Question

Solve for x: 12x^2 – 6 (a^2 + b^2)x + 3a^2b^2 = 0

Answer 1

Answer: x₁=6b²,  x₂=6a².

Step-by-step explanation:

$12x^2-6*(a^2+b^2)+3a^2b^2=0\\D=(6*(a^2+b^2))^2-4*12*3a^2b^2\\D=6^2*(a^2+b^2)^2-144a^2b^2\\D=36*(a^4+2a^2b^2+b^4)-144a^2b^2\\D=36a^4+72a^2b^2+b^4-144a^2b^2\\D=36a^4-72a^2b^2+36b^4\\D=(6a^2)^2-2*6a^2*6b^2+(6b^2)^2\\D=(6a^2-6b^2)^2\\\sqrt{D}=\sqrt{(6a^2-6b^2)^2}=|6a^2-6b^2|=б(6a^2-6b^2).\\\displaystyle x_{1,2}=\frac{-(-6*(a^2+b^2))б(6a^2-6b^2)}{2} .\\x_1=\frac{6a^2+6b^2-6a^2+6b^2}{2} \\x_1=\frac{12b^2}{2} \\x_1=6b^2.\\x_2=\frac{6a^2+6b^2+6a^2-6b^2}{2} \\x_2=\frac{12a^2}{2} \\x_2=6a^2.$

Answer 2

Step-by-step explanation:

$12 {x}^{2} - 6( {a}^{2} + {b}^{2} )x + 3 {a}^{2} {b}^{2} = 0$

Using quadratic formula, we get

$x = 6( {a}^{2} + {b}^{2} )± \frac{ \sqrt{( - 6) {}^{2}( {a}^{2} + {b}^{2}) {}^{2} - 4(12)(3 {a}^{2} {b}^{2} )} }{24}$

$x = 6( {a}^{2} + {b}^{2} )± \frac{ \sqrt{36( {a}^{2} + {b}^{2} ) {}^{2} - 144 {a}^{2} {b}^{2} } }{24}$

$x = 6( {a}^{2} + {b}^{2} )± \frac{ \sqrt{36( ({a}^{2} + {b}^{2}) {}^{2} - 4 {a}^{2} {b}^{2} }) }{24}$

$x = 6( {a}^{2} + {b}^{2} )± \frac{6 \sqrt{ ({a}^{2} + {b}^{2} ) {}^{2} - 4 {a}^{2} {b}^{2} } }{24}$

$x = 6( {a}^{2} + {b}^{2} )± \frac{ \sqrt{( {a}^{2} + {b}^{2} ) {}^{2} - 4 {a}^{2} {b}^{2} } }{4}$