Algebra II? Help with Quadratic equation
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Algebra II? Help with Quadratic equation

[From: ] [author: ] [Date: 12-05-21] [Hit: ]
y = 4x^2 + 21x - 18-since the roots are -6 and 3/4,that means that those two values are the solutions to the equation i.x=-6 and x=3/4. just make the right side zero (x+6=0, and 4x-3=0) and multiply them.-(x+6)(4x-3) is such an equation.......
The question is :
Write a quadratic equation with -6 and 3/4 as its roots. Write the equation in the form ax^2 + bx + c = 0, where a, b, and c, are integers.



Please explain to me how you came up with the equation. it would be so helpful! Thank you!

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Roots: (x + 6)(4x - 3) = 0
Equation:
y = 4x^2 + 21x - 18

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since the roots are -6 and 3/4,that means that those two values are the solutions to the equation i.e
x=-6 and x=3/4. just make the right side zero (x+6=0, and 4x-3=0) and multiply them.

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(x+6)(4x-3) is such an equation.
expand the brackets

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(4x - 3)(x + 6)
4x² - 3x + 24x - 18

4x² + 21x - 18
1
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