A Nice Algebra Problem | Math Olympiad x = ?

Añadido: 2026-05-31

[00:00:00]Hello everyone. Welcome to how to solve this very nice quadratic equation. 3 * x ^ 4 + x cubed + 3x squared - 10x - 12 = 0. Our job is to find all possible values of x. So, let's start. If we divide this constant in -12 by this leading coefficient 3, if we divide -12 by 3, we get -4.

[00:00:31]And factors of -4 are plus minus 1, plus minus 2, and plus minus 4. If we check a plus minus 1 in this equation, plus minus 1 is not a solution. If we check plus minus 2, this is not a solution. And this a plus minus 4 is not a solution. So, let's use another trick to find a solution of this quadratic equation.

[00:01:01]We suppose that this quadratic expression at the left-hand side is a product of two quadratic factors. Let's say the first factor is x squared plus a times x plus b. And the second factor is 3 times x squared plus c times x plus d equal to 0.

[00:01:29]Now, expand these expressions. x squared times 3x squared will become 3x to the power 4. And x squared times 3x will become plus cx cubed. And x squared times d plus dx squared.

[00:01:49]And ax times 3x squared will become plus 3 times ax cubed.

[00:01:58]ax * cx will become plus a * c x squared.

[00:02:06]And ax * d will become plus adx.

[00:02:13]b * 3x squared will become 3 * bx squared.

[00:02:19]And b * cx will become plus bc x.

[00:02:25]And b * d will become plus bd equal to zero. If we combine like terms, then this will become 3x to the power four plus this 3a plus c 3a plus c times x cubed.

[00:02:48]Plus this ac plus 3b ac plus 3b plus b.

[00:02:58]a * c plus 3 * b plus d times x squared.

[00:03:07]Plus this ad and plus bc.

[00:03:12]a * d plus bc times x.

[00:03:19]Plus bd equal to zero.

[00:03:26]Now, we compare the coefficients of this equation by the coefficients of this equation, we'll get the first equation 3a plus c equal to one.

[00:03:39]3 * a plus c equal to one.

[00:03:44]And we get the second equation ac plus 3b plus d equal to three.

[00:03:51]a * c plus 3 * B + D = 3 and we get the third equation AD + BC = -10.

[00:04:05]A * D + B * C = -10.

[00:04:12]10 And the fourth equation will be BD = -12.

[00:04:22]Fourth equation will be equal to BD = 12. For this -12, we have six combinations. The first one is 1 * -12.

[00:04:37]12 Second is 12 * -1.

[00:04:43]Third is 2 * -6.

[00:04:47]And fourth is uh 6 * -2.

[00:04:52]And fifth is 3 * -4.

[00:04:56]Sixth is uh 4 [snorts] * - 3.

[00:05:02]Since in the original equation we have the leading coefficient we have the leading coefficient 3. So [snorts] So first we try this fifth combination.

[00:05:19]3 * 4.

[00:05:24]If B * D = 3 * -4 it means that B is equal to 3 and uh D is equal to 4.

[00:05:40]And from this equation, C will be equal to 1 - 3 * A. So this is second equation will become a times c is 1 minus three times a plus three times a three times a b is this a three and d is -4 -4 equal to three.

[00:06:12]a times 1 is a a times -3a -3a squared three times three will become plus nine and this is -4 -4.

[00:06:26]Move this three to the left hand side this will become negative three equal to zero.

[00:06:33]Further simplify this will become -3a squared plus a and 9 minus 4 5 5 minus 3 plus 2 equal to zero. If we divide the whole equation by -1 this will become three times a squared minus a minus 2 equal to zero.

[00:07:00]>> [snorts] >> Now, this is a quadratic equation and it's factorable we write this a squared break this negative a as negative three times a plus two times a minus two equal to zero.

[00:07:15]From these two terms we can factor out three times a in bracket left a minus one. From these two terms we can factor out plus two in bracket left a minus one equal to zero.

[00:07:30]And this a minus one is a common factor so I factor out this a minus one.

[00:07:35]In bracket left three times a plus two equal to zero. Either this expression a minus one equal to zero or this expression three times a plus two equal to zero.

[00:07:54]From this equation we get the value of a equal to one and from this equation we get the value of a equal to negative two over three.

[00:08:07]Now first we try this value of a one.

[00:08:12]And using this equation we can find the value of c.

[00:08:18]C is equal to one minus three a and b is equal to three. B is equal to negative four.

[00:08:32]Since c is equal to one minus three a and a is one, c will be equal to one minus three and c will be equal to negative two.

[00:08:48]And the first we found that b is equal to three and d is equal to negative four.

[00:08:57]So [snorts] we have a is equal to one, b is equal to three c negative two and d is equal to four.

[00:09:08]Now we check these four values of a, b, c and d in the equation.

[00:09:17]In this equation ad plus bc equal to negative 10.

[00:09:28]We check these four values in the equation a times d plus b times c equal to 10.

[00:09:39]Since A is 1 and D is -4, this will become 1 * - 4.

[00:09:48]Plus B is 3 times C is -2 -2.

[00:09:56]Is this equal to 10?

[00:10:00]And 1 * -4 is -4.

[00:10:03]3 * -2 -6.

[00:10:06]Is this equal to 10?

[00:10:10]And -4 -6 is -10.

[00:10:14]This is equal to 10.

[00:10:18]Since the left-hand side is equal to the right-hand side, it means that these values of A, B, C, and D will work perfectly. So, we copy these values of A, B, C, and D.

[00:10:32]A is equal to 1 and B is equal to 3 and C is equal to -2 and D is equal to 4.

[00:10:48]And recall that recall that this quartic factor is a product of two quadratic factors x² + ax + b times the 3x² + cx + d.

[00:11:17]Recall that our quartic factor is a product of two quadratic factors x² + a times x + b times 3x² + cx + the d equal to zero.

[00:11:38]Since a is one and b is three.

[00:11:43]So, this will become x squared plus x plus the three times the And this will become 3 x squared. c is negative two, so this will become negative two x and d is negative four, this will become negative four. equal to zero. Either this expression x squared plus x plus three equal to zero or this expression 3 x squared minus 2 x minus 4 equal to zero.

[00:12:27]From this quadratic equation of According to quadratic formula, x will be equal to negative one plus minus i times root 11 over two.

[00:12:43]And according to quadratic formula this uh x will be equal to one plus minus root 13 over three.

[00:12:58]This is first and second and this is uh third and fourth. So, we have uh four [snorts] solutions for this equation. These two solutions are uh complex and these two solutions are uh real.