Question

Which equation has no real roots? a. x2 – 6x + 12= 0 b. x2 – 25 = 0 c. x2 + 11x = 0 d. x2 + 12x + 11= 0

Answer 1

Answer:

A

Step-by-step explanation:

To find the number of real roots for a quadratic, we apply the discriminate. The discriminate is the inside portion of the square root from the quadratic formula.

• $b^2-4ac>0$ yields 2 real roots
• $b^2-4ac=0$ yields 1 real root
• [tx]b^2-4ac<0[/tex] yields no real roots

a. $x^2-6x+12=0$ where a=1, b=-6, and c=12

    $b^2-4ac=(-6)^2-4(1)(12)=36-48=-12<0)$ has no real roots

b. $x^2-25=0$ where a=1, b=0, and c=-25

    $b^2-4ac=(0)^2-4(1)(-25)=0+100=100>0)$ has 2 real roots

c. $x^2+11x=0$ where a=1, b=11, and c=0

    $b^2-4ac=(11)^2-4(1)(0)=121-0=-121>0)$ has 2 real roots

d. $x^2+12x+11=0$ where a=1, b=12, and c=11

    $b^2-4ac=(12)^2-4(1)(11)=144-44=100>0)$ has 2 real roots