ASK
Okay, so I need to figure out how to solve this problem: "Find the number of real solutions to the equation ( x^4 - 4x^2 + 3 = 0 )." Hmm, let's start by understanding what the problem is asking. It wants to know how many real numbers x satisfy this equation. Since it's a quartic equation (degree 4), there could be up to four real solutions, but it might have fewer.
First, maybe I can simplify the equation. Let me see... The equation is ( x^4 - 4x^2 + 3 = 0 ). I notice that all the terms are even powers of x. That makes me think of substituting ( y = x^2 ), which is a common technique for equations with even exponents. Let's try that.
If I let ( y = x^2 ), then ( x^4 = y^2 ). Substituting into the original equation, it becomes:
( y^2 - 4y + 3 = 0 ).
Okay, so now we have a quadratic equation in terms of y. That should be easier to solve. Let's factor this quadratic. Looking for two numbers that multiply to 3 and add up to -4. Hmm, factors of 3 are 1 and 3. Since the middle term is -4y, maybe both numbers are negative. So, -1 and -3 multiply to 3 and add up to -4. Yes, that works.
So, factoring the quadratic: ( (y - 1)(y - 3) = 0 ). Therefore, the solutions for y are y = 1 and y = 3.
But wait, y was substituted for ( x^2 ), so now we have to go back and solve for x. Let's take each solution for y and set them equal to ( x^2 ).
First, for y = 1: ( x^2 = 1 ). Solving for x, we take the square root of both sides. So, ( x = \sqrt{1} ) or ( x = -\sqrt{1} ), which gives x = 1 and x = -1. That's two solutions.
Next, for y = 3: ( x^2 = 3 ). Similarly, taking the square root of both sides, ( x = \sqrt{3} ) or ( x = -\sqrt{3} ). So, two more solutions: x = √3 and x = -√3.
Adding those up, we have four solutions in total: 1, -1, √3, -√3. All of these are real numbers because the square roots of 1 and 3 are real. Therefore, the equation has four real solutions.
Wait a second, let me double-check to make sure I didn't make a mistake. Let's plug each of these solutions back into the original equation to verify.
Starting with x = 1: ( (1)^4 - 4*(1)^2 + 3 = 1 - 4 + 3 = 0. Correct.
x = -1: ( (-1)^4 - 4*(-1)^2 + 3 = 1 - 4 + 3 = 0. Correct.
x = √3: ( (√3)^4 - 4*(√3)^2 + 3. Let's compute each term:
(√3)^4 = (3)^2 = 9
https://www.18122.com/url/usdt/usdt3491.jpeg
4*(√3)^2 = 4*3 = 12
So, 9 - 12 + 3 = 0. Correct.
x = -√3: Same as above because even powers make it positive. So, same result: 9 - 12 + 3 = 0. Correct.
All four solutions check out. So, the equation indeed has four real solutions.
Alternatively, maybe I can graph the function ( f(x) = x^4 - 4x^2 + 3 ) to visualize how many times it crosses the x-axis. Since it's a quartic with a positive leading coefficient, the ends go to positive infinity as x approaches both positive and negative infinity. The function will have a "W" shape or possibly two minima and one maximum, depending on the coefficients.
To find critical points, take the derivative: ( f'(x) = 4x^3 - 8x ). Setting this equal to zero: ( 4x^3 - 8x = 0 ). Factor out 4x: ( 4x(x^2 - 2) = 0 ). So, critical points at x = 0 and x = ±√2.
Evaluating f(x) at these critical points:
At x = 0: ( 0 - 0 + 3 = 3. So, the point (0, 3).
At x = √2: ( (√2)^4 - 4*(√2)^2 + 3 = (4) - 4*(2) + 3 = 4 - 8 + 3 = -1. Similarly, at x = -√2, same value: -1.
So, the function has a local maximum at x=0 (value 3) and local minima at x=±√2 (value -1). Since the minima are below the x-axis and the function tends to infinity on both ends, the graph must cross the x-axis four times: twice between the left end and -√2, twice between √2 and the right end. Wait, but actually, since the minima are at -1, which is below the x-axis, and the function comes down from +infinity to -1, then goes back up to +infinity. So, it should cross the x-axis twice on the left of -√2, and twice on the right of √2? Wait, no. Wait, the critical points are at x=0 (local max) and x=±√2 (local min). Let me sketch mentally:
From the left (x approaching -infty), f(x) approaches +infty. Then, the function decreases to a local minimum at x=-√2 (value -1), then increases to a local maximum at x=0 (value 3), then decreases again to a local minimum at x=√2 (value -1), and finally increases to +infty as x approaches +infty.
So, between -infty and -√2, the function goes from +infty down to -1. So, it must cross the x-axis once on the left of -√2. Then, from -√2 to 0, it goes from -1 up to 3, crossing the x-axis once. Then, from 0 to √2, it goes from 3 down to -1, crossing the x-axis once. Then, from √2 to +infty, it goes from -1 up to +infty, crossing the x-axis once. Wait, that would be four crossings total. But wait, each crossing between two critical points? Wait, actually, if the function is going from +infty to -1 (at x=-√2), crosses the x-axis once. Then from -1 to 3 (at x=0), crosses once. Then from 3 to -1 (at x=√2), crosses once. Then from -1 to +infty, crosses once. So, total four crossings. That matches our algebraic solution. So, four real roots.
Alternatively, maybe using the Intermediate Value Theorem. Let's check the sign of f(x) at various points.
For example, at x = -2: f(-2) = 16 - 16 + 3 = 3 > 0.
At x = -√2 ≈ -1.414: f(-√2) = -1 < 0.
At x = -1: f(-1) = 1 - 4 + 3 = 0.
At x = 0: f(0) = 3 > 0.
At x = 1: f(1) = 0.
At x = √2 ≈ 1.414: f(√2) = -1 < 0.
At x = 2: f(2) = 16 - 16 + 3 = 3 > 0.
So, between x=-2 and x=-√2, f(x) goes from 3 to -1, crossing zero once (which we already found at x=-√3 ≈ -1.732, which is between -2 and -√2 ≈ -1.414). Wait, but x=-√3 is approximately -1.732, which is between -2 and -√2. So, that's one crossing. Then between x=-√2 and x=0, f(x) goes from -1 to 3, crossing zero once at x=-1. Then between x=0 and x=√2, f(x) goes from 3 to -1, crossing zero once at x=1. Then between x=√2 and x=2, f(x) goes from -1 to 3, crossing zero once at x=√3. So, four crossings in total. That confirms again four real solutions.
Another way to check is using Descartes' Rule of Signs. For the polynomial ( x^4 - 4x^2 + 3 ), let's count the number of sign changes in the coefficients. The terms are +x^4, -4x^2, +3. So, from +x^4 to -4x^2: one sign change. From -4x^2 to +3: another sign change. So, two sign changes. Descartes' Rule says the number of positive real roots is either equal to the number of sign changes or less by an even number. So, two or zero positive real roots. Similarly, for negative real roots, we substitute x with -x: ( (-x)^4 - 4(-x)^2 + 3 = x^4 -4x^2 +3, which is the same as the original polynomial. So, same number of sign changes: two. Therefore, two or zero negative real roots. But we already found two positive (1 and √3) and two negative (-1 and -√3), so total four, which fits Descartes' Rule as two sign changes for positive and two for negative, each giving two real roots.
Alternatively, using the discriminant for quartic equations, but that might be more complicated. Since we already factored it into quadratics and found real roots, that's sufficient.
Wait, another thought: maybe using substitution z = x^2, then the equation becomes z^2 -4z +3 =0, which we solved as z=1 and z=3. Since both z=1 and z=3 are positive real numbers, each gives two real roots for x (positive and negative square roots). If z had been negative, those would not contribute real solutions. But since both are positive, we get four real solutions. So that's another way to see it.
So, all methods point to four real solutions. Therefore, the answer should be 4.
The equation ( x^4 - 4x^2 + 3 = 0 ) can be solved by substituting ( y = x^2 ), transforming it into the quadratic equation ( y^2 - 4y + 3 = 0 ). Factoring gives ( (y - 1)(y - 3) = 0 ), yielding solutions ( y = 1 ) and ( y = 3 ). Substituting back:
For ( y = 1 ): ( x^2 = 1 ) leads to ( x = \pm 1 ).
For ( y = 3 ): ( x^2 = 3 ) leads to ( x = \pm \sqrt{3} ).
Each substitution yields two real solutions, resulting in a total of four real solutions: ( x = 1, -1, \sqrt{3}, -\sqrt{3} ).
Answer: The equation has (\boxed{4}) real solutions.
页:
[1]