Question

F(x)=6x^2+10x−1 How many distinct real number zeros does f have?

Answer 1

Answer:

Here, $D>0$, hence the quadratic equation has two distinct real roots.

Step-by-step explanation:

Given quadratic equation is $6x^{2} +10x-1$.

Let, the quadratic equation is $ax^{2} +bx+c$    [where, $a,b,c$ are the constants]

The Discriminant $(D)=b^{2}-4ac$

Case $1$:   $b^{2}-4ac>0$, if the discriminant is greater than $0$, it means the quadratic equation has two real distinct roots.

Case $2$: $b^{2}-4ac$, if the discriminant is less than $0$, it means the quadratic equation has no real roots.

Case $3$: $b^{2}-4ac=0$, if the discriminants is equal to $0$, it means the quadratic equation has two real identical roots.

Now,

      we have $6x^{2} +10x-1$, where $a=6,b=10,\ and\ c=-1$

∴$D=b^{2}-4ac$

     $=(10)^{2}-(4\times 6\times\ -1)$

     $=100+24$

     $= 124$

Here, $D>0$, hence the quadratic equation has two distinct real roots.