Question

6x square - 6(a+b)x + (4/3a square + 10/3ab + 4/3b square) = 0 solve for x.

Answer 1

hope it will help you

Answer 2

$x=\dfrac{2a+b}{3} ,\dfrac{a+2b}{3}$

Step-by-step explanation:

We have,

$6x^{2} -6(a+b)x+(\dfrac{4}{3}b^{2} +\dfrac{10}{3} ab+\dfrac{4}{3}b^{2})=0$

To find, the value of x = ?

Divided by 2, we get

$3x^{2} -3(a+b)x+(\dfrac{2}{3}b^{2} +\dfrac{5}{3} ab+\dfrac{2}{3}b^{2})=0$

∴ $D=b^{2} -4ac$

$=[3(a+b)]^{2} -4(3)(\dfrac{2}{3}b^{2} +\dfrac{5}{3} ab+\dfrac{2}{3}b^{2})$

$=9(a^{2} +2ab+b^{2})-4(2a^{2} +5ab+2b^{2})$

$=9a^{2} +18ab+9b^{2}-8a^{2} -10ab-2b^{2}$

$=a^{2} +b^{2} -2ab$

=$=(a-b)^{2}$

$x=\frac{-b±\sqrt{D}}{2a}$

$x=\frac{-3(a+b)±\sqrt{(a-b)^2}}{2\times 3}$

$=\dfrac{3a+3b+a-b}{6} ,\dfrac{3a+3b-a+b}{6}$

$=\dfrac{4a+2b}{6} ,\dfrac{2a+4b}{6}$

$=\dfrac{2(2a+b)}{6} ,\dfrac{2(a+2b)}{6}$

$=\dfrac{2a+b}{3} ,\dfrac{a+2b}{3}$

Hence, $x=\dfrac{2a+b}{3} ,\dfrac{a+2b}{3}$