Example Question - roots of quadratic equations
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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 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 Quadratic Equations for Specific Function Values
The image shows a problem from a textbook asking to graph the function \( h(x) = x^2 - 6x + 5 \) and solve the equations for the following conditions: a) \( h(x) = 5 \) b) \( h(x) = 0 \) c) \( h(x) = -5 \) Let's solve these equations one by one. a) To solve \( h(x) = 5 \), we substitute \( 5 \) for \( h(x) \) in the original function: \( x^2 - 6x + 5 = 5 \) Subtracting \( 5 \) from both sides of the equation gives: \( x^2 - 6x = 0 \) Factoring out an \( x \) gives: \( x(x - 6) = 0 \) So \( x = 0 \) or \( x = 6 \). These are the values of \( x \) when \( h(x)=5 \). b) To solve \( h(x) = 0 \), we need to find the roots of the quadratic equation: \( x^2 - 6x + 5 = 0 \) Factoring the quadratic gives: \( (x - 5)(x - 1) = 0 \) Hence, the values of \( x \) that make \( h(x) = 0 \) are \( x = 5 \) and \( x = 1 \). c) To solve \( h(x) = -5 \), we substitute \( -5 \) for \( h(x) \) in the original function: \( x^2 - 6x + 5 = -5 \) Adding \( 5 \) to both sides gives: \( x^2 - 6x + 10 = 0 \) This quadratic equation does not factor nicely, so we will need to use the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1 \), \( b = -6 \), and \( c = 10 \). Plugging these values in gives: \( x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(10)}}{2(1)} \) \( x = \frac{6 \pm \sqrt{36 - 40}}{2} \) \( x = \frac{6 \pm \sqrt{-4}}{2} \) Since there is a negative number inside the square root, the solutions for \( x \) will be complex numbers. Hence, there are no real solutions for \( x \) when \( h(x) = -5 \).
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