QuestionSeptember 19, 2025

Solve. 6x^wedge 4-7x^wedge 2+2=0

Solve. 6x^wedge 4-7x^wedge 2+2=0
Solve. 6x^wedge 4-7x^wedge 2+2=0

Solution
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Answer

x = \pm \sqrt{\frac{2}{3}},\ \pm \frac{1}{\sqrt{2}} Explanation 1. Substitute y = x^2 Let y = x^2, so equation becomes 6y^2 - 7y + 2 = 0. 2. Use quadratic formula Apply \displaystyle y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} with a=6, b=-7, c=2. 3. Calculate discriminant (-7)^2 - 4(6)(2) = 49 - 48 = 1. 4. Solve for y y = \frac{7 \pm 1}{12} \implies y_1 = \frac{8}{12} = \frac{2}{3},\ y_2 = \frac{6}{12} = \frac{1}{2}. 5. Back-substitute x^2 = y and solve for x x^2 = \frac{2}{3} \implies x = \pm \sqrt{\frac{2}{3}}; x^2 = \frac{1}{2} \implies x = \pm \frac{1}{\sqrt{2}}.

Explanation

1. Substitute $y = x^2$<br /> Let $y = x^2$, so equation becomes $6y^2 - 7y + 2 = 0$.<br />2. Use quadratic formula<br /> Apply $\displaystyle y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ with $a=6$, $b=-7$, $c=2$.<br />3. Calculate discriminant<br /> $(-7)^2 - 4(6)(2) = 49 - 48 = 1$.<br />4. Solve for $y$<br /> $y = \frac{7 \pm 1}{12} \implies y_1 = \frac{8}{12} = \frac{2}{3},\ y_2 = \frac{6}{12} = \frac{1}{2}$.<br />5. Back-substitute $x^2 = y$ and solve for $x$<br /> $x^2 = \frac{2}{3} \implies x = \pm \sqrt{\frac{2}{3}}$; $x^2 = \frac{1}{2} \implies x = \pm \frac{1}{\sqrt{2}}$.
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