Photo AI

Show that $x^2 + 6x + 11$ can be written as $(x + p)^2 + q$ where $p$ and $q$ are integers to be found - Edexcel - A-Level Maths Pure - Question 8 - 2010 - Paper 1

Question 8

Show that $x^2 + 6x + 11$ can be written as $(x + p)^2 + q$ where $p$ and $q$ are integers to be found. In the space at the top of page 7, sketch the curve with equ... show full transcript

Worked Solution & Example Answer:Show that $x^2 + 6x + 11$ can be written as $(x + p)^2 + q$ where $p$ and $q$ are integers to be found - Edexcel - A-Level Maths Pure - Question 8 - 2010 - Paper 1

Step 1

Show that $x^2 + 6x + 11$ can be written as $(x + p)^2 + q$ where $p$ and $q$ are integers to be found.

96%

114 rated

Answer

To write the expression in the required form, we will complete the square for the quadratic expression $x^2 + 6x + 11$ .

Start with the quadratic: $x^2 + 6x + 11$
Determine the value needed to complete the square. Take half of the coefficient of $x$ (which is 6), square it to get: $\left(\frac{6}{2}\right)^2 = 9$
Now rewrite the quadratic as follows: $x^2 + 6x + 9 + 2$
This is equivalent to: $(x + 3)^2 + 2$
Therefore, we have $p = 3$ and $q = 2$ .

Step 2

In the space at the top of page 7, sketch the curve with equation $y = x^2 + 6x + 11$, showing clearly any intersections with the coordinate axes.

99%

104 rated

Answer

The curve represents a quadratic function. We know that a quadratic opens upwards and has a U shape.

Identify the vertex of the quadratic: The vertex form derived from the completed square is $(x + 3)^2 + 2$ . This indicates that the vertex is at the point (-3, 2).
Determine the y-intercept by setting $x = 0$ : $y = 0^2 + 6(0) + 11 = 11$ So, the y-intercept is (0, 11).
Determine the x-intercepts by solving: $x^2 + 6x + 11 = 0$ The discriminant ( $b^2 - 4ac$ ) will show that there are no real x-intercepts, as calculated in part (c).
In the sketch, you must illustrate the U shape with a vertex at (-3, 2) and a y-intercept at (0, 11), ensuring it does not cross the x-axis.

Step 3

Find the value of the discriminant of $x^2 + 6x + 11$.

96%

101 rated

Answer

The discriminant for a quadratic equation of the form $ax^2 + bx + c$ is given by the formula: $D = b^2 - 4ac$ Here, $a = 1$ , $b = 6$ , and $c = 11$ .

Substitute the values into the formula: $D = 6^2 - 4 \times 1 \times 11$
Calculate: $D = 36 - 44 = -8$
Therefore, the value of the discriminant is $-8$ , indicating that there are no real roots.

Show that $x^2 + 6x + 11$ can be written as $(x + p)^2 + q$ where $p$ and $q$ are integers to be found - Edexcel - A-Level Maths Pure - Question 8 - 2010 - Paper 1

Worked Solution & Example Answer:Show that $x^2 + 6x + 11$ can be written as $(x + p)^2 + q$ where $p$ and $q$ are integers to be found - Edexcel - A-Level Maths Pure - Question 8 - 2010 - Paper 1

Show that $x^2 + 6x + 11$ can be written as $(x + p)^2 + q$ where $p$ and $q$ are integers to be found.

In the space at the top of page 7, sketch the curve with equation $y = x^2 + 6x + 11$, showing clearly any intersections with the coordinate axes.

Find the value of the discriminant of $x^2 + 6x + 11$.

Join the A-Level students using SimpleStudy...