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

2 Answers

5 votes

Answer: x₁=6b², x₂=6a².

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\\√(D)=√((6a^2-6b^2)^2)=|6a^2-6b^2|=б(6a^2-6b^2).\\\displaystyle x_(1,2)=(-(-6*(a^2+b^2))б(6a^2-6b^2))/(2) .\\x_1=(6a^2+6b^2-6a^2+6b^2)/(2) \\x_1=(12b^2)/(2) \\x_1=6b^2.\\x_2=(6a^2+6b^2+6a^2-6b^2)/(2) \\x_2=(12a^2)/(2) \\x_2=6a^2.

User Peterxwl
by
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3 votes

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}

User Arise
by
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