Solving Quadratic Equations Practice

1. Solve for $x$.

   $2x^2 - 28x + 98 = 0$

2. Solve for $x$. Enter the solutions from least to greatest.

   $5x^2 + 15x - 50 = 0$

   Lesser $x = \text{______}$, greater $x=\text{______}$.

3. Solve for $x$.

   $2x^2 - 40x + 200 = 0$

4. Solve for $x$. Enter the solutions from least to greatest.

   $5x^2 + 15x - 140 = 0$

   Lesser $x =\text{______}$, greater $x= \text{______}$.

Source: Khan Academy, https://www.khanacademy.org/math/college-algebra/xa5dd2923c88e7aa8:quadratic-functions-and-equations/xa5dd2923c88e7aa8:solving-quadratics-by-factoring/e/solving_quadratics_by_factoring_2
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1. Dividing both sides by ‍2 gives:

   $\qquad x^2 {-14}x + {49} = 0$

   The coefficient on the ‍$x$ term is ‍-14 and the constant term is ‍49, so we need to find two numbers that add up to -14 and multiply to ‍49.

   The number ‍-7 used twice satisfies both conditions:

   $\qquad{-7} +{-7} = {-14}$

   $\qquad {-7} \times {-7} = {49}$

   So $(x - {7})^2 = 0$.

   $x - 7 = 0$

   Thus, ‍$x = 7$ is the solution.

   ---

2. $\begin{aligned}5x^2 + 15x - 50 &= 0\\\\5(x^2+3x-10)&=0\end{aligned}$

   Now let's factor the expression in the parentheses.

   $x^2+3x-10$ can be factored as $(x+5)(x-2)$.

   $\begin{aligned}5(x+5)(x-2)&=0\\\\x+5=0&\text{ or }x-2=0\\\\x=-5&\text{ or }x=2\end{aligned}$
   

   In conclusion,
   $\text{lesser }x = -5\\ \text{greater }x = 2$

   ---

3. Dividing both sides by ‍2 gives:

   $\qquad x^2 {-20}x + {100} = 0$

   The coefficient on the ‍$x$ term is ‍-20 and the constant term is 100, so we need to find two numbers that add up to -20 and multiply to 100.

   The number ‍-10 used twice satisfies both conditions:

   $\qquad {-10} + {-10} = {-20}$

   $\qquad {-10} \times {-10} = {100}$

   So $(x - {10})^2 = 0$.

   $x - 10 = 0$

   Thus, ‍$x = 10$ is the solution.

   ---

4. $\begin{aligned}5x^2 + 15x - 140 &= 0\\\\5(x^2+3x-28)&=0\end{aligned}$

   Now let's factor the expression in the parentheses.
   

   $x^2+3x-28$ can be factored as $(x+7)(x-4)$

   $\begin{aligned}5(x+7)(x-4)&=0\\\\x+7=0&\text{ or }x-4=0\\\\x=-7&\text{ or }x=4 \end{aligned}$

   In conclusion,
   $\text{lesser }x = -7\\\text{greater }x = 4$