Solving Quadratic Equations Practice

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.

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

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

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.

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

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