Answer:
First Picture ⇒ [B] a = -2, b = 1, and c = -5
Second Picture ⇒ First step: Identify a = 1, b = 3, c = -4
[tex]x=\frac{-3\pm \sqrt{3^2-4\cdot \:1\cdot \left(-4\right)}}{2\cdot \:1}[/tex]
[tex]x=\frac{-3\pm 5\sqrt{25}}{2}[/tex]
[tex]x=\frac{-3\pm \:5}{2}[/tex]
[tex]x=\frac{-3+5}{2},\:x=\frac{-3-5}{2\c}[/tex]
Third Picture ⇒ Missing Info
Fourth Picture ⇒ [A] [tex]x=-\frac{5}{2},\:x=-3[/tex]
Fifth Picture ⇒ [tex]x=\frac{5\pm\sqrt{1} }{2}[/tex]
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
[tex]x=\frac{-3\pm \sqrt{3^2-4\cdot \:1\cdot \left(-4\right)}}{2\cdot \:1}[/tex]
[tex]x=\frac{-3\pm 5\sqrt{25}}{2}[/tex]
[tex]x=\frac{-3\pm \:5}{2}[/tex]
[tex]x_1=\frac{-3+5}{2\cdot \:1},\:x_2=\frac{-3-5}{2\cdot \:1}[/tex]
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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..
[tex]x_{1,\:2}=\frac{-11\pm \sqrt{11^2-4\cdot \:2\cdot \:15}}{2\cdot \:2}[/tex]
[tex]x_{1,\:2}=\frac{-11\pm \:1}{2\cdot \:2}[/tex]
[tex]x_1=\frac{-11+1}{2\cdot \:2},\:x_2=\frac{-11-1}{2\cdot \:2}[/tex]
[tex]x=-\frac{5}{2},\:x=-3[/tex]
Hence, [A] [tex]x=-\frac{5}{2},\:x=-3[/tex]
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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:
[tex]x=\frac{-\left(-5\right)\pm \sqrt{\left(-5\right)^2-4\cdot \:1\cdot \:6}}{2\cdot \:1}[/tex]
[tex]x=\frac{-\left(-5\right)\pm \:1}{2\cdot \:1}[/tex]
[tex]x=\frac{5\pm\sqrt{1} }{2}[/tex]
[tex]x_1=\frac{-\left(-5\right)+1}{2\cdot \:1},\:x_2=\frac{-\left(-5\right)-1}{2\cdot \:1}[/tex]
[tex]x=3,x=2[/tex]
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Kavinsky
