Question

Match each equation with its solution set.

Answer 1

Answer:

Part 1) $a^{2} -9a+14=0$ -----> solution set {7,2}

Part 2) $a^{2} +9a+14=0$ -----> solution set {-2,-7}

Part 3) $a^{2} +3a-10=0$  -----> solution set {2,-5}

Part 4) $a^{2} +5a-14=0$   ----> solution set {2,-7}

Part 5) $a^{2} -5a-14=0$ ----> solution set {-2,7}

Step-by-step explanation:

we know that

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

$x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}$

Part 1)

in this problem we have

$a^{2} -9a+14=0$  

so

$a=1\\b=-9\\c=14$

substitute in the formula

$a=\frac{9(+/-)\sqrt{-9^{2}-4(1)(14)}} {2(1)}$

$a=\frac{9(+/-)\sqrt{25}} {2}$

$a=\frac{9(+/-)5} {2}$

$a=\frac{9(+)5} {2}=7$

$a=\frac{9(-)5} {2}=2$

The solution set is {7,2}

Part 2)

in this problem we have

$a^{2} +9a+14=0$  

so

$a=1\\b=9\\c=14$

substitute in the formula

$a=\frac{-9(+/-)\sqrt{9^{2}-4(1)(14)}} {2(1)}$

$a=\frac{-9(+/-)\sqrt{25}} {2}$

$a=\frac{-9(+/-)5} {2}$

$a=\frac{-9(+)5} {2}=-2$

$a=\frac{-9(-)5} {2}=-7$

The solution set is {-2,-7}

Part 3)

in this problem we have

$a^{2} +3a-10=0$  

so

$a=1\\b=3\\c=-10$

substitute in the formula

$a=\frac{-3(+/-)\sqrt{3^{2}-4(1)(-10)}} {2(1)}$

$a=\frac{-3(+/-)\sqrt{49}} {2}$

$a=\frac{-3(+/-)7} {2}$

$a=\frac{-3(+)7} {2}=2$

$a=\frac{-3(-)7} {2}=-5$

The solution set is {2,-5}

Part 4)

in this problem we have

$a^{2} +5a-14=0$  

so

$a=1\\b=5\\c=-14$

substitute in the formula

$a=\frac{-5(+/-)\sqrt{5^{2}-4(1)(-14)}} {2(1)}$

$a=\frac{-5(+/-)\sqrt{81}} {2}$

$a=\frac{-5(+/-)9} {2}$

$a=\frac{-5(+)9} {2}=2$

$a=\frac{-5(-)9} {2}=-7$

The solution set is {2,-7}

Part 5)

in this problem we have

$a^{2} -5a-14=0$  

so

$a=1\\b=-5\\c=-14$

substitute in the formula

$a=\frac{5(+/-)\sqrt{-5^{2}-4(1)(-14)}} {2(1)}$

$a=\frac{5(+/-)\sqrt{81}} {2}$

$a=\frac{5(+/-)\sqrt{81}} {2}$

$a=\frac{5(+)9} {2}=7$

$a=\frac{5(-)9} {2}=-2$

The solution set is {-2,7}