Example Question - finding roots
Here are examples of questions we've helped users solve.
Solving Quadratic Equation
Chúng ta cần giải phương trình sau: \[ \frac{x^2}{16} = \frac{25}{4} \] Đầu tiên, ta nhân cả hai vế của phương trình với 16 để hủy mẫu số ở vế trái: \[ x^2 = \frac{25}{4} \times 16 \] Tính tiếp tục, ta có: \[ x^2 = 25 \times 4 \] \[ x^2 = 100 \] Tiếp theo, ta sẽ lấy căn bậc hai cho cả hai vế của phương trình: \[ x = \pm\sqrt{100} \] \[ x = \pm10 \] Vậy nghiệm của phương trình là \( x = 10 \) hoặc \( x = -10 \).
Solving a Cubic Polynomial Equation by Finding Roots
Sure, let's solve the equation \( h(x) = 3x^3 + 12x^2 + 3x - 18 \) by finding its roots. First, we try to identify if there is any common factor among the terms. Since \( 3 \) is a common factor, we can factor it out: \[ h(x) = 3(x^3 + 4x^2 + x - 6) \] Now, we need to factor the cubic polynomial inside the parentheses. To factor a cubic polynomial, we can try to find at least one real root by inspecting the polynomial or by using the rational root theorem, which suggests that any rational root, written in the form \( p/q \), should be a divisor of the constant term (-6) divided by a divisor of the leading coefficient (1). The possible rational roots are therefore the divisors of -6: \( \pm 1, \pm 2, \pm 3, \pm 6 \). We can perform synthetic division or use direct substitution to check each of these possible roots to find an actual root. If we find one root, say \( r \), then we can factor the cubic polynomial as \( (x - r) \) times a quadratic polynomial. Let's try some values: For \( x = 1 \): \[ 1^3 + 4(1)^2 + 1 - 6 = 1 + 4 + 1 - 6 = 0 \] This means that \( x = 1 \) is a root of the polynomial. We can perform synthetic division or use polynomial long division to divide the cubic term by \( (x - 1) \) to find the remaining quadratic. \[ x^3 + 4x^2 + x - 6 = (x - 1)(x^2 + 5x + 6) \] Next, we factor the quadratic polynomial: \[ x^2 + 5x + 6 = (x + 2)(x + 3) \] So now, we have: \[ h(x) = 3(x - 1)(x + 2)(x + 3) \] Therefore, the roots of \( h(x) \) are \( x = 1 \), \( x = -2 \), and \( x = -3 \). These are the values of \( x \) that will make the original equation \( h(x) = 0 \).
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