Example Question - factoring quadratic expressions
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Solving Quadratic Equations
Кешіріңіз, суретте көрсетілген теңдеуді келесі тәртіппен шешеміз: \( (k^2 - 16) \cdot (2k + 32) = 0 \) Бұл екілік көбейтінді түріндегі теңдеу, сондықтан біз көбейткіштердің әрқайсысын жеке-жеке тең нөлге теңестіреміз: 1) \( k^2 - 16 = 0 \) Бұл бірінші дәрежелі теңдеу, оны шешу үшін: \( k^2 = 16 \) \( k = \pm\sqrt{16} \) \( k = \pm4 \) 2) \( 2k + 32 = 0 \) Бұл екінші бірінші дәрежелі теңдеу, оны шешу үшін: \( 2k = -32 \) \( k = -32 / 2 \) \( k = -16 \) Демек, бұл теңдеудің үш шешімі бар: \( k = 4, k = -4 \), және \( k = -16 \).
Solving a Quadratic Equation by Factoring
The equation in the image is a quadratic equation, which can be solved by factoring, completing the square, or using the quadratic formula. We will try to solve the equation by factoring first. The equation is: \[ 15x^2 - 11x - 38 = 0 \] For factoring, we need to find two numbers that multiply to \( 15 \times (-38) = -570 \) and add up to \( -11 \). The two numbers that fit these requirements are \( -30 \) and \( 19 \), since \( -30 \times 19 = -570 \) and \( -30 + 19 = -11 \). Now we can express \( -11x \) as \( -30x + 19x \): \[ 15x^2 - 30x + 19x - 38 = 0 \] Group the terms: \[ (15x^2 - 30x) + (19x - 38) = 0 \] Factor out common factors from each group: \[ 15x(x - 2) + 19(x - 2) = 0 \] Now factor out \( (x - 2) \) which is common: \[ (x - 2)(15x + 19) = 0 \] Finally, we set each factor equal to zero to find the solutions for \( x \): \[ x - 2 = 0 \quad \text{or} \quad 15x + 19 = 0 \] Solving each equation for \( x \) gives us: \[ x = 2 \quad \text{or} \quad x = -\frac{19}{15} \] So the solutions to the quadratic equation are \( x = 2 \) and \( x = -\frac{19}{15} \).
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