Find the critical values: x^2 - 9 = 0 \;\; (x - 3) (x + 3) = 0 \;\; x = -3 or x = 3 Sketch the parabola and pick the side you need: y = x^2 - 9 State the solution: -3 < x < 3 Answer: -3 < x < 3

Find the critical values: x^2 - 25 = 0 \;\; (x - 5) (x + 5) = 0 \;\; x = -5 or x = 5 Sketch the parabola and pick the side you need: y = x^2 - 25 State the solution: x ≤ -5 or x ≥ 5 Answer: x ≤ -5 or x ≥ 5

Find the critical values: x^2 - 16 = 0 \;\; (x - 4) (x + 4) = 0 \;\; x = -4 or x = 4 Sketch the parabola and pick the side you need: y = x^2 - 16 State the solution: -4 ≤ x ≤ 4 Answer: -4 ≤ x ≤ 4

Find the critical values: x^2 - 49 = 0 \;\; (x - 7) (x + 7) = 0 \;\; x = -7 or x = 7 Sketch the parabola and pick the side you need: y = x^2 - 49 State the solution: x 7 Answer: x 7

Find the critical values: x^2 - 4 = 0 \;\; (x - 2) (x + 2) = 0 \;\; x = -2 or x = 2 Sketch the parabola and pick the side you need: y = x^2 - 4 State the solution: x 2 Answer: x 2

Quadratic & 2variable inequalities Worked Solutions — GCSE Maths

Fully worked, step-by-step solutions to GCSE Quadratic & 2variable inequalities questions. See exactly how to solve problems on difference of two squares, critical values, between the roots, outside the roots.

difference of two squarescritical valuesbetween the rootsoutside the rootsinclusive endpointsstrict endpoints

GCSE Higher70 questionsStep-by-step solutions

Question 1

2 markseasy

Solve

x^{2} < 9

Worked solution

Find the critical values
$x^{2} - 9 = 0 \;\Rightarrow\; \left(x - 3\right) \left(x + 3\right) = 0 \;\Rightarrow\; x = -3 \quad \text{or} \quad x = 3$
Start by finding where the quadratic is zero — these are the critical values.
Sketch the parabola and pick the side you need
$y = x^{2} - 9$
The $x^2$ coefficient is positive, so the curve is a U-shape cutting the axis at $-3$ and $3$ . We want the curve below the $x$ -axis, which is the part BETWEEN the critical values.
State the solution
$-3 < x < 3$
This is the complete solution set: every $x$ that satisfies the original inequality, and no others.

Answer

-3 < x < 3

Question 2

2 markseasy

Solve

x^{2} \ge 25

Worked solution

Find the critical values
$x^{2} - 25 = 0 \;\Rightarrow\; \left(x - 5\right) \left(x + 5\right) = 0 \;\Rightarrow\; x = -5 \quad \text{or} \quad x = 5$
Start by finding where the quadratic is zero — these are the critical values.
Sketch the parabola and pick the side you need
$y = x^{2} - 25$
The $x^2$ coefficient is positive, so the curve is a U-shape cutting the axis at $-5$ and $5$ . We want the curve above the $x$ -axis, which is the part OUTSIDE the critical values.
State the solution
$x \le -5 \quad \text{or} \quad x \ge 5$
This is the complete solution set: every $x$ that satisfies the original inequality, and no others.

Answer

x \le -5 \quad \text{or} \quad x \ge 5

Question 3

2 markseasy

Solve

x^{2} \le 16

Worked solution

Find the critical values
$x^{2} - 16 = 0 \;\Rightarrow\; \left(x - 4\right) \left(x + 4\right) = 0 \;\Rightarrow\; x = -4 \quad \text{or} \quad x = 4$
Start by finding where the quadratic is zero — these are the critical values.
Sketch the parabola and pick the side you need
$y = x^{2} - 16$
The $x^2$ coefficient is positive, so the curve is a U-shape cutting the axis at $-4$ and $4$ . We want the curve below the $x$ -axis, which is the part BETWEEN the critical values.
State the solution
$-4 \le x \le 4$
This is the complete solution set: every $x$ that satisfies the original inequality, and no others.

Answer

-4 \le x \le 4

Question 4

2 markseasy

Solve

x^{2} > 49

Worked solution

Find the critical values
$x^{2} - 49 = 0 \;\Rightarrow\; \left(x - 7\right) \left(x + 7\right) = 0 \;\Rightarrow\; x = -7 \quad \text{or} \quad x = 7$
Start by finding where the quadratic is zero — these are the critical values.
Sketch the parabola and pick the side you need
$y = x^{2} - 49$
The $x^2$ coefficient is positive, so the curve is a U-shape cutting the axis at $-7$ and $7$ . We want the curve above the $x$ -axis, which is the part OUTSIDE the critical values.
State the solution
$x < -7 \quad \text{or} \quad x > 7$
This is the complete solution set: every $x$ that satisfies the original inequality, and no others.

Answer

x < -7 \quad \text{or} \quad x > 7

Question 5

2 markseasy

Solve

x^{2} - 4 > 0

Worked solution

Find the critical values
$x^{2} - 4 = 0 \;\Rightarrow\; \left(x - 2\right) \left(x + 2\right) = 0 \;\Rightarrow\; x = -2 \quad \text{or} \quad x = 2$
Start by finding where the quadratic is zero — these are the critical values.
Sketch the parabola and pick the side you need
$y = x^{2} - 4$
The $x^2$ coefficient is positive, so the curve is a U-shape cutting the axis at $-2$ and $2$ . We want the curve above the $x$ -axis, which is the part OUTSIDE the critical values.
State the solution
$x < -2 \quad \text{or} \quad x > 2$
This is the complete solution set: every $x$ that satisfies the original inequality, and no others.

Answer

x < -2 \quad \text{or} \quad x > 2

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