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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

#1
+117
+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
+117
+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.

cryptoaops Dec 29, 2020
edited by cryptoaops  Dec 29, 2020
#2
+117546
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Nice, cryptoaops   !!!!

CPhill  Dec 29, 2020