Question

solve by factoring.
$2x^2 + 3x - 35 = 0$
a $left{-\frac{7}{2}, 5
ight}$
b $left{-5, \frac{7}{2}
ight}$
c ${-10, 7}$
d ${-7, 10}$

Explanation:

Step1: Factor the quadratic equation

We have the quadratic equation \(2x^{2}+3x - 35 = 0\). To factor it, we need to find two numbers that multiply to \(2\times(-35)=-70\) and add up to \(3\). The numbers are \(10\) and \(-7\) since \(10\times(-7)=-70\) and \(10 + (-7)=3\).
So we can rewrite the middle term:
\(2x^{2}+10x-7x - 35 = 0\)
Now, factor by grouping:
\(2x(x + 5)-7(x + 5)=0\)
\((2x - 7)(x + 5)=0\)

Step2: Solve for x

Set each factor equal to zero:

1. \(2x-7 = 0\)

Solving for \(x\), we add \(7\) to both sides: \(2x=7\), then divide by \(2\): \(x=\frac{7}{2}\)

1. \(x + 5 = 0\)

Solving for \(x\), we subtract \(5\) from both sides: \(x=-5\)

So the solutions are \(x=-5\) and \(x = \frac{7}{2}\), which means the solution set is \(\{-5,\frac{7}{2}\}\)

Answer:

B. \(\{-5,\frac{7}{2}\}\)