Question

name: camillna period: algebra unit 7 homework 7 - quadratic formula day 1 directions: show all work in order to receive full credit. find the solution to each quadratic equation below using the quadratic formula. express all answers in simplest radical form. 1. ( m^2 - 5m - 14 = 0 ) 2. ( b^2 - 4b = -4 ) 3. ( x^2 + 4x + 3 = 0 ) 4. ( 2x^2 + 3x - 20 = 0 )

Explanation:

For problem 1: $m^2 - 5m - 14 = 0$

Step1: Identify a, b, c

$a=1,\ b=-5,\ c=-14$

Step2: Substitute into quadratic formula

Quadratic formula: $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
$m=\frac{-(-5)\pm\sqrt{(-5)^2-4(1)(-14)}}{2(1)}$

Step3: Simplify discriminant

$m=\frac{5\pm\sqrt{25+56}}{2}=\frac{5\pm\sqrt{81}}{2}$

Step4: Solve for m

$m=\frac{5\pm9}{2}$
$m_1=\frac{5+9}{2}=7,\ m_2=\frac{5-9}{2}=-2$

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For problem 2: $b^2 - 4b = -4$

Step1: Rewrite in standard form

$b^2 - 4b + 4 = 0$

Step2: Identify a, b, c

$a=1,\ b=-4,\ c=4$

Step3: Substitute into quadratic formula

$b=\frac{-(-4)\pm\sqrt{(-4)^2-4(1)(4)}}{2(1)}$

Step4: Simplify discriminant

$b=\frac{4\pm\sqrt{16-16}}{2}=\frac{4\pm0}{2}$

Step5: Solve for b

$b=\frac{4}{2}=2$

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For problem 3: $x^2 + 4x + 3 = 0$

Step1: Identify a, b, c

$a=1,\ b=4,\ c=3$

Step2: Substitute into quadratic formula

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

Step3: Simplify discriminant

$x=\frac{-4\pm\sqrt{16-12}}{2}=\frac{-4\pm\sqrt{4}}{2}$

Step4: Solve for x

$x=\frac{-4\pm2}{2}$
$x_1=\frac{-4+2}{2}=-1,\ x_2=\frac{-4-2}{2}=-3$

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For problem 4: $2x^2 + 3x - 20 = 0$

Step1: Identify a, b, c

$a=2,\ b=3,\ c=-20$

Step2: Substitute into quadratic formula

$x=\frac{-(3)\pm\sqrt{(3)^2-4(2)(-20)}}{2(2)}$

Step3: Simplify discriminant

$x=\frac{-3\pm\sqrt{9+160}}{4}=\frac{-3\pm\sqrt{169}}{4}$

Step4: Solve for x

$x=\frac{-3\pm13}{4}$
$x_1=\frac{-3+13}{4}=\frac{10}{4}=\frac{5}{2},\ x_2=\frac{-3-13}{4}=\frac{-16}{4}=-4$

Answer:

1. $m=7$ and $m=-2$
2. $b=2$
3. $x=-1$ and $x=-3$
4. $x=\frac{5}{2}$ and $x=-4$