Example Question - factoring quadratics
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Solving Quadratic Equations with Quadratic Formula
The image contains a mathematical expression: \[ 3x^2 + x - 10 \] To solve this quadratic equation, we would typically first try to factor it, if possible. However, if the quadratic doesn't factor easily or at all, then we can use the quadratic formula to find the roots of the equation, which is given by: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] In this case, the coefficients for our quadratic expression \(3x^2 + x - 10\) are \(a = 3\), \(b = 1\), and \(c = -10\). Let's apply the quadratic formula: \[ x = \frac{-1 \pm \sqrt{1^2 - 4(3)(-10)}}{2(3)} \] \[ x = \frac{-1 \pm \sqrt{1 + 120}}{6} \] \[ x = \frac{-1 \pm \sqrt{121}}{6} \] \[ x = \frac{-1 \pm 11}{6} \] This gives us two solutions: \[ x = \frac{-1 + 11}{6} = \frac{10}{6} = \frac{5}{3} \] \[ x = \frac{-1 - 11}{6} = -2 \] Therefore, the two solutions to the equation \(3x^2 + x - 10 = 0\) are \(x = \frac{5}{3}\) and \(x = -2\).
Factoring Trinomials
The question is asking to factor the trinomial \(2x^2 + 11x + 14\). To factor it, one of the methods is to look for two numbers that both add up to the coefficient of the \(x\) term (which is 11 in this case) and multiply to the product of the coefficient of \(x^2\) term and the constant term (which is \(2 \times 14 = 28\)). So, we need two numbers that add up to 11 and multiply to 28. These two numbers are 4 and 7, because: \[4 + 7 = 11\] \[4 \times 7 = 28\] Now we can rewrite the middle term (11x) using 4 and 7: \[2x^2 + 4x + 7x + 14\] Next, let's factor by grouping: \[2x(x + 2) + 7(x + 2)\] Now, we can take out the common factor \((x + 2)\): \[(2x + 7)(x + 2)\] Therefore, the factored form of \(2x^2 + 11x + 14\) is \((2x + 7)(x + 2)\).
Solving Quadratic Equations by Factoring
The equation provided in the image is a quadratic equation of the form \( ax^2 + bx + c = 0 \), where \( a = 6 \), \( b = -7 \), and \( c = -10 \). To solve this equation, we can either factor the quadratic, complete the square, or use the quadratic formula. Let's attempt factoring first. We need two numbers that multiply to \( a \times c = 6 \times (-10) = -60 \) and that add to \( b = -7 \). Looking for such numbers, we find that -12 and +5 multiply to -60 and add to -7. Now we can write: \(6x^2 - 12x + 5x - 10 = 0\) Group the terms to factor by grouping: \( (6x^2 - 12x) + (5x - 10) = 0 \) Factor out common factors: \( 6x(x - 2) + 5(x - 2) = 0 \) Now we can see that \( (x - 2) \) is a common factor: \( (6x + 5)(x - 2) = 0 \) Using the zero product property, we can set each factor equal to zero: \( 6x + 5 = 0 \) or \( x - 2 = 0 \) Solving for \( x \) gives us: \( x = -\frac{5}{6} \) or \( x = 2 \) So the solutions to the quadratic equation are \( x = -\frac{5}{6} \) and \( x = 2 \).
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