Question

Match each equation with its solution set.

Answer 1

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}

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(+/-)√(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(+/-)√(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(+/-)√(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(+/-)√(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(+/-)√(81)} {2}`

`a=\frac{5(+/-)√(81)} {2}`

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

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

The solution set is {-2,7}