Question

if the roots of the equation 12x^2+mx+5=0 are real and equal then m is.

Answer 1

Let the roots be α,β

αβ=32

α+β=−m12

αβ=512

α=32β⇒32β2=512

⇒β2=1036⇒β=−10−−√6

α=32β=−10−−√4

m=−12(α+β)

=12(10−−√6+10−−√4)

=510−−√

Hence (D) is the correct answer.

Answer 2

Answer:

The correct answer is $&\pm 4 \sqrt{15}$.

Step-by-step explanation:

Given:

The roots of the given equation is $12 x^{2}+m x+5=0$ are real and equal.

To find:

the value of m.

Step 1

Given that roots of the equation $12 x^{2}+m x+5=0$ are real and equal.

Therefore,  $m^{2}-4 \times 12 \times 5=0$

$&\Rightarrow m^{2}=240$

$&\Rightarrow m^{2}=16 \times 15$

Simplify $m^{2}=240$

Step 2

For $x^{2}=f(a)$ the solutions are $x=\sqrt{f(a)},-\sqrt{f(a)}$

Where, $m=\sqrt{240}$  and $m=-\sqrt{240}$

$m^{2}-4 \times 12 \times 5=0$

Add $4 \times 12 \times 5$ to both sides

$m^{2}-4 \times 12 \times 5+4 \times 12 \times 5=0+4 \times 12 \times 5$

Simplifying the above equation as

$\sqrt{240}=4 \sqrt{15}$

$-\sqrt{240}=-4 \sqrt{15}$

$&m=\pm 4 \sqrt{15}$

$m=4 \sqrt{15}$  and $m=-4 \sqrt{15}$

Therefore, the correct answer is $&\pm 4 \sqrt{15}$.

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