The Quadratic Formula

1.

$\text { lesser x} = 1$

$\text { greater x} = 7$

To factor $x^{2}-8 x+7 \text { as }(x+a)(x+b)$, we need to find numbers $a$ and $b$ such that $a+b=−8$ and $ab = 7$.

$a=−7$ and $b=-1$ satisfy both conditions, so our equation can be re-written:

$(x−7)(x−1)=0$

According to the zero-product property, we know that

$x−7=0$ or $x - 1 = 0$

which means

$x=7x$ or $x=1$

In conclusion,

$\text { lesser x} = 1$

$\text { greater x} = 7$

2. $x = -8$

Both $x^{2}$ and $64$ are perfect squares, since $x^{2}=(x)^{2}$ and $64=(8)^{2}$.

Additionally, $16x$ is twice the product of the roots of $x^{2}$ and $64$, since $16 x=2(x)(8)$.

$x^{2}+16 x+64=(x)^{2}+2(x)(8)+(8)^{2}$

So we can use the square of a sum pattern to factor:

$a^{2}+2(a)(b)+b^{2}=(a+b)^{2}$

In this case, $a = x$ and $b = 8$:

$(x)^{2}+2(x)(8)+(8)^{2}=(x+8)^{2}$

So our equation can be re-written:

$(x+8)^{2}=0$

The only possible solution is when $x+8 = 0$, which is

$x = -8$

3.

$\text { lesser x} = -9$

$\text { greater x} = -3$

To factor $x^{2}+12 x+27$ as $(x+a)(x+b)$, we need to find numbers $a$ and $b$ such that

$a+b=12$ and $a b=27$.

$a = 3$ and $b = 9$ satisfy both conditions, so our equation can be re-written:

$(x+3)(x+9)=0$

According to the zero-product property, we know that

$x+3=0$ or $x+9 = 0$

which means

$x=−3$ or $x=-9$

In conclusion,

$\text { lesser x} = -9$

$\text { greater x} = -3$

4. $x = -5$

Both $x^{2}$ and $25$ are perfect squares, since $x^{2}=(x)^{2} \text { and } 25=(5)^{2}$.

Additionally, $10x$ is twice the product of the roots of $x^{2} \text { and } 25, \text { since } 10 x=2(x)(5)$.

$x^{2}+10 x+25=(x)^{2}+2(x)(5)+(5)^{2}$

So we can use the square of a sum pattern to factor:

$a^{2}+2(a)(b)+b^{2}=(a+b)^{2}$

In this case, $a = x$ and $b = 5$:

$(x)^{2}+2(x)(5)+(5)^{2}=(x+5)^{2}$

So our equation can be re-written:

$(x+5)^{2}=0$

The only possible solution is when $x + 5 = 0$, which is

$x = -5$