For what values of a does the equation
a+x=1+a^2x
have no solution?

Never. I don't know maybe this is a trick question.

Step-by-step explanation:

a + x = 1 + a²x

Rearrange to look like standard form : ax² + bx + c = 0

0 = a + a²x - a - x

0 = a²x - x  <= This is a quadratic equation.

Let's state the values of a, b and c.

a=a; b = -1; c = 0

INTERPRET THE FORMULA:

c represents the y-intercept.

When c is 0, the y-intercept is 0. The graph always passes through the origin.

Since the origin is (0,0) and is on the x-axis, the equation always has a solution: 0.

(solution = root = zero. It's when the graph passes through the x-axis.)

ALGEBRAICALLY:

The discriminant formula to find the number of solutions or roots in a graph is:

b² - 4ac.

When a quadratic has no solution, the discriminant equates to a negative number: .

If it equates to 0, there is one solution.

If it equates to a number greater than 0, there are 2 roots.

Substitute the values into the discriminant formula.  ( a=a; b = -1; c = 0 )

b² - 4ac

= (-1)² - 4(a)(0)

= 1 - 0

= 1

1 is greater than 0. It doesn't matter what a is.

Because a is multiplied with c=0, all of 4ac equates to 0, and is never greater than b², thus the entire formula will never equate to less than 0 to give the graph no solutions.

62.5 paisa

step-by-step explanation:

4000*70 = 280000

280000-(2000*90) = 100000

4000-2400 = 1600

100000/1600 = 62.5

( give brainliest)

answer

step-by-step explanation: