SL 2.7—Solutions of quadratic equations and inequalities, discriminant and nature of roots

For a quadratic $a{x}^2+bx+c=0$ to have exactly one (repeated) real root, the discriminant must equal zero: $\Delta =b^2-4ac=0$.

Here $a=1$, the middle coefficient is $b$, and $c=16$. Setting $\Delta =0$: $b^2-4(1)(16)=0⟹b^2=64⟹b=\pm 8.$

From the options given, $b=8$ satisfies $b=\pm 8$. Therefore $b=8$ is the correct answer.

We need the value of $b$ that makes $\Delta =b^2-4(1)(16)=0$, i.e. $b^2=64$, giving $b=\pm 8$.

$\Delta =6^2-64=36-64=-28<0$. This gives no real roots, not one repeated root. Eliminated.

$\Delta =10^2-64=100-64=36>0$. This gives two distinct real roots. Eliminated.

$\Delta =12^2-64=144-64=80>0$. This also gives two distinct real roots. Eliminated.

$b=8$: $\Delta =64-64=0$. This confirms exactly one repeated real root. The correct answer is $b=8$.

For $f(x)=3x^2+kx+12$ to have no real roots, the discriminant must be strictly negative: $\Delta =k^2-4(3)(12)<0$.

$k^2-144<0⟹k^2<144⟹-12<k<12.$

Among the options, only $k=10$ satisfies $-12<k<12$. Values $k=14$, $k=12$, and $k=-12$ are on or outside the boundary.

We need $\Delta =k^2-144<0$, i.e. $k^2<144$, which means $|k|<12$.

$14^2=196>144$, so $\Delta =196-144=52>0$. This gives two distinct real roots. Eliminated.

$12^2=144$, so $\Delta =0$. This gives one repeated real root, not no real roots. Eliminated.

$(-12{)}^2=144$, so $\Delta =0$. Same as above — one repeated root. Eliminated.

$k=10$: $\Delta =100-144=-44<0$. This confirms no real roots. The correct answer is $k=10$.