How do you use the quadratic formula to solve $3/4(x-1)^2=-4/3x+4/5$ ?

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Answer 1 · 2017-06-20T14:20:10

Before one uses the quadratic formula, one must write the equation in the form:

$ax^2+bx+c=0$

Then substitute, a, b, and c into the formula:

$x = (-b+-sqrt(b^2-4(a)(c)))/(2a)$

Given: $3/4(x-1)^2=-4/3x+4/5$

Multiply both sides of the equation by the factors: $(3)(4)(5)$

$45(x-1)^2=-80x+48$

Expand the square:

$45(x^2-2x+1) = -80x +48$

Distribute the 45:

$45x^2-90x+45 = -80x +48$

Combine like terms:

$45x^2-10x-3 = 0$

By observation, $a =45, b = -10, and c = -3$

Substitute into the formula:

$x = (10+-sqrt((-10)^2-4(45)(-3)))/(2(45))$

$x = (10+-sqrt(640))/90$

$x = (10+-8sqrt(10))/90$

$x = (10-8sqrt10)/90$ and $x = (10+8sqrt10)/90$

How do you use the quadratic formula to solve 3/4(x-1)^2=-4/3x+4/53/4(x-1)^2=-4/3x+4/5?