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All Common Core: High School - Algebra Resources

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Flashcards: Solve Simple System of Two Variable Linear and Quadratic Equations: CCSS.Math.Content.HSA-REI.C.7

Find the points of intersection of $y = 51 x^{2} - 46$ and y = 26 x + 27 . Round your answers to the nearest hundredth.

$\left( -4.39 , -17.43 \right) , \left( 2.5 , 10.63 \right)$

$\left( 1.12 , -3.84 \right) , \left( 8.17 , -15.05 \right)$

$\left( 1.48 , 102.48 \right), \left( -0.97 , -22.47 \right)$

$\left( -4.02 , 5.3 \right) , \left( -2.85 , 0.66 \right)$

$\left( 8.96 , 5.74 \right) , \left( -8.69 , 7.97 \right)$

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Flashcards: Solve Simple System of Two Variable Linear and Quadratic Equations: CCSS.Math.Content.HSA-REI.C.7

Find the points of intersection of $y = 51 x^{2} - 46$ and y = 26 x + 27 . Round your answers to the nearest hundredth.

Your answer:

$\left( -4.39 , -17.43 \right) , \left( 2.5 , 10.63 \right)$

Your answer:

$\left( 1.12 , -3.84 \right) , \left( 8.17 , -15.05 \right)$

Your answer:

$\left( -4.02 , 5.3 \right) , \left( -2.85 , 0.66 \right)$

Your answer:

$\left( 8.96 , 5.74 \right) , \left( -8.69 , 7.97 \right)$

Correct answer:

$\left( 1.48 , 102.48 \right), \left( -0.97 , -22.47 \right)$

Explanation:

The first step to solving this problem is to set the equations equal to each other.

$51 x^{2} - 46 = 26 x + 27$

Now subtract the right hand side, to make it zero, so we can use the quadratic equation.

$51 x^{2} - 46 - 26 x - 27 = 0$

$51 x^{2} - 26 x - 73 = 0$

Recall the quadratic equation.

$x =\frac{ -b \pm\sqrt{ b^{2} - 4 a c }}{ 2 a }$

Where $\uptext{a}$ , $\uptext{b}$ , and $\uptext{c}$ correspond to coefficients in the following quadratic equation.

$a x^{2} + b x + c = 0$

In this case, a = 51 , b = -26 , c = -73 .

Now plug these values in.

$x =\frac{ 26 \pm\sqrt{\left( -26 \right)^2- 4 \cdot 51 \cdot -73 }}{ 2 \cdot 51 }$

$x =\frac{ 26 \pm 4 \sqrt{973} }{ 102 }$

Split this up into equations.

$x =\frac{ 26 + 4 \sqrt{973} }{ 102 }$

x = 1.48

$x = - \frac{2 \sqrt{973}}{51} + \frac{13}{51}$

x = -0.97

Since we want points, we need to plug these values into one of the original equations.

$p_{1} = 51 \cdot 1.48 + 27$

$p_{1} = 75.48 + 27$

$p_{1} = 102.48$

So the first intersection point is at $\left( 1.48 , 102.48 \right)$

Now we need to find the last point of intersection.

$p_{2} = 51 \cdot -0.97 + 27$

$p_{2} = -49.47 + 27$

$p_{2} = -22.47$

So the second intersection point is at $\left( -0.97 , -22.47 \right)$

Flashcards: Solve Simple System of Two Variable Linear and Quadratic Equations: CCSS.Math.Content.HSA-REI.C.7

Find the points of intersection of $y = 24 x^{2} - 70$ and y = 48 x + 93 . Round your answers to the nearest hundredth.

$\left( -2.8 , 3.57 \right) , \left( -8.72 , -7.49 \right)$

$\left( 3.79 , 183.96 \right), \left( -1.79 , 50.04 \right)$

$\left( 8.87 , -6.51 \right) , \left( -2.84 , -3.34 \right)$

$\left( 7.61 , 14.19 \right) , \left( -6.06 , -16.25 \right)$

$\left( 8.38 , 18.56 \right) , \left( -2.93 , 4.26 \right)$

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Flashcards: Solve Simple System of Two Variable Linear and Quadratic Equations: CCSS.Math.Content.HSA-REI.C.7

Find the points of intersection of $y = 24 x^{2} - 70$ and y = 48 x + 93 . Round your answers to the nearest hundredth.

Your answer:

$\left( -2.8 , 3.57 \right) , \left( -8.72 , -7.49 \right)$

Your answer:

$\left( 8.87 , -6.51 \right) , \left( -2.84 , -3.34 \right)$

Your answer:

$\left( 7.61 , 14.19 \right) , \left( -6.06 , -16.25 \right)$

Your answer:

$\left( 8.38 , 18.56 \right) , \left( -2.93 , 4.26 \right)$

Correct answer:

$\left( 3.79 , 183.96 \right), \left( -1.79 , 50.04 \right)$

Explanation:

The first step to solving this problem is to set the equations equal to each other.

$24 x^{2} - 70 = 48 x + 93$

Now subtract the right hand side, to make it zero, so we can use the quadratic equation.

$24 x^{2} - 70 - 48 x - 93 = 0$

$24 x^{2} - 48 x - 163 = 0$

Recall the quadratic equation.

$x =\frac{ -b \pm\sqrt{ b^{2} - 4 a c }}{ 2 a }$

Where $\uptext{a}$ , $\uptext{b}$ , and $\uptext{c}$ correspond to coefficients in the following quadratic equation.

$a x^{2} + b x + c = 0$

In this case, a = 24 , b = -48 , c = -163 .

Now plug these values in.

$x =\frac{ 48 \pm\sqrt{\left( -48 \right)^2- 4 \cdot 24 \cdot -163 }}{ 2 \cdot 24 }$

$x =\frac{ 48 \pm 4 \sqrt{1122} }{ 48 }$

Split this up into equations.

$x =\frac{ 48 + 4 \sqrt{1122} }{ 48 }$

x = 3.79

$x = - \frac{\sqrt{1122}}{12} + 1$

x = -1.79

Since we want points, we need to plug these values into one of the original equations.

$p_{1} = 24 \cdot 3.79 + 93$

$p_{1} = 90.96 + 93$

$p_{1} = 183.96$

So the first intersection point is at $\left( 3.79 , 183.96 \right)$

Now we need to find the last point of intersection.

$p_{2} = 24 \cdot -1.79 + 93$

$p_{2} = -42.96 + 93$

$p_{2} = 50.04$

So the second intersection point is at $\left( -1.79 , 50.04 \right)$

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All Common Core: High School - Algebra Resources

Flashcards: Solve Simple System of Two Variable Linear and Quadratic Equations: CCSS.Math.Content.HSA-REI.C.7

Flashcards: Solve Simple System of Two Variable Linear and Quadratic Equations: CCSS.Math.Content.HSA-REI.C.7

Flashcards: Solve Simple System of Two Variable Linear and Quadratic Equations: CCSS.Math.Content.HSA-REI.C.7

Flashcards: Solve Simple System of Two Variable Linear and Quadratic Equations: CCSS.Math.Content.HSA-REI.C.7

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