Question

Find the least value of k for which the quadratic equation -3x^2 -6x +k=0 has real roots​

Answer 1

The least value of k would be -3

Step-by-step explanation:

Given qudratic equation,

$-3x^2-6x+k=0$

A quadratic equation $ax^2+bx+c=0$ has real roots,

If $b^2-4ac\geq 0$

$\implies b^2 \geq 0$

Here, b = -6, a = -3, c = k,

If $(-6)^2 \geq 4\times -3\times k$

$36\geq -12k$

Multiply both sides by -1/12,

We get

$-3\leq k$    ( a > b ⇒ ca < cb if c < 1 ∀ a, b, c ∈ R )

Hence, the least value of k for which the given quadratic equation has real roots would be -3.

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Answer 2

Answer:

Step-by-step explanation:

Given qudratic equation,

-3x^2-6x+k=0

A quadratic equation ax^2+bx+c=0 has real roots,

If b^2-4ac\geq 0

\implies b^2 \geq 0

Here, b = -6, a = -3, c = k,

If (-6)^2 \geq 4\times -3\times k

36\geq -12k

Multiply both sides by -1/12,

We get

-3\leq k