Photo AI

L is the straight line with equation $y = 2x - 5$ C is a graph with equation $y = 6x^2 - 25x - 8$ Using algebra, find the coordinates of the points of intersection of L and C - Edexcel - GCSE Maths - Question 2 - 2022 - Paper 3

Question icon

Question 2

L-is-the-straight-line-with-equation-$y-=-2x---5$--C-is-a-graph-with-equation-$y-=-6x^2---25x---8$--Using-algebra,-find-the-coordinates-of-the-points-of-intersection-of-L-and-C-Edexcel-GCSE Maths-Question 2-2022-Paper 3.png

L is the straight line with equation $y = 2x - 5$ C is a graph with equation $y = 6x^2 - 25x - 8$ Using algebra, find the coordinates of the points of intersection... show full transcript

Worked Solution & Example Answer:L is the straight line with equation $y = 2x - 5$ C is a graph with equation $y = 6x^2 - 25x - 8$ Using algebra, find the coordinates of the points of intersection of L and C - Edexcel - GCSE Maths - Question 2 - 2022 - Paper 3

Step 1

Using algebra, find the coordinates of the points of intersection of L and C.

96%

114 rated

Answer

To find the points of intersection, we need to set the equations of L and C equal to each other:

2x5=6x225x82x - 5 = 6x^2 - 25x - 8

Rearranging the equation gives:

0=6x227x30 = 6x^2 - 27x - 3

Next, we can use the quadratic formula to solve for x, where the coefficients are a=6a = 6, b=27b = -27, and c=3c = -3:

x=b±b24ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Calculating the discriminant:

b24ac=(27)246(3)=729+72=801b^2 - 4ac = (-27)^2 - 4 \cdot 6 \cdot (-3) = 729 + 72 = 801

Then substituting back into the quadratic formula:

x=27±80112x = \frac{27 \pm \sqrt{801}}{12}

We can simplify 801\sqrt{801} to find its approximate value:

80128.3\sqrt{801} \approx 28.3

Thus,

x1=27+28.3124.6x_1 = \frac{27 + 28.3}{12} \approx 4.6

x2=2728.3120.1x_2 = \frac{27 - 28.3}{12} \approx -0.1

Next, we substitute these x-values back into the equation of L to find the corresponding y-values: For x14.6x_1 \approx 4.6:

y=2(4.6)54.2y = 2(4.6) - 5 \approx 4.2

For x20.1x_2 \approx -0.1:

y=2(0.1)55.2y = 2(-0.1) - 5 \approx -5.2

So, the points of intersection are approximately:

(4.6,4.2) and (0.1,5.2)(4.6, 4.2) \text{ and } (-0.1, -5.2)

Join the GCSE students using SimpleStudy...

97% of Students

Report Improved Results

98% of Students

Recommend to friends

100,000+

Students Supported

1 Million+

Questions answered

;