Question Video: Solving a Disguised Quadratic | Nagwa Question Video: Solving a Disguised Quadratic | Nagwa

# Question Video: Solving a Disguised Quadratic Mathematics • Second Year of Preparatory School

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Solve 36π₯β΄ β 97π₯Β² + 36 = 0.

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### Video Transcript

Solve 36π₯ to the power four minus 97π₯ squared plus 36 equals zero.

Notice the presence of a fourth power here. This tells us that this is a fourth-degree, or quartic, equation. There does exist an algorithm analogous to the quadratic formula for solving quartic equations, but it is extremely complicated and difficult. Rather, when presented with higher-degree equations such as this, one should be on the lookout for some kind of trick to simplify the situation.

In particular, one thing to look out for is situations in which all of the terms have even degree. That is the case here. In such cases, one can use the laws of exponents to rewrite the equation in terms of π₯ squared. Making the substitution π’ equals π₯ squared, we have reduced the problem to a quadratic, which we can solve. You can use the quadratic formula or stare at factor pairs of 36 until you see how it factors. Either way, we find the two solutions π’ equals nine over four and π’ equals four over nine.

Beware though! To solve the original equation, we need to find the values of π₯ which make it true, and we havenβt done that yet. Since we made the substitution π’ equals π₯ squared, we have that π₯ equals the square root of π’. Thus, we have the four solutions root nine over four, negative root nine over four, root four over nine, and negative root four over nine. Observe that these are all ratios of perfect squares and simplify down to the solutions three over two, negative three over two, two over three, and negative two over three.

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