Example Question - quadratic equation solution
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Quadratic Equation Solution Using the Quadratic Formula
The provided image shows the quadratic formula, which is used to find the solutions of a quadratic equation \( ax^2 + bx + c = 0 \). The solutions using the quadratic formula are: \[ x = \frac{{-b \pm \sqrt{b^2 - 4ac}}}{{2a}} \] This formula calculates the roots of any quadratic equation where \( a \neq 0 \). The term inside the square root, \( b^2 - 4ac \), is known as the discriminant, and it determines the nature of the roots: - If \( b^2 - 4ac > 0 \), there are two distinct real roots. - If \( b^2 - 4ac = 0 \), there is one real root (a repeated root). - If \( b^2 - 4ac < 0 \), there are two complex roots. To solve a specific quadratic equation using this formula, one should substitute the values of \( a \), \( b \), and \( c \) from the equation into the formula and simplify.
Solving Quadratic Equation
确定 a=3，b=5 和 c=-2。 使用二次公式：x = [-b ± √(b²-4ac)] / (2a) 计算 b² - 4ac = 5² - 4×3×(-2) = 25 + 24 = 49 求根 √(b²-4ac) = √49 = 7 代入公式：x = [-5 ± 7] / (2×3) 解得两个解： x1 = (-5 + 7) / 6 = 2 / 6 = 1/3 x2 = (-5 - 7) / 6 = -12 / 6 = -2 所以，方程的解为 x = 1/3 或 x = -2。
Quadratic Equation Solution
The given equation is a quadratic equation of the form ax^2 + bx + c = 0. To solve for x, we can use the quadratic formula: x = [-b ± sqrt(b^2 - 4ac)] / (2a) For our given equation: 3x^2 + 5x - 2 = 0 Here, a = 3, b = 5, and c = -2. Plugging these values into the quadratic formula gives us: x = [-5 ± sqrt(5^2 - 4 * 3 * (-2))] / (2 * 3) x = [-5 ± sqrt(25 + 24)] / 6 x = [-5 ± sqrt(49)] / 6 x = [-5 ± 7] / 6 This gives us two possible solutions for x: x = (-5 + 7) / 6 = 2 / 6 = 1/3 or x = (-5 - 7) / 6 = -12 / 6 = -2 Therefore, the solutions for x are x = 1/3 or x = -2.
Solving a Right-Angled Triangle Question Using Algebra
The question in the image involves a right-angled triangle, and it gives you two expressions for the lengths of the sides: \( x - 1 \) and \( 4x \), and for the hypotenuse, the expression is \( 5x - 9 \). According to the Pythagorean theorem, in a right-angled triangle, the sum of the squares of the two shorter sides is equal to the square of the longest side (the hypotenuse). The equation is often expressed as \( a^2 + b^2 = c^2 \), where \( a \) and \( b \) are the lengths of the two shorter sides and \( c \) is the length of the hypotenuse. Using the expressions provided for the lengths, we can create the following equation: \[ (x - 1)^2 + (4x)^2 = (5x - 9)^2 \] Now we'll expand and simplify this equation to solve for \( x \): \[ x^2 - 2x + 1 + 16x^2 = 25x^2 - 90x + 81 \] Combine like terms: \[ 17x^2 - 2x + 1 = 25x^2 - 90x + 81 \] Bringing all terms to one side gives us: \[ 0 = 25x^2 - 17x^2 - 90x + 2x + 81 - 1 \] Simplify and combine like terms: \[ 0 = 8x^2 - 88x + 80 \] To make it easier to solve, divide the entire equation by 8: \[ x^2 - 11x + 10 = 0 \] Now we can factor this quadratic equation: \[ (x - 10)(x - 1) = 0 \] Setting each factor equal to zero gives us the possible values of \( x \): \[ x - 10 = 0 \quad \Rightarrow \quad x = 10 \] \[ x - 1 = 0 \quad \Rightarrow \quad x = 1 \] Since \( x = 1 \) would result in one of the sides being zero length (since \( x - 1 = 0 \)), which is not possible for the side of a triangle, the only valid solution for \( x \) is: \[ x = 10 \]
Solving a Quadratic Equation by Factoring
I see an equation written on a piece of paper: \[ g = 5x^2 - 10x + 5 \] To solve this equation for \( x \), first, let's see if we can factor it. A quadratic equation in the form \( ax^2 + bx + c = 0 \) can sometimes be factored into the form: \[ (dx + e)(fx + h) = 0 \] The given equation \( g = 5x^2 - 10x + 5 \) is not set equal to zero, so we can't solve for \( x \) until we do so. Let's assume you meant to solve for \( x \) when \( g = 0 \). That would give us: \[ 0 = 5x^2 - 10x + 5 \] Now, we clearly see that each term is divisible by 5. Let's factor out a 5: \[ 0 = 5(x^2 - 2x + 1) \] The quadratic inside the parentheses is a perfect square trinomial and factors as: \[ 0 = 5(x - 1)^2 \] Now we can find the value of \( x \) that makes the equation true by setting \( (x - 1)^2 \) equal to zero: \[ (x - 1)^2 = 0 \] Taking the square root of both sides gives us: \[ x - 1 = 0 \] So the solution for \( x \) is: \[ x = 1 \] Thus, the equation \( 5x^2 - 10x + 5 = 0 \) has a repeated real solution, which is \( x = 1 \).
Solving a Quadratic Equation
The equation in the image is: \[ 2r + 2 \sqrt{r^2} + \frac{r}{2} = 10 \] To solve this equation for \( r \), follow these steps: 1. Simplify the square root term. Since \( \sqrt{r^2} \) is the square root of a square, it simplifies to \( r \) if \( r \geq 0 \) (assuming \( r \) is non-negative, as you cannot take the square root of a negative number in real numbers). So, you have: \[ 2r + 2r + \frac{r}{2} = 10 \] \[ 4r + \frac{r}{2} = 10 \] 2. Combine like terms by finding a common denominator for the \( r \)-terms. Multiply each term by 2 to get rid of the fraction: \[ 2 \cdot 4r + 2 \cdot \frac{r}{2} = 2 \cdot 10 \] \[ 8r + r = 20 \] \[ 9r = 20 \] 3. Divide both sides by 9 to solve for \( r \): \[ r = \frac{20}{9} \] Therefore, \( r = \frac{20}{9} \) or approximately 2.2222.
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