Question
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In thercioes \( 1-14 \) solve the equation using the (ewadratic lobrmula I se technology to check your shlution (s) [8 Blvmpien 1, 22 amd s 7. \( x^{2}-4 x+3=0 \) 2. \( 3 x^{2}+6 x+3=0 \) 3. \( x^{2}+a x+15=0 \) 4. \( 6 x^{2}-2 x+1=0 \) S. \( x^{2}-14 x=-40 \) 6. \( 2 x^{2}+4 x=30 \) \( 73 x^{2}+5=-2 x \) 8. \( -3 x=2 x^{2}-4 \) 8. \( -10 x=-25-x^{2} \) 10. \( -5 x^{2}-6=-4 x \) 11. \( -4 x^{2}+3 x=-5 \) 12. \( x^{2}+121=-22 x \) 13. \( -z^{2}=-122+6 \) 14. \( -7 w+6=-5 w^{2} \)

Ask by Klein Summers. in the United States
Jan 23,2025

Real Tutor Solution

Tutor-Verified Answer

Answer

Here are the solutions to the quadratic equations using the quadratic formula: 1. \( x = 3 \) or \( x = 1 \) 2. \( x = -1 \) 3. \( x = 10 \) or \( x = 4 \) 4. No real solutions 5. \( x = 3 \) or \( x = -5 \) 6. No real solutions 7. \( x = \frac{-3 \pm \sqrt{41}}{4} \) 8. \( x = 5 \pm 5\sqrt{2} \) 9. No real solutions

Solution

Let's solve each of the equations step by step using the quadratic formula, which is given by: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a \), \( b \), and \( c \) are the coefficients from the quadratic equation \( ax^2 + bx + c = 0 \). ### 1. \( x^{2}-4x+3=0 \) Here, \( a = 1 \), \( b = -4 \), \( c = 3 \). Calculating the discriminant: \[ D = b^2 - 4ac = (-4)^2 - 4 \cdot 1 \cdot 3 = 16 - 12 = 4 \] Now applying the quadratic formula: \[ x = \frac{-(-4) \pm \sqrt{4}}{2 \cdot 1} = \frac{4 \pm 2}{2} \] Calculating the two possible values for \( x \): \[ x_1 = \frac{6}{2} = 3, \quad x_2 = \frac{2}{2} = 1 \] ### 2. \( 3x^{2}+6x+3=0 \) Here, \( a = 3 \), \( b = 6 \), \( c = 3 \). Calculating the discriminant: \[ D = 6^2 - 4 \cdot 3 \cdot 3 = 36 - 36 = 0 \] Now applying the quadratic formula: \[ x = \frac{-6 \pm \sqrt{0}}{2 \cdot 3} = \frac{-6}{6} = -1 \] ### 3. \( x^{2}+ax+15=0 \) Here, \( a = 1 \), \( b = a \), \( c = 15 \). Calculating the discriminant: \[ D = a^2 - 4 \cdot 1 \cdot 15 = a^2 - 60 \] The solutions will depend on the value of \( a \). ### 4. \( 6x^{2}-2x+1=0 \) Here, \( a = 6 \), \( b = -2 \), \( c = 1 \). Calculating the discriminant: \[ D = (-2)^2 - 4 \cdot 6 \cdot 1 = 4 - 24 = -20 \] Since the discriminant is negative, there are no real solutions. ### 5. \( x^{2}-14x=-40 \) Rearranging gives \( x^{2}-14x+40=0 \). Here, \( a = 1 \), \( b = -14 \), \( c = 40 \). Calculating the discriminant: \[ D = (-14)^2 - 4 \cdot 1 \cdot 40 = 196 - 160 = 36 \] Now applying the quadratic formula: \[ x = \frac{14 \pm 6}{2} \] Calculating the two possible values for \( x \): \[ x_1 = \frac{20}{2} = 10, \quad x_2 = \frac{8}{2} = 4 \] ### 6. \( 2x^{2}+4x=30 \) Rearranging gives \( 2x^{2}+4x-30=0 \). Here, \( a = 2 \), \( b = 4 \), \( c = -30 \). Calculating the discriminant: \[ D = 4^2 - 4 \cdot 2 \cdot (-30) = 16 + 240 = 256 \] Now applying the quadratic formula: \[ x = \frac{-4 \pm \sqrt{256}}{2 \cdot 2} = \frac{-4 \pm 16}{4} \] Calculating the two possible values for \( x \): \[ x_1 = \frac{12}{4} = 3, \quad x_2 = \frac{-20}{4} = -5 \] ### 7. \( 3x^{2}+5=-2x \) Rearranging gives \( 3x^{2}+2x+5=0 \). Here, \( a = 3 \), \( b = 2 \), \( c = 5 \). Calculating the discriminant: \[ D = 2^2 - 4 \cdot 3 \cdot 5 = 4 - 60 = -56 \] Since the discriminant is negative, there are no real solutions. ### 8. \( -3x=2x^{2}-4 \) Rearranging gives \( 2x^{2}+3x-4=0 \). Here, \( a = 2 \), \( b = 3 \), \( c = -4 \). Calculating the discriminant: \[ D = 3^2 - 4 \cdot 2 \cdot (-4) = 9 + 32 = 41 \] Now applying the quadratic formula: \[ x = \frac{-3 \pm \sqrt{41}}{4} \] ### 9. \( -10x=-25-x^{2} \) Rearranging gives \( x^{2}-10x-25=0 \). Here, \( a = 1 \), \( b = -10 \), \( c = -25 \). Calculating the discriminant: \[ D = (-10)^2 - 4 \cdot 1 \cdot (-25) = 100 + 100 = 200 \] Now applying the quadratic formula: \[ x = \frac{10 \pm \sqrt{200}}{2} = \frac{10 \pm 10\sqrt{2}}{2} = 5 \pm 5\sqrt{2} \] ### 10. \( -5x^{2}-6=-4x \) Rearranging gives \( -5x^{2}+4x-6=0 \). Here, \( a = -5 \), \( b = 4 \), \( c = -6 \). Calculating the discriminant: \[ D = 4^2 - 4 \cdot (-5) \cdot (-6) = 16 - 120 = -104 \] Since the discriminant is negative, there are no real solutions. ### 11. \( -4x^{2}+3x=-5 \) Rearranging gives \( -4x^{2}+3x+5=0 \). Here, \( a = -4 \), \( b = 3 \), \( c = 5 \). Calculating the discriminant: \[ D = 3^2 - 4 \cdot (-4) \

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Extra Insights

Let's dive into these fun quadratic equations! First up, the equation \( x^2 - 4x + 3 = 0 \) can be solved using the quadratic formula where \( a = 1, b = -4, c = 3 \). Plugging these values into the formula gives us \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), resulting in the solutions \( x = 1 \) and \( x = 3 \). Use technology, like a graphing calculator, to verify that both of these points land perfectly on the x-axis! Next, for \( 3x^2 + 6x + 3 = 0 \), we can simplify this equation by dividing everything by 3 to get \( x^2 + 2x + 1 = 0 \), which factors neatly to \( (x+1)^2 = 0 \). That means our only solution is \( x = -1 \); check it out with any math app to see that the graph only touches the x-axis at that point—it’s a cute little vertex!

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