16$ $4x^2-12x=16$ using any method?

2 Answers

Impact of this question

11981 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-09-18T09:53:39

$x = 4$ $x=4$
$x = - 1$ $x=-1$

Explanation:

Given -

$4 x^{2} - 12 x = 16$ $4x^2-12x=16$

$4 x^{2} - 12 x - 16 = 0$ $4x^2-12x-16=0$

$4 x^{2} + 4 x - 16 x - 16 = 0$ $4x^2+4x-16x-16=0$

$4 x (x + 1) - 16 (x + 1) = 0$ $4x(x+1)-16(x+1)=0$

$(4 x - 16) (x + 1) = 0$ $(4x-16)(x+1)=0$

$4 x - 16 = 0$ $4x-16=0$

$x = \frac{16}{4} = 4$ $x=16/4=4$

$x + 1 = 0$ $x+1=0$

$x = - 1$ $x=-1$

Answer 2 · 2016-09-18T10:03:42

$x = 4$ $x=4$ or $x = - 1$ $x = -1$

Explanation:

$4 x^{2} - 12 x = 16$ $4x^2-12x=16$

$4 x^{2} - 12 x - 16 = 0 \leftarrow \div 4$ $4x^2 -12x-16 =0" "larr"div4$

$x^{2} - 3 x - 4 = 0 \leftarrow$ $x^2-3x-4=0" "larr$ factorise

Find factors of 4 which subtract to give 3.

$(x - 4) (x + 1) = 0$ $(x-4)(x+1) =0$

If $x - 4 = 0, then x = 4$ $x-4 = 0, " then " x = 4$

If $x + 1 = 0, then x = - 1$ $x+1 = 0, " then " x = -1$

Science

Math

Humanities

... and beyond

How do you solve the quadratic $4 x^{2} - 12 x = 16$ $4x^2-12x=16$ using any method?

2 Answers

Explanation:

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you solve the quadratic 4x^2-12x=164x2−12x=164x^2-12x=16 using any method?

2 Answers

Explanation:

Explanation:

Related questions

Impact of this question

How do you solve the quadratic $4 x^{2} - 12 x = 16$ $4x^2-12x=16$ using any method?