Question

Match each equation with its solution set. Tiles a2 − 9a + 14 = 0 a2 + 9a + 14 = 0 a2 + 3a − 10 = 0 a2 + 5a − 14 = 0 a2 − 5a − 14 = 0 Pairs {-2, 7} {2, -7} {-2, -7} {7, 2}

Answer 1

For this case we will solve the equations step by step:

 a2 - 9a + 14 = 0
 (a-7) * (a-2) = 0
 a1 = 7
 a2 = 2
 {7, 2}

 a2 + 9a + 14 = 0
 (a + 7) * (a + 2) = 0
 a1 = -7
 a2 = -2
 {-2, -7}

 a2 + 5a - 14 = 0
 (a + 7) * (a-2) = 0
 a1 = -7
 a2 = 2
 {2, -7}

 a2 - 5a - 14 = 0
 (a-7) * (a + 2) = 0 
 a1 = 7
 a2 = -2
 {-2, 7}

Answer 2

Answer:

The solution set for the first equation is {2,7}.

The solution set for the second equation is {-2,-7}.

The solution set for the third equation is {-5,2}.

The solution set for the fourth equation is {2,-7}.

The solution set for the fifth equation is {-2,7}.

Step-by-step explanation:

The first equation is

$a^2-9a+14=0$

$a^2-7a-2a+14=0$

$a(a-7)-2(a-7)=0$

$(a-7)(a-2)=0$

$a-7=0\Rightarrow x=7$

$a-2=0\Rightarrow x=2$

Therefore the solution set for the first equation is {2,7}.

The second equation is

$a^2+9a+14=0$

$a^2+7a+2a+14=0$

$a(a+7)+2(a+7)=0$

$(a+7)(a+2)=0$

$a+7=0\Rightarrow x=-7$

$a+2=0\Rightarrow x=-2$

Therefore the solution set for the second equation is {-2,-7}.

The third equation is

$a^2+3a-10=0$

$a^2+5a-2a-10=0$

$a(a+5)-2(a+5)=0$

$(a+5)(a-2)=0$

$a+5=0\Rightarrow x=-5$

$a-2=0\Rightarrow x=2$

Therefore the solution set for the third equation is {-5,2}.

The fourth equation is

$a^2+5a-14=0$

$a^2+7a-2a-14=0$

$a(a+7)-2(a+7)=0$

$(a+7)(a-2)=0$

$a+7=0\Rightarrow x=-7$

$a-2=0\Rightarrow x=2$

Therefore the solution set for the fourth equation is {2,-7}.

The fifth equation is

$a^2-5a-14=0$

$a^2-7a+2a-14=0$

$a(a-7)+2(a-7)=0$

$(a-7)(a+2)=0$

$a-7=0\Rightarrow x=7$

$a+2=0\Rightarrow x=-2$

Therefore the solution set for the fifth equation is {-2,7}.