Solve the System of Inequalities D>0 , b^2-4ac>0 , (-2)^2-4*(1(a-3))>0

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$D > 0$ , $b^{2} - 4 a c > 0$ , ${(- 2)}^{2} - 4 \cdot (1 (a - 3)) > 0$

Solve $b^{2} - 4 a c > 0$ for $b$ .

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Add $4 a c$ to both sides of the inequality.

$b^{2} > 4 a c$

Take the square root of both sides of the inequality to eliminate the exponent on the left side.

$b > \pm \sqrt{4 a c}$

The complete solution is the result of both the positive and negative portions of the solution.

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Simplify the right side of the equation.

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Rewrite $4 a c$ as $2^{2} (a c)$ .

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Rewrite $4$ as $2^{2}$ .

$b > \pm \sqrt{2^{2} a c}$

Add parentheses.

$b > \pm \sqrt{2^{2} (a c)}$

Pull terms out from under the radical.

$b > \pm 2 \sqrt{a c}$

The complete solution is the result of both the positive and negative portions of the solution.

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First, use the positive value of the $\pm$ to find the first solution.

$b > 2 \sqrt{a c}$

Next, use the negative value of the $\pm$ to find the second solution. Since this is an inequality, flip the direction of the inequality sign on the $-$ portion of the solution.

$b < - 2 \sqrt{a c}$

The complete solution is the result of both the positive and negative portions of the solution.

$b > 2 \sqrt{a c}$ or $b < - 2 \sqrt{a c}$

Solve ${(- 2)}^{2} - 4 \cdot (1 (a - 3)) > 0$ for $a$ .

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Simplify ${(- 2)}^{2} - 4 \cdot (1 (a - 3))$ .

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Simplify each term.

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Raise $- 2$ to the power of $2$ .

$4 - 4 \cdot (1 (a - 3)) > 0$

Multiply $a - 3$ by $1$ .

$4 - 4 \cdot (a - 3) > 0$

Apply the distributive property.

$4 - 4 a - 4 \cdot - 3 > 0$

Multiply $- 4$ by $- 3$ .

$4 - 4 a + 12 > 0$

Add $4$ and $12$ .

$- 4 a + 16 > 0$

Subtract $16$ from both sides of the inequality.

$- 4 a > - 16$

Divide each term by $- 4$ and simplify.

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Divide each term in $- 4 a > - 16$ by $- 4$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- 4 a}{- 4} < \frac{- 16}{- 4}$

Cancel the common factor of $- 4$ .

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Cancel the common factor.

$\frac{- 4 a}{- 4} < \frac{- 16}{- 4}$

Divide $a$ by $1$ .

$a < \frac{- 16}{- 4}$

Divide $- 16$ by $- 4$ .

$a < 4$

Find the intersection of $D > 0$ and $b > 2 \sqrt{a c} or b < - 2 \sqrt{a c}$ .

$a > 0$ and $0 < a < 4$

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