Question

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Answer 1

Answer:

First Picture ⇒ [B] a = -2, b = 1, and c = -5

Second Picture ⇒ First step: Identify a = 1, b = 3, c = -4

                          $x=\frac{-3\pm \sqrt{3^2-4\cdot \:1\cdot \left(-4\right)}}{2\cdot \:1}$

                             $x=\frac{-3\pm 5\sqrt{25}}{2}$

                           $x=\frac{-3\pm \:5}{2}$

                        $x=\frac{-3+5}{2},\:x=\frac{-3-5}{2\c}$

Third Picture ⇒ Missing Info

Fourth Picture ⇒ [A] $x=-\frac{5}{2},\:x=-3$

Fifth Picture ⇒ $x=\frac{5\pm\sqrt{1} }{2}$

Step-by-step explanation:

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Identify the a,b, and c-values for this equation:

y = -2x²  + x -5  

Quadratic Formula:

ax²+bx+c=0

−2x²+1x−5=0

a = -2, b = 1, and c = -5

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Order the steps for solving this equation using the quadratic formula.

x²  + 3x - 4 = 0

Solving to see what are the steps:

First step: Identify a = 1, b = 3, c = -4

$x=\frac{-3\pm \sqrt{3^2-4\cdot \:1\cdot \left(-4\right)}}{2\cdot \:1}$

$x=\frac{-3\pm 5\sqrt{25}}{2}$

$x=\frac{-3\pm \:5}{2}$

$x_1=\frac{-3+5}{2\cdot \:1},\:x_2=\frac{-3-5}{2\cdot \:1}$

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Use the quadratic formula to find the solution set for 2x² + 15 = -11x.

Add 11x to both sides:

2x² + 15+11x = -11x+11x

Simplify

2x² + 11x+15 = 0

Now solve with quadratic formula..

$x_{1,\:2}=\frac{-11\pm \sqrt{11^2-4\cdot \:2\cdot \:15}}{2\cdot \:2}$

$x_{1,\:2}=\frac{-11\pm \:1}{2\cdot \:2}$

$x_1=\frac{-11+1}{2\cdot \:2},\:x_2=\frac{-11-1}{2\cdot \:2}$

$x=-\frac{5}{2},\:x=-3$

Hence, [A] $x=-\frac{5}{2},\:x=-3$

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Suppose you are solving a quadratic equation and this is your work so far...

x  ²  - 5x + 6 = 0

Thus, we have:

$x=\frac{-\left(-5\right)\pm \sqrt{\left(-5\right)^2-4\cdot \:1\cdot \:6}}{2\cdot \:1}$

$x=\frac{-\left(-5\right)\pm \:1}{2\cdot \:1}$

$x=\frac{5\pm\sqrt{1} }{2}$

$x_1=\frac{-\left(-5\right)+1}{2\cdot \:1},\:x_2=\frac{-\left(-5\right)-1}{2\cdot \:1}$

$x=3,x=2$

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Kavinsky