If the quadratic equation a x2+b x+6=0 does not have two distinct real roots, then the least value of Za + b is -

Question

If the quadratic equation a x2+b x+6=0 does not have two distinct real roots, then the least value of Za + b is - (A) 2 (B) -3 (C) -6 (D) 1

Tardigrade Admin · Accepted Answer

Put x=0 we get f( x )=f(0)>6 graph is shown as ⇒ f(2) ≥ 0 4 a +2 b +6 ≥ 0 2 a + b ≥-3 ⇒ text least value =-3 <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/80dff272549d10d48644b791d916fd65-.png" />

Q. If the quadratic equation $a x^{2} + b x + 6 = 0$ does not have two distinct real roots, then the least value of $Z a + b$ is -

2

-3

-6

1

Put $x = 0$
we get $f (x) = f (0) > 6$
graph is shown as
$\Rightarrow f (2) \geq 0$
$4 a + 2 b + 6 \geq 0$
$2 a + b \geq - 3$
$\Rightarrow least value = - 3$

Q. If the quadratic equation ax2+bx+6=0 does not have two distinct real roots, then the least value of Za+b is -

2

-3

-6

1

Put x=0 we get f(x)=f(0)>6 graph is shown as ⇒f(2)≥0 4a+2b+6≥0 2a+b≥−3 ⇒ least value =−3

Q. If the quadratic equation $a x^{2} + b x + 6 = 0$ does not have two distinct real roots, then the least value of $Z a + b$ is -

Put $x = 0$
we get $f (x) = f (0) > 6$
graph is shown as
$\Rightarrow f (2) \geq 0$
$4 a + 2 b + 6 \geq 0$
$2 a + b \geq - 3$
$\Rightarrow least value = - 3$