Question

3a(4a-8)=-2 find 27a2+3/4a2+4=?​

Answer 1

$\frac{27a^2+3}{4a^2+4}=\frac{3(105+18\sqrt{30}}{2(34+4\sqrt{30}}$ or $\frac{27a^2+3}{4a^2+4}=\frac{3(105-18\sqrt{30}}{2(34-4\sqrt{30}}$

Step-by-step explanation:

$3a(4a-8)=-2$

$12a^2-24a=-2$

$12a^2-24a+2=0$

$6a^2-12a+1=0$

It is quadratic equation in a.

Discriminant,$D=b^2-4ac$ for equation $ax^2+bx+c=0$

Where a= coefficient of $x^2$

b=Coefficient of x

c=Constant term

$D=(-12)^2-4(6)(1)=144-24=120$

Quadratic formula:

$x=\frac{-b\pm \sqrt D}{2a}$

Using this formula then we get

$a=\frac{12\pm\sqrt{120}}{2(6)}$

$a=\frac{12\pm 2\sqrt{30}}{12}$

$a=1+\frac{\sqrt{30}}{6}$ or $a=1-\frac{\sqrt{30}}{6}$

Substitute $a=1+\frac{\sqrt{30}}{6}$

$\frac{27a^2+3}{4a^2+4}=\frac{27(1+\frac{\sqrt{30}}{6})^2+3}{4(1+\frac{\sqrt{30}}{6})^2+4}$

$\frac{27a^2+3}{4a^2+4}=\frac{27(1+\frac{30}{36}+\frac{\sqrt{30}}{3})+3}{4(1+\frac{30}{36}+\frac{\sqrt{30}}{3})+4}=\frac{27+\frac{45}{2}+9\sqrt{30}+3}{4+\frac{10}{3}+\frac{4\sqrt{30}}{3}+4}$

$\frac{27a^2+3}{4a^2+4}=\frac{\frac{105+18\sqrt{30}}{2}}{\frac{34+4\sqrt{30}}{3}}=\frac{105+18\sqrt{30}}{2}\times\frac{3}{34+4\sqrt{30}}$

$\frac{27a^2+3}{4a^2+4}=\frac{3(105+18\sqrt{30})}{2(34+4\sqrt{30})}$

Now, substitute $a=1-\frac{\sqrt{30}}{6}$

$\frac{27a^2+3}{4a^2+4}=\frac{27(1+\frac{30}{36}-\frac{\sqrt{30}}{3})+3}{4(1+\frac{30}{36}-\frac{\sqrt{30}}{3})+4}$

$\frac{27a^2+3}{4a^2+4}=\frac{27+\frac{45}{2}-9\sqrt{30}+3}{4+\frac{10}{3}-\frac{4\sqrt{30}}{3}+4}$

$\frac{27a^2+3}{4a^2+4}=\frac{\frac{105-18\sqrt{30}}{2}}{\frac{34-4\sqrt{30}}{3}}$

$\frac{27a^2+3}{4a^2+4}=\frac{3(105-18\sqrt{30})}{2(34-4\sqrt{30})}$

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Answer 2

Step-by-step explanation:

Answer is 103. See this image.