Question

If the roots of the equation 12x2 + mx +5 = 0 are in the ratio 5:4, then m
1) 9√3
2) +-9√3
3) 6√3
4)+- 6√3​

Answer 1

Answer:

The correct option is (2) $\pm9\sqrt{3}$

Step-by-step explanation:

The quadratic equation is: $12x^{2}+mx+5=0$

The roots of a quadratic equation are:

$x=[\frac{-b+\sqrt{b^{2}-4ac} }{2a} , \frac{-b-\sqrt{b^{2}-4ac} }{2a}]$

Here $a=12,\ b=m\ and\ c=5$

Compute the roots as follows:

$x=[\frac{-m+\sqrt{m^{2}-(4\times12\times5)} }{2\times12} , \frac{-m-\sqrt{m^{2}-(4\times12\times5)} }{2\times12}]\\=[\frac{-m+\sqrt{m^{2}-240} }{24}, \frac{-m-\sqrt{m^{2}-240} }{24}]$

It is provided that the roots are in the ration 5:4

Compute the value of m as follows:

$\frac{(\frac{-m+\sqrt{m^{2}-240} }{24})}{(\frac{-m-\sqrt{m^{2}-240} }{24})} =\frac{5}{4} \\\frac{-m+\sqrt{m^{2}-240}}{-m-\sqrt{m^{2}-240}} =\frac{5}{4} \\-4m+4\sqrt{m^{2}-240}=-5m-5\sqrt{m^{2}-240}\\m=-9\sqrt{m^{2}-240}$

Squaring both sides:

$(m)^{2}=(-9\sqrt{m^{2}-240})^{2}\\m^{2}=81(m^{2}-240)\\m^{2}=81m^{2}-19440\\80m^{2}=19440\\m^{2}=243\\m=\sqrt{243} \\=\sqrt{81\times3} \\=\pm9\sqrt{3}$

Thus, the correct value of m is given by option (2).