Find all values of x such that 2x^2 + 9x - 6 = 0. If you find more than one value, then list your solutions in increasing order, separated by commas.
+122 Answer: \(\boxed{x = -5, \frac12}\)
Step-by-step explanation: The given equation is \(2x^2 + 9x - 5 = 0\)
We have to find the values of \(x\).
We know that if in a quadratic equation \(ax^2+ bx + c = 0, b^2- 4ac > 0\) then there will be two real roots of the equation.
In the given equation \(b^2- 4ac = (9)^2-4(2)(-5) = 81 + 40\)
\(= 212 > 0\)
So there will be two real roots.
Now we will factorize the given equation.
\(2x^2 + 9x - 5 = 0\)
\(2x^2 + 10x - x - 5 = 0\)
\(2x(x + 5) - 1(x + 5) = 0\)
\((2x - 1)(x + 5) = 0\)
So two roots will be \(2x - 1 = 0\)
\(x = \frac12\)
and \(x + 5 = 0\)
\(x = -5\)
Therefore, \(x = -5, \frac12\), are the solutions.
+122 Answer: \(\boxed{x = -5, \frac12}\)
Step-by-step explanation: The given equation is \(2x^2 + 9x - 5 = 0\)
We have to find the values of \(x\).
We know that if in a quadratic equation \(ax^2+ bx + c = 0, b^2- 4ac > 0\) then there will be two real roots of the equation.
In the given equation \(b^2- 4ac = (9)^2-4(2)(-5) = 81 + 40\)
\(= 212 > 0\)
So there will be two real roots.
Now we will factorize the given equation.
\(2x^2 + 9x - 5 = 0\)
\(2x^2 + 10x - x - 5 = 0\)
\(2x(x + 5) - 1(x + 5) = 0\)
\((2x - 1)(x + 5) = 0\)
So two roots will be \(2x - 1 = 0\)
\(x = \frac12\)
and \(x + 5 = 0\)
\(x = -5\)
Therefore, \(x = -5, \frac12\), are the solutions.
