Answer:
You did not post the options, but i will try to answer this in a general way.
Because we have two solutions, i know that we are talking about quadratic equations, of the form of:
0 = a*x^2 + b*x + c.
There are two easy ways to see if the solutions of this equation are real or not.
1) look at the graph, if the graph touches the x-axis, then we have real solutions (if the graph does not touch the x-axis, we have complex solutions).
2) look at the determinant.
The determinant of a quadratic equation is:
D = b^2 - 4*a*c.
if D > 0, we have two real solutions.
if D = 0, we have one real solution (or two real solutions that are equal)
if D < 0, we have two complex solutions.
Answer:
![1\frac{7}{9} =\frac{16}{9} =[\frac{4}{3}] ^{2}](https://tex.z-dn.net/?f=1%5Cfrac%7B7%7D%7B9%7D%20%3D%5Cfrac%7B16%7D%7B9%7D%20%3D%5B%5Cfrac%7B4%7D%7B3%7D%5D%20%5E%7B2%7D)
Step-by-step explanation:
Answer:
Step-by-step explanation:
Given expression is,
(2x - 1)² + 2(2x - 1) = (2x - 1)(2x + 1)
To prove this identity we will take the left hand side of the equation and will prove equal to the right side.
(2x - 1)² + 2(2x - 1) = (2x - 1)(2x + 1)
4x² - 4x + 1 + 4x - 2 = (2x - 1)(2x + 1)
4x² - 1 = (2x - 1)(2x + 1)
(2x - 1)(2x + 1) = (2x - 1)(2x + 1) [Since a² - b² = (a - b)(a + b)]
