Question - Solving a Quadratic Equation
Solution:
$$\begin{align*}2x^2 - 4x - 6 &= 0\\x^2 - 2x - 3 &= 0 \quad \text{(Divide by 2)}\\(x - 3)(x + 1) &= 0\\x - 3 &= 0 \ \text{or} \ x + 1 = 0\\x &= 3 \ \text{or} \ x = -1\end{align*}$$
Tutor Solution:
We are given a quadratic equation in the form $$ ax^2 + bx + c = 0 $$. The equation is $$ 2x^2 - 4x - 6 = 0 $$.To solve this quadratic equation, we can use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]In our quadratic equation, $$ a = 2 $$, $$ b = -4 $$, and $$ c = -6 $$.Let's compute the discriminant ($$ b^2 - 4ac $$) first:\[ b^2 - 4ac = (-4)^2 - 4(2)(-6) = 16 + 48 = 64 \]Now that we have the discriminant, we can solve for $$ x $$ by plugging $$ a $$, $$ b $$, and the discriminant into the quadratic formula:\[ x = \frac{-(-4) \pm \sqrt{64}}{2(2)} = \frac{4 \pm 8}{4} \]We then solve for $$ x $$ by considering the plus and minus cases separately:For the plus case:\[ x_1 = \frac{4 + 8}{4} = \frac{12}{4} = 3 \]For the minus case:\[ x_2 = \frac{4 - 8}{4} = \frac{-4}{4} = -1 \]So, the solutions to the equation $$ 2x^2 - 4x - 6 = 0 $$ are $$ x_1 = 3 $$ and $$ x_2 = -1 $$. These are the values of $$ x $$ that satisfy the original quadratic equation.
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