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Answer 1 · 2018-01-18T07:36:50

See below

Explanation:

The quadratic formula is $x = \frac{- b \pm \sqrt{D}}{2 a}$ $x=(-b+-sqrtD)/(2a)$
Here $D = b^{2} - 4 a c$ $D = b^2 - 4ac$
Only to need to put the values in the formula.
a = 6
b = 5
c = -6

$x = \frac{- 5 \pm \sqrt{5^{2} - 4 (6) (- 6)}}{2 \cdot 6}$ $x = [-5+-sqrt (5^2-4(6)(-6))]/(2*6)$
$x = \frac{- 5 \pm \sqrt{25 + 144}}{12}$ $x = [-5+-sqrt(25 + 144)]/12$
$x = \frac{- 5 \pm \sqrt{169}}{12}$ $x = [-5+-sqrt169]/12$
$x = \frac{- 5 \pm (13)}{12}$ $x = [-5+-(13)]/12$
So x is either,
$\frac{- 5 - 13}{12}$ $(-5-13)/12$
= $- \frac{18}{12}$ $-18/12$
= $- \frac{3}{2}$ $-3/2$
Or
$\frac{- 5 + 13}{12}$ $(-5+13)/12$
= $\frac{8}{12}$ $8/12$
= $\frac{2}{3}$ $2/3$
Hope it helps you

Answer 2 · 2018-01-18T07:44:40

See explanation.

Explanation:

1) $f (x) = 6 x^{2} + 5 x - 6$ $f(x)=6x^2+5x-6$
$= 6 x^{2} + 9 x - 4 x - 6$ $=6x^2+9x-4x-6$
$= 3 x (2 x + 3) - 2 (2 x + 3)$ $=3x(2x+3)-2(2x+3)$
$= (2 x + 3) (3 x - 2)$ $=(2x+3)(3x-2)$

That's it for part1

2)
$f (x) = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $f(x)=(-b+- sqrt(b^2-4ac))/(2a)$
Here, a=6, b=5, c=-6
Plugging in the values, the roots of the equation will be:
$\frac{- 5 \pm \sqrt{5^{2} - 4 \cdot 6 \cdot (- 6)}}{2 \cdot 6}$ $(-5+- sqrt(5^2-4*6*(-6)))/(2*6$
Simplify the equation, and the roots will be
$\frac{- 5 \pm \sqrt{169}}{12}$ $(-5+- sqrt169)/12$
$= \frac{- 5 + \sqrt{169}}{12} or \frac{- 5 - \sqrt{169}}{12}$ $=(-5+sqrt169)/12 or (-5-sqrt169)/12$
$= \frac{- 5 + 13}{12} or \frac{- 5 - 13}{12}$ $=(-5+13)/12 or (-5-13)/12$
$= \frac{8}{12} or - \frac{18}{12}$ $=8/12 or -18/12$
$= \frac{2}{3} or - \frac{3}{2}$ $=2/3 or -3/2$

therefore, the equation will be:

$(x - \frac{2}{3}) (x + \frac{3}{2}) = 0$ $(x-2/3)(x+3/2)=0$

Thus, your final equation will be:
$(2 x + 3) (3 x - 2)$ $(2x+3)(3x-2)$

$T h a n k s .$ $Thanks.$
Hope you got it.

Answer 3 · 2018-01-18T08:05:57

Factoring Method

$f (x) = 6 x^{2} + 5 x - 6 = (3 x - 2) (2 x + 3)$ $color(blue)(f(x) = 6x^2+5x-6=(3x-2)(2x+3)$

Quadratic Formula

$x = \frac{2}{3}, x = - \frac{3}{2}$ $color(blue)(x = 2/3, x= -3/2$

Explanation:

Given:

$f (x) = 6 x^{2} + 5 x - 6$ $color(green)(f(x) = 6x^2+5x-6$

The Standard Form of a Quadratic Equation:

$y = f (x) = a x^{2} + b x + c = 0$ $color(red)(y = f(x) = ax^2+bx+c = 0$

From our problem:

$a = 6; b = 5; and c = - 6$ $a = 6; b = 5; and c = -6$

$M e t h o d .1$ $color(brown)(Method.1)" "$ Factoring Method

Using the Standard Form

$y = f (x) = a x^{2} + b x + c$ $y = f(x) = ax^2+bx+c$

we find $u$ $color(blue)u$ and $v$ $color(blue)v$ such that

$u \cdot v = a \cdot c and u + v = b$ $color(green)(u *v = a*c and u + v = b$

Then we need to group them as shown below:

$a x^{2} + u x + v x + c$ $ax^2 + ux + vx + c$

We have

$f (x) = 6 x^{2} + 5 x - 6 = 0$ $color(green)(f(x) = 6x^2+5x-6=0$

we find $u$ $color(blue)u$ and $v$ $color(blue)v$ as:

$u = [- 4] and v = [9]$ $color(green)(u = [-4] and v = [9]$

So, the middle term $5 x$ $color(blue)(5x)$ can be written as $[- 4 x + 9 x]$ $color(blue)([-4x+9x]$

We can now write our $f (x)$ $f(x)$ as

$f (x) = 6 x^{2} - 4 x + 9 x - 6 = 0$ $color(green)(f(x) = 6x^2-4x+9x-6=0$

$\Rightarrow 6 x^{2} - 4 x + 9 x - 6 = 0$ $rArr 6x^2-4x+9x-6=0$

$\Rightarrow 2 x (3 x - 2) + 3 (3 x - 2) = 0$ $rArr 2x(3x-2)+3(3x-2)=0$

$\Rightarrow (3 x - 2) (2 x + 3) = 0$ $rArr (3x-2)(2x+3)=0$

We get

$(3 x - 2) = 0, (2 x + 3) = 0$ $(3x-2) = 0, (2x+3) = 0$

$3 x - 2 \Rightarrow 3 x = 2$ $3x-2 rArr 3x = 2$ hence $x = \frac{2}{3}$ $x= 2/3$

$2 x + 3 = 0 \Rightarrow 2 x = - 3$ $2x+3 = 0 rArr 2x = -3$ hence $x = - \frac{3}{2}$ $x = -3/2$

Hence, $x = \frac{2}{3}, x = - \frac{3}{2}$ $color(blue)(x=2/3, x = -3/2)$

$M e t h o d .2$ $color(brown)(Method.2)" "$ Using Quadratic Formula

Quadratic Formula is given by

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $color(blue)(x = [-b +- sqrt(b^2 - 4ac)]/(2a)$

From our problem:

$a = 6; b = 5; and c = - 6$ $a = 6; b = 5; and c = -6$

Substituting these values of $a, b and c$ $a,b and c$ in our formula

$x = \frac{- 5 \pm \sqrt{5^{2} - 4 \cdot 6 \cdot (- 6)}}{2 \cdot 6}$ $x = (-5+-sqrt(5^2 - 4*6*(-6)))/(2*6)$

$\Rightarrow \frac{- 5 \pm \sqrt{25 + 144}}{12}$ $rArr (-5+- sqrt(25+144))/12$

$\Rightarrow \frac{- 5 \pm \sqrt{169}}{12}$ $rArr (-5+- sqrt(169))/12$

$\Rightarrow \frac{- 5 \pm 13}{12}$ $rArr (-5+- 13)/12$

Hence,

$x = \frac{- 5 + 13}{12}, x = \frac{- 5 - 13}{12}$ $x = (-5+13)/12, x = (-5-13)/12$

$x = \frac{8}{12}, x = - \frac{18}{12}$ $x = 8/12, x = -18/12$

$x = \frac{2}{3}, x = - \frac{3}{2}$ $x = 2/3, x = -3/2$

Hence, $x = \frac{2}{3}, x = - \frac{3}{2}$ $color(blue)(x=2/3, x = -3/2)$

We can observe that both the methods yield the same values for $x$ $x$

Hope you find this solution helpful.

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