How to Tackle a Nice Radical Equation with Ease?

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How to Tackle a Nice Radical Equation with Ease?

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Күн бұрын

How to Tackle a Nice Radical Equation with Ease?
Unlock the secrets of radical equation solving with our comprehensive guide! 🧠💡 In this tutorial, we break down the steps to tackle even the nicest (and trickiest) radical equations with ease. Whether you're a math enthusiast or looking to ace your next test, join us on this journey to mastering radical equations. No more confusion - just clarity! 🎓✨
Topics Covered:
1. Understanding the basics of radical equations.
2. Analyzing the unique properties and substitution of the given equation.
3. Step-by-step approach to solving the radical equation.
4. Tips and tricks for handling tricky radicals with ease.
5. Algebraic identities and manipulations while solving equations.
Timestamps:
0:00 Introduction
0:38 Algebraic manipulations
1:40 Substitution
3:22 Solving Quartic equation
7:46 Finding 'a'
9:02 Real Solutions
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Thanks for Watching !!
@infyGyan

Пікірлер: 8

@tunneloflight Ай бұрын

Nice. The other added constraint {which turns out not to be needed} is that x < 4 => a < 16 => y

@gkwugqbfig2vjg332 7 ай бұрын

Effectivement ! C'est pas fassil! Merci.

@alibhukoo5400 7 ай бұрын

X=3-2squareroot2 and X=9-4squareroot2

@user-ji5su2uq9m 6 ай бұрын

Another method From 1:35 sqrt(x) * sqrt(x + 4) * (sqrt(x) + 2) * (sqrt(x) - 2) = -7 [sqrt(x) * (sqrt(x) + 2)] * [sqrt(x+4) * (sqrt(x) - 2)] = -7 [x + 2 * sqrt(x)] * [x + 2 * sqrt(x) - 8] = -7 let t = x + 2*sqrt(x) => t * (t - 8) = -7 => t=1 or t=7 (case t=1) x + 2*sqrt(x) = 1 => x = 3-2*sqrt(2) (case t=7) x + 2*sqrt(x) = 7 => x = 9-4*sqrt(2)

@libardouribe7617 7 ай бұрын

Is perfect...

@alibhukoo5400 7 ай бұрын

This question is too tricky ❤❤😊

@NadiehFan 7 ай бұрын

No, merely standard procedures used in math competitions. Just get rid of the radical by substituting √x = a, then factor to get a(a + 4)(a + 2)(a − 2) = −7 From this point you can also proceed in a slightly different way than the method shown in the video. Note that the linear factors at the left hand side have a second term −2, 0, 2, 4, so we can try to create symmetry. By rewriting this as (a − 2)(a − 0)(a + 2)(a + 4) = −7 and multiplying the outer two factors and the inner two factors we have (a² + 2a − 8)(a² + 2a) = −7 which we can rewrite as (a² + 2a − 4 − 4)(a² + 2a − 4 + 4) = −7 and applying the difference of two squares identity to the left hand side this gives (a² + 2a − 4)² − 4² = −7 so (a² + 2a − 4)² = 9 Therefore, we have a² + 2a − 4 = 3 ⋁ a² + 2a − 4 = −3 and so a² + 2a − 7 = 0 ⋁ a² + 2a − 1 = 0 or (a + 1)² = 8 ⋁ (a + 1)² = 2 Then it is just a matter of solving these quadratic eqations. Of course, since a = √x, you only want the positive roots, which are a = −1 + 2√2 and a = −1 + √2 and then, since x = a², it is just a matter of squaring these to find x = 9 − 4√2 ⋁ x = 3 − 2√2 as the solutions of the original equation. As you can see, this can be solved with much simpler algebra compared to what is done in the video.