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
Accepted Solution
A:
Answer:Option D [tex]x^{2} -x-6=0[/tex]Step-by-step explanation:Verify each quadratic equationcase A) we have[tex]x^{2} +x+6=0[/tex]This quadratic equation has a leading coefficient of 1Substitute the value of x=3 and x=-2 in the equationFor x=3[tex](3)^{2} +(3)+6=0[/tex][tex]18=0[/tex] ----> is not truethereforex=3 is not a solution of the quadratic equationcase B) we have[tex]x^{2} -x-5=0[/tex]This quadratic equation has a leading coefficient of 1Substitute the value of x=3 and x=-2 in the equationFor x=3[tex](3)^{2} -(3)-5=0[/tex][tex]1=0[/tex] ----> is not truethereforex=3 is not a solution of the quadratic equationcase C) we have[tex]x^{2} +x+5=0[/tex]This quadratic equation has a leading coefficient of 1Substitute the value of x=3 and x=-2 in the equationFor x=3[tex](3)^{2} +(3)+5=0[/tex][tex]17=0[/tex] ----> is not truethereforex=3 is not a solution of the quadratic equationcase D) we have[tex]x^{2} -x-6=0[/tex]This quadratic equation has a leading coefficient of 1Substitute the value of x=3 and x=-2 in the equationFor x=3[tex](3)^{2} -(3)-6=0[/tex][tex]0=0[/tex] ----> is truethereforex=3 is a solution of the quadratic equationFor x=-2[tex](-2)^{2} -(-2)-6=0[/tex][tex]4+2-6=0[/tex][tex]0=0[/tex] ----> is truethereforex=-2 is a solution of the quadratic equation