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Common Core: High School - Algebra : Solve Simple System of Two Variable Linear and Quadratic Equations: CCSS.Math.Content.HSA-REI.C.7

Study concepts, example questions & explanations for Common Core: High School - Algebra

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All Common Core: High School - Algebra Resources

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Example Questions

Example Question #121 : Reasoning With Equations & Inequalities

Find the points of intersection of $y = 39 x^{2} - 97$ and y = 14 x + 66 . Round your answers to the nearest hundredth.

Possible Answers:

$\left( -4.2 , 18.15 \right) , \left( 2.68 , -19.5 \right)$

$\left( 2.23 , 152.97 \right), \left( -1.87 , -6.93 \right)$

$\left( -0.73 , -16.91 \right) , \left( -0.96 , 16.36 \right)$

$\left( 7.46 , -19.52 \right) , \left( -1.09 , -16.79 \right)$

$\left( 8.95 , -17.41 \right) , \left( 4.01 , 10.85 \right)$

Correct answer:

$\left( 2.23 , 152.97 \right), \left( -1.87 , -6.93 \right)$

Explanation:

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

$39 x^{2} - 97 = 14 x + 66$

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

$39 x^{2} - 97 - 14 x - 66 = 0$

$39 x^{2} - 14 x - 163 = 0$

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 = 39 , b = -14 , c = -163 .

Now plug these values in.

$x =\frac{ 14 \pm\sqrt{\left( -14 \right)^2- 4 \cdot 39 \cdot -163 }}{ 2 \cdot 39 }$

$x =\frac{ 14 \pm 2 \sqrt{6406} }{ 78 }$

Split this up into equations.

$x =\frac{ 14 + 2 \sqrt{6406} }{ 78 }$

x = 2.23

$x = - \frac{\sqrt{6406}}{39} + \frac{7}{39}$

x = -1.87

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

$p_{1} = 39 \cdot 2.23 + 66$

$p_{1} = 86.97 + 66$

$p_{1} = 152.97$

So the first intersection point is at $\left( 2.23 , 152.97 \right)$

Now we need to find the last point of intersection.

$p_{2} = 39 \cdot -1.87 + 66$

$p_{2} = -72.93 + 66$

$p_{2} = -6.93$

So the second intersection point is at $\left( -1.87 , -6.93 \right)$

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Example Question #122 : Reasoning With Equations & Inequalities

Find the points of intersection of $y = 49 x^{2} - 76$ and y = 25 x + 33 . Round your answers to the nearest hundredth.

Possible Answers:

$\left( 7.51 , -18.71 \right) , \left( -6.11 , -0.05 \right)$

$\left( 1.77 , 119.73 \right), \left( -1.26 , -28.74 \right)$

$\left( -0.75 , 0.71 \right) , \left( -5.82 , -6.31 \right)$

$\left( 8.38 , -2.92 \right) , \left( -9.33 , -2.63 \right)$

$\left( -0.9 , 3.49 \right) , \left( 4.07 , -10.53 \right)$

Correct answer:

$\left( 1.77 , 119.73 \right), \left( -1.26 , -28.74 \right)$

Explanation:

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

$49 x^{2} - 76 = 25 x + 33$

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

$49 x^{2} - 76 - 25 x - 33 = 0$

$49 x^{2} - 25 x - 109 = 0$

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 = 49 , b = -25 , c = -109 .

Now plug these values in.

$x =\frac{ 25 \pm\sqrt{\left( -25 \right)^2- 4 \cdot 49 \cdot -109 }}{ 2 \cdot 49 }$

$x =\frac{ 25 \pm \sqrt{21989} }{ 98 }$

Split this up into equations.

$x =\frac{ 25 + \sqrt{21989} }{ 98 }$

x = 1.77

$x = - \frac{\sqrt{21989}}{98} + \frac{25}{98}$

x = -1.26

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

$p_{1} = 49 \cdot 1.77 + 33$

$p_{1} = 86.73 + 33$

$p_{1} = 119.73$

So the first intersection point is at $\left( 1.77 , 119.73 \right)$

Now we need to find the last point of intersection.

$p_{2} = 49 \cdot -1.26 + 33$

$p_{2} = -61.74 + 33$

$p_{2} = -28.74$

So the second intersection point is at $\left( -1.26 , -28.74 \right)$

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Common Core: High School - Algebra : Solve Simple System of Two Variable Linear and Quadratic Equations: CCSS.Math.Content.HSA-REI.C.7

Study concepts, example questions & explanations for Common Core: High School - Algebra

All Common Core: High School - Algebra Resources

Example Questions

Example Question #121 : Reasoning With Equations & Inequalities

Example Question #122 : Reasoning With Equations & Inequalities

All Common Core: High School - Algebra Resources

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