Solve quadratic equation by completing the square 9x²=4 please help me answer this

Mathematics

2 answers:

Sveta_85 [38]3 years ago

6 0

Answer:

$9 {x}^{2} - 4 = 0$

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

$3x - 2 = 0$

$3x = 2$

$x = \frac{2}{3}$

AND

$3x + 2 = 0$

$3x = - 2$

$\times = - \frac{2}{3}$

andrew11 [14]3 years ago

4 0

9x²-4=0

or,(3x)² +2*3x*3 +(3)² -5 = 18x -5

or,(3x+3)² = 18x

either,3x+3= 18x or, 3x+3=-18x

or, 15x=3 or, 21x=-3

or, x=1/5 or, x=-1/7

You might be interested in

Suppose x is a real number and epsilon > 0. Prove that (x - epsilon, x epsilon) is a neighborhood of each of its members; in

kaheart [24]

Answer:

See proof below

Step-by-step explanation:

We will use properties of inequalities during the proof.

Let $y\in (x-\epsilon,x+\epsilon)$ . then we have that $x-\epsilon$ . Hence, it makes sense to define the positive number delta as $\delta=\min\{x+\epsilon-y,y-(x-\epsilon)\}$ (the inequality guarantees that these numbers are positive).

Intuitively, delta is the shortest distance from y to the endpoints of the interval. Now, we claim that $(y-\delta,y+\delta)\subseteq (x-\epsilon,x+\epsilon)$ , and if we prove this, we are done. To prove it, let $z\in (y-\delta,y+\delta)$ , then $y-\delta$ . First, $\delta \leq y-(x-\epsilon)$ then $-\delta \geq -y+x-\epsilon$ hence $z>y-\delta \geq x-\epsilon$

On the other hand, $\delta \leq x+\epsilon-y$ then $z$ hence $z$ . Combining the inequalities, we have that $x-\epsilon$ , therefore $(y-\delta,y+\delta)\subseteq (x-\epsilon,x+\epsilon)$ as required.