Question

If a + b=12 and ab= 6 find the
value of a2 +6² .​

Answer 1

Answer:

So the equation is a^2 + 6^2

We know, 6^2 is 36

, SO, a^2 + 36

Here,

a+b=12

ab=6.

We know,

a+b = 12,

so a = 12-b

Now,

Substitute the value of 'a' in the second equation, ab = 6

SO,

(12-b)*b = 6

b^2-12b = 6

SO,

b^2 - 12b - 6 = 0

using the Quadratic Formula here,

a = 1, b = -12, and c = -6

b = −b ± b^2 − 4ac/2a

x=−(−12)±(−12)2−4(1)(−6)/2(1)

x=12±144−−24/2

x = 12 ± 168/2

The discriminant b2−4ac>0

so, there are two real roots.

x = 12 ± 2$\sqrt{42\\}$

 

x = 12 ± 2$\sqrt{42\\}$

Simplify fractions and/or signs:

x=6 ± 2$\sqrt{42\\}$

which becomes

b = 12.4807

b = −0.480741

ab = 6

a*6+2$\sqrt{42\\}$ = 6

a = 6 ÷ 6+2$\sqrt{42\\}$

a = 0.31265

a = -0.86

Hope it Helps,

I think the question is a^2 + b^2, maybe you typed wrong?

Byeeee