Quadratics by factoring - Questions

Answers

1.

\(\text { lesser } x=-7\)

\(\text { lesser } x=-3\)

\(4 x^{2}+40 x+84=0\)

\(4\left(x^{2}+10 x+21\right)=0\)

Now let's factor the expression in the parentheses.

\(x^{2}+10 x+21\) can be factored as \((x+7)(x+3)\).

\(4(x+7)(x+3)=0\)

\(x+7=0\) or \(x+3=0\)

\(x = -7\) or \(x=-3\)

In conclusion,

\(\text { lesser } x=-7\)

\(\text { lesser } x=-3\)


2. \(x = 6\)

Dividing both sides by \(4\) gives

\(x^{2}-12 x+36=0\)

The coefficient on the \(x\) term is \(-12\) and the constant term is \(36\), so we need to find two numbers that add up to \(-12\) and multiply to \(36\).

The number \(-6\) used twice satisfies both conditions:

\(−6+−6=−12\)

\(−6×−6=36\)

So \((x-6)^{2}=0\).

\(x−6=0\)

Thus, \(x = 6\) is the solution.


3.

\(\text { lesser } x=1\)

\(\text { lesser } x=3\)

\(5 x^{2}-20 x+15=0\)

\(5\left(x^{2}-4 x+3\right)=0\)

Now let's factor the expression in the parentheses.

\(x^{2}-4 x+3\) can be factored as \((x-1)(x-3)\).

\(5(x−1)(x−3) = 0\)

\( x-1 = 0\) or \(x -3 = 0\)

\(x = 1\) or \(x= 3\)

In conclusion,

\(\text { lesser } x=1\)

\(\text { lesser } x=3\)


4. \(x =9\)

Dividing both sides by \(3\) gives \(x^{2}-18 x+81=0\)

The coefficient on the \(x\) term is \(-18\) and the constant term is \(81\), so we need to find two numbers that add up to \(-18\) and multiply to \(81\).

The number \(-9\) used twice satisfies both conditions:

\(−9+−9=−18\)

\(−9×−9=81\)

So \((x-9)^{2}=0\).

\(x−9=0\)

Thus, \(x =9\) is the solution.