QUESTION IMAGE
Question
2 solve for the values of x by factoring.
-3x²+4x = -15
o x = -3; x = -3/5
o x = 4; x = -4/3
o x = -3; x = 5/3
o x = 3; x = -5/3
Step1: Rearrange the equation
First, we want to set the equation to standard quadratic form \(ax^2 + bx + c = 0\). Start with \(-3x^2 + 4x=-15\). Add \(15\) to both sides:
\(-3x^2 + 4x + 15 = 0\)
Multiply both sides by \(-1\) to make the coefficient of \(x^2\) positive:
\(3x^2 - 4x - 15 = 0\)
Step2: Factor the quadratic
We need two numbers that multiply to \(3\times(-15)= -45\) and add up to \(-4\). The numbers are \(-9\) and \(5\) because \(-9\times5=-45\) and \(-9 + 5=-4\).
Rewrite the middle term using these numbers:
\(3x^2 - 9x + 5x - 15 = 0\)
Factor by grouping:
\(3x(x - 3)+5(x - 3)=0\)
Now, factor out \((x - 3)\):
\((3x + 5)(x - 3)=0\)
Step3: Solve for \(x\)
Set each factor equal to zero:
- \(3x + 5 = 0\)
Subtract \(5\) from both sides: \(3x=-5\)
Divide by \(3\): \(x =-\frac{5}{3}\)
- \(x - 3 = 0\)
Add \(3\) to both sides: \(x = 3\)
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\(x = 3; x=-\frac{5}{3}\) (the last option: \(x = 3;x=-\frac{5}{3}\))