Solve the equation 12x-4+-9x^2=0

-9x^2+12x-4=0

Go!
1
2
3
4
5
6
7
8
9
0
x
y
(◻)
◻/◻
2

e
π
ln
log
lim
d/dx
d/dx
>
<
>=
<=
sin
cos
tan
cot
sec
csc

asin
acos
atan
acot
asec
acsc

sinh
cosh
tanh
coth
sech
csch

asinh
acosh
atanh
acoth
asech
acsch

Answer

$x_1=0.6667,\:x_2=0.6667$

Step by step solution

Problem

$-9x^2+12x-4=0$
1

To find the roots of a polynomial of the form $ax^2+bx+c$ we use the quadratic formula, where $a=-9$, $b=12$ and $c=-4$

$x =\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
2

Substituting the values of the coefficients of the equation in the quadratic formula

$x=\frac{12\left(-1\right)\pm \sqrt{12^2-4\left(-9\right)\left(-4\right)}}{-9\cdot 2}$
3

Multiply $-1$ times $12$

$x=\frac{-12\pm \sqrt{12^2-144}}{-18}$
4

Calculate the power

$x=\frac{-12\pm \sqrt{144-144}}{-18}$
5

Add the values $144$ and $-144$

$x=\frac{-12\pm \sqrt{0}}{-18}$
6

Calculate the power

$x=\frac{-12\pm 0}{-18}$
7

To obtain the two solutions, divide the equation in two equations, one when $\pm$ is positive ($+$), and another when $\pm$ is negative ($-$)

$x_1=\frac{-12+ 0}{-18}\:\:,\:\:x_2=\frac{-12- 0}{-18}$
8

Simplifying

$x_1=0.6667,\:x_2=0.6667$
9

We found that the two real solutions of the equation are

$x_1=0.6667,\:x_2=0.6667$

Answer

$x_1=0.6667,\:x_2=0.6667$

Problem Analysis

Main topic:

Quadratic formula

Time to solve it:

0.21 seconds

Views:

95