How many real solutions does −15x2−3=2(7x2−1.5) have?(1 point)

real solution(s)

1 answer

To find the number of real solutions for the equation

\[ -15x^2 - 3 = 2(7x^2 - 1.5), \]

first, let's simplify the right side of the equation:

\[ 2(7x^2 - 1.5) = 14x^2 - 3. \]

Now we can rewrite the equation as:

\[ -15x^2 - 3 = 14x^2 - 3. \]

Next, we'll move all terms to one side of the equation by adding \(15x^2\) and 3 to both sides:

\[ 0 = 14x^2 - 3 + 15x^2 + 3. \]

This simplifies to:

\[ 0 = (14x^2 + 15x^2) + (-3 + 3), \] \[ 0 = 29x^2. \]

Now, we can solve for \(x^2\):

\[ 29x^2 = 0. \]

Dividing both sides by 29 gives:

\[ x^2 = 0. \]

This implies:

\[ x = 0. \]

Therefore, there is exactly one real solution to the equation, which is \(x = 0\).

In conclusion, the number of real solutions to the equation is

\[ \boxed{1}. \]