How do you solve $3 b^{2} - 6 b + 51 = 0$ $3b^2-6b+51=0$ ?

Question

How do you solve $3 b^{2} - 6 b + 51 = 0$ $3b^2-6b+51=0$ ?

1 Answer

Impact of this question

1835 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-29T12:30:24

There are no Real solutions;
as Complex solutions: $b = 1 \pm 4 i$ $b=1+-4i$

Explanation:

Given $3 b^{2} - 6 b + 51 = 0$ $3b^2-6b+51=0$

We could apply the quadratic formula directly to obtain solutions,
but, noticing that all terms are divisible by $3$ $3$ , we can reduce the complexity of calculations by replacing the given equation with
$XXX b^{2} - 2 b + 17 = 0$ $color(white)("XXX")b^2-2b+17=0$

The quadratic formula tells us that for an equation in the form:
$XXX a x^{2} + b x + c = 0$ $color(white)("XXX")color(red)ax^2+color(blue)bx+color(green)c=0$
the solution(s) are given by:
$XXX x = \frac{- b \pm \sqrt{b - 4 a c}}{2 a}$ $color(white)("XXX")x=(-color(blue)b+-sqrt(color(blue)b-4color(red)acolor(green)c))/(2color(red)a)$

The equation we have been given has complicated this situation (probably intentionally) by using $b$ $b$ instead of $x$ $x$ as it's variable. This can cause some confusion with the constant $b$ $color(blue)b$ in the standard form of the quadratic formula.

Let's re-write the quadratic formula with $b$ $b$ as the variable and different letters as the place-holder variables:
$XXX p b^{2} + q b + r = 0$ $color(white)("XXX")color(red)pb^2+color(blue)qb+color(green)r=0$
has solutions:
$XXX b = \frac{- q \pm \sqrt{q^{2} - 4 p r}}{2 p}$ $color(white)("XXX")b=(-color(blue)q+-sqrt(color(blue)q^2-4color(red)pcolor(green)r))/(2color(red)p)$

Using $b^{2} - 2 b + 17 = 0$ $b^2color(blue)(-2)bcolor(green)(+17)=0$ as our equation
we have
$XXX p = 1$ $color(white)("XXX")color(red)p=color(red)1$ (by default)
$XXX q = (- 2)$ $color(white)("XXX")color(blue)q=color(blue)(""(-2))$
$XXX r = 17$ $color(white)("XXX")color(green)r=color(green)(17)$

So the solutions are
$XXX b = \frac{2 \pm \sqrt{{(- 2)}^{2} - 4 \cdot 1 \cdot 17}}{2 \cdot 1}$ $color(white)("XXX")b=(color(blue)2+-sqrt((color(blue)(""-2))^2-4 * color(red) 1 * color(green)(17)))/(2 * color(red)1)$

$XXX = \frac{2 \pm \sqrt{- 64}}{2} = \frac{2 \pm 8 \sqrt{- 1}}{2} = 1 \pm 4 i$ $color(white)("XXX")=(2+-sqrt(-64))/2=(2+-8sqrt(-1))/2 = 1+-4i$

Science

Math

Humanities

... and beyond

How do you solve $3 b^{2} - 6 b + 51 = 0$ $3b^2-6b+51=0$ ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you solve 3b2−6b+51=03b^2-6b+51=0?

1 Answer

Explanation:

Related questions

Impact of this question

How do you solve $3 b^{2} - 6 b + 51 = 0$ $3b^2-6b+51=0$ ?