Question

Quick algebra 1 question for 10 points!

Answer 1

Answer:

$\left\{\dfrac{3}{2},\ -\dfrac{5}{3}\right\}$

Step-by-step explanation:

Given quadratic equation: $6x^2+x-15=0$

$\implies a=\textsf{6},b=\textsf{1},c=\textsf{-15}$

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${\large \textsf{ Quadratic Formula: }}x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}\ \Bigg{\|}\ \textsf{when }ax^2+bx+c=0$

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Step 1: Substitute the given values into the formula and simplify.

$\implies x=\dfrac{-{(\textsf1})\pm \sqrt{\textsf{(1)}^2-4(\textsf{6}){(\textsf{-15})}}}{2{(\textsf{6})}}$

$\implies x=\dfrac{-1\pm \sqrt{1-4(-90)}}{12}$

$\implies x=\dfrac{-1\pm \sqrt{1+360}}{12}$

$\implies x=\dfrac{-1\pm \sqrt{361}}{12}$

$\implies x=\dfrac{-1\pm 19}{12}$

Step 2: Separate into two cases.

$x_1=\dfrac{-1+19}{2}$

$x_2=\dfrac{-1-19}{2}$

Step 3: Solve both cases.

$\implies x_1=\dfrac{-1+19}{12}=\dfrac{18}{12}=\dfrac{6\times3}{6\times2}=\boxed{\dfrac{3}{2}}$

$\implies x_2=\dfrac{-1-19}{12}=\dfrac{-20}{12}=\dfrac{-4\times5}{4\times3}=\boxed{-\dfrac{5}{3}}$

The solutions to this quadratic are: $x_1=\dfrac{3}{2},\ x_2=-\dfrac{5}{3}$

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Learn more about quadratic equations here:

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Answer 2

Answer:

B {3/2, -5/3}

Step-by-step explanation:

6x² + x - 15 = 0

6x² + 10x - 9x - 15 = 0

2x(3x + 5) - 3(3x + 5) = 0

(3x + 5)(2x - 3) = 0

3x + 5 = 0  or  2x - 3 = 0

3x = -5  or  2x = 3

x = -5/3  or  x = 3/2

Answer: B {3/2, -5/3}