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10x2+23x+12=010x^2 + 23x + 12 = 0 \newline a. 23,54\frac{2}{3}, \frac{5}{4} \newline b. 32,45\frac{3}{2}, \frac{4}{5} \newline c. βˆ’23,βˆ’45-\frac{2}{3}, -\frac{4}{5} \newline d. βˆ’32,βˆ’45-\frac{3}{2}, -\frac{4}{5}

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Q. 10x2+23x+12=010x^2 + 23x + 12 = 0 \newline a. 23,54\frac{2}{3}, \frac{5}{4} \newline b. 32,45\frac{3}{2}, \frac{4}{5} \newline c. βˆ’23,βˆ’45-\frac{2}{3}, -\frac{4}{5} \newline d. βˆ’32,βˆ’45-\frac{3}{2}, -\frac{4}{5}
  1. Identify coefficients: Identify the coefficients of the quadratic equation.\newlineThe quadratic equation is in the form ax2+bx+c=0ax^2 + bx + c = 0. For the given equation, 10x2+23x+12=010x^2 + 23x + 12 = 0, the coefficients are:\newlinea = 1010, b = 2323, and c = 1212.
  2. Use quadratic formula: Use the quadratic formula to find the solutions for xx. The quadratic formula is x=βˆ’bΒ±b2βˆ’4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}. We will use this formula to find the values of xx.
  3. Calculate discriminant: Calculate the discriminant b2βˆ’4acb^2 - 4ac.\newlineDiscriminant = b2βˆ’4ac=(23)2βˆ’4(10)(12)=529βˆ’480=49b^2 - 4ac = (23)^2 - 4(10)(12) = 529 - 480 = 49.
  4. Calculate possible x values: Calculate the two possible values for x using the quadratic formula.\newlineFirst, we find the square root of the discriminant: 49=7\sqrt{49} = 7.\newlineNow, we can calculate x:\newlinex=βˆ’bΒ±b2βˆ’4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\newlinex=βˆ’(23)Β±72Γ—10x = \frac{-(23) \pm 7}{2 \times 10}
  5. Calculate solutions for x: Calculate the two solutions for x.\newlineFirst solution:\newlinex=(βˆ’23+7)/20x = (-23 + 7) / 20\newlinex=βˆ’16/20x = -16 / 20\newlinex=βˆ’4/5x = -4 / 5\newlinex=βˆ’0.8x = -0.8\newlineSecond solution:\newlinex=(βˆ’23βˆ’7)/20x = (-23 - 7) / 20\newlinex=βˆ’30/20x = -30 / 20\newlinex=βˆ’3/2x = -3 / 2\newlinex=βˆ’1.5x = -1.5
  6. Verify solutions: Verify the solutions by plugging them back into the original equation.\newlineFor x=βˆ’0.8x = -0.8:\newline10(βˆ’0.8)2+23(βˆ’0.8)+12=10(0.64)βˆ’18.4+12=6.4βˆ’18.4+12=010(-0.8)^2 + 23(-0.8) + 12 = 10(0.64) - 18.4 + 12 = 6.4 - 18.4 + 12 = 0\newlineFor x=βˆ’1.5x = -1.5:\newline10(βˆ’1.5)2+23(βˆ’1.5)+12=10(2.25)βˆ’34.5+12=22.5βˆ’34.5+12=010(-1.5)^2 + 23(-1.5) + 12 = 10(2.25) - 34.5 + 12 = 22.5 - 34.5 + 12 = 0\newlineBoth solutions satisfy the original equation.

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