Question

Choose the quadratic equation that has a leading coefficient of 1 and solutions 3 and -2.

Options:

A.) x^2+x+6=0
B.) x^2-x-5=0
C.) x^2+x+5=0
D.) x^2-x-6=0

Answer 1

Answer:

Option D $x^{2} -x-6=0$

Step-by-step explanation:

Verify each quadratic equation

case A) we have

$x^{2} +x+6=0$

This quadratic equation has a leading coefficient of 1

Substitute the value of x=3 and x=-2 in the equation

For x=3

$(3)^{2} +(3)+6=0$

$18=0$ ----> is not true

therefore

x=3 is not a solution of the quadratic equation

case B) we have

$x^{2} -x-5=0$

This quadratic equation has a leading coefficient of 1

Substitute the value of x=3 and x=-2 in the equation

For x=3

$(3)^{2} -(3)-5=0$

$1=0$ ----> is not true

therefore

x=3 is not a solution of the quadratic equation

case C) we have

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

This quadratic equation has a leading coefficient of 1

Substitute the value of x=3 and x=-2 in the equation

For x=3

$(3)^{2} +(3)+5=0$

$17=0$ ----> is not true

therefore

x=3 is not a solution of the quadratic equation

case D) we have

$x^{2} -x-6=0$

This quadratic equation has a leading coefficient of 1

Substitute the value of x=3 and x=-2 in the equation

For x=3

$(3)^{2} -(3)-6=0$

$0=0$ ----> is true

therefore

x=3 is a solution of the quadratic equation

For x=-2

$(-2)^{2} -(-2)-6=0$

$4+2-6=0$

$0=0$ ----> is true

therefore

x=-2 is a solution of the quadratic equation