18 (a) (i) Write $x^2 + 4x - 16$ in the form $(x + a)^2 - b$ - OCR - GCSE Maths - Question 18 - 2018 - Paper 5
Question 18
18 (a) (i) Write $x^2 + 4x - 16$ in the form $(x + a)^2 - b$.
(a) (ii) Solve the equation $x^2 + 4x - 16 = 0$.
Give your answers in surd form as simply as possible.... show full transcript
Worked Solution & Example Answer:18 (a) (i) Write $x^2 + 4x - 16$ in the form $(x + a)^2 - b$ - OCR - GCSE Maths - Question 18 - 2018 - Paper 5
Step 1
Write $x^2 + 4x - 16$ in the form $(x + a)^2 - b$
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Answer
To express x2+4x−16 in the required form, we first complete the square.
Start with the quadratic expression:
x2+4x
Identify the coefficient of x, which is 4. Half of this is 2, hence:
Squaring it gives us 22=4.
Rewrite the expression by adding and subtracting this square:
x2+4x+4−4−16
This simplifies to:
(x+2)2−20
Thus, the final expression in the required form is:
(x+2)2−20
Step 2
Solve the equation $x^2 + 4x - 16 = 0$
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Answer
To solve the equation, we can use the previously derived expression:
(x+2)2−20=0
Rearranging gives:
(x+2)2=20
Taking the square root of both sides, we find:
x+2=extpm20
Simplifying sqrt20 yields:
20=4×5=25
Hence, we have:
x+2=25extorx+2=−25
This leads us to:
x=−2+25extorx=−2−25
Step 3
Sketch the graph of $y = x^2 + 4x - 16$, showing clearly the coordinates of any turning points
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Answer
To sketch the graph:
We previously found the vertex using the complete square method; the vertex point is (−2,−20). This is a turning point of the parabola.
The function is a U-shaped parabola (as the coefficient of x2 is positive), opening upwards.
Plot the vertex at (−2,−20) on the graph.
To find the x-intercepts, set y=0:
x2+4x−16=0
Which we already solved. Here, the intercepts approximate to:
x=−2+25extandx=−2−25
Mark these intercepts on the x-axis.
Finally, sketch the full graph, ensuring it reflects the U-shape with the noted turning point at (−2,−20) and x-intercepts correctly placed.
