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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.

 Dec 29, 2020

Best Answer 

 #1
avatar+80 
+1

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.

 Dec 29, 2020
edited by cryptoaops  Dec 29, 2020
 #1
avatar+80 
+1
Best Answer

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.

cryptoaops Dec 29, 2020
edited by cryptoaops  Dec 29, 2020
 #2
avatar+114171 
0

Nice, cryptoaops   !!!!

 

cool cool cool

CPhill  Dec 29, 2020

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