Question

X^2-|3x-4|=0 через дискриминант

Answer 1

Ответ:

Объяснение:

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Answer 2

Відповідь:за оберненою теоремою вієта

X1+x2=3

X1×x2=4

X1=-4

X2=1

Пояснення:

Answer 3

To solve the equation $x^2 - |3x-4| = 0$ using the discriminant, we first need to understand that the equation has two cases to consider: one where $|3x-4| = 3x-4$ and another where $|3x-4| = -(3x-4)$.

1. $|3x-4| = 3x-4$: In this case, we can rewrite the equation as $x^2 - (3x-4) = 0$. Simplify it: $x^2 - 3x + 4 = 0$.

2. $|3x-4| = -(3x-4)$: In this case, we can rewrite the equation as $x^2 + (3x-4) = 0$. Simplify it: $x^2 + 3x - 4 = 0$.

Now, both of these equations are quadratic equations in the form $ax^2 + bx + c = 0$, where: $a = 1$, $b = -3$, and $c = 4$.

The discriminant of a quadratic equation $ax^2 + bx + c = 0$ is given by $D = b^2 - 4ac$.

Let's calculate the discriminant for both cases:

1. For the equation $x^2 - 3x + 4 = 0$: $D_1 = (-3)^2 - 4(1)(4) = 9 - 16 = -7$.

2. For the equation $x^2 + 3x - 4 = 0$: $D_2 = (3)^2 - 4(1)(-4) = 9 + 16 = 25$.

Now, based on the values of the discriminant $D_1$ and $D_2$, we can determine the number of solutions for each case:

1. For $D_1 = -7$: Since the discriminant $D_1$ is negative, this quadratic equation has no real solutions. Thus, there are no solutions for the first case $|3x-4| = 3x-4$.

2. For $D_2 = 25$: Since the discriminant $D_2$ is positive, this quadratic equation has two real solutions. Thus, there are two solutions for the second case $|3x-4| = -(3x-4)$.

To find the specific values of $x$ for the second case, solve the quadratic equation $x^2 + 3x - 4 = 0$ using any appropriate method (factoring, quadratic formula, etc.).