Factorise:

\(x^2+4x+3=(x+1)(x+3)\).

So the roots are \(x=-3\) and \(x=-1\).

Because the coefficient of \(x^2\) is positive, the parabola opens upwards.

An upward-opening parabola is below the x-axis between the roots and above the x-axis outside them.

So the solution is \(-3 < x < -1\).

Factorise:

\(x^2-5x+6=(x-2)(x-3)\).

So the roots are \(x=2\) and \(x=3\).

Because the coefficient of \(x^2\) is positive, the parabola opens upwards.

An upward-opening parabola is below the x-axis between the roots and above the x-axis outside them.

So the solution is \(x \le 2\) or \(x \ge 3\).

Factorise:

\(x^2+x-6=(x+3)(x-2)\).

So the roots are \(x=-3\) and \(x=2\).

Because the coefficient of \(x^2\) is positive, the parabola opens upwards.

An upward-opening parabola is below the x-axis between the roots and above the x-axis outside them.

So the solution is \(x < -3\) or \(x > 2\).

Factorise:

\(x^2-7x+10=(x-2)(x-5)\).

So the roots are \(x=2\) and \(x=5\).

Because the coefficient of \(x^2\) is positive, the parabola opens upwards.

An upward-opening parabola is below the x-axis between the roots and above the x-axis outside them.

So the solution is \(2 \le x \le 5\).

Factorise:

\(2x^2-8x+6=2(x-1)(x-3)\).

So the roots are \(x=1\) and \(x=3\).

Because the coefficient of \(x^2\) is positive, the parabola opens upwards.

An upward-opening parabola is below the x-axis between the roots and above the x-axis outside them.

So the solution is \(1 < x < 3\).

Factorise:

\(3x^2+3x-18=3(x+3)(x-2)\).

So the roots are \(x=-3\) and \(x=2\).

Because the coefficient of \(x^2\) is positive, the parabola opens upwards.

An upward-opening parabola is below the x-axis between the roots and above the x-axis outside them.

So the solution is \(x \le -3\) or \(x \ge 2\).

Factorise:

\(5x^2-5x-30=5(x+2)(x-3)\).

So the roots are \(x=-2\) and \(x=3\).

Because the coefficient of \(x^2\) is positive, the parabola opens upwards.

An upward-opening parabola is below the x-axis between the roots and above the x-axis outside them.

So the solution is \(x < -2\) or \(x > 3\).

Factorise:

\(2x^2+10x+12=2(x+2)(x+3)\).

So the roots are \(x=-3\) and \(x=-2\).

Because the coefficient of \(x^2\) is positive, the parabola opens upwards.

An upward-opening parabola is below the x-axis between the roots and above the x-axis outside them.

So the solution is \(-3 \le x \le -2\).

Factorise:

\(x^2-2x-8=(x+2)(x-4)\).

So the roots are \(x=-2\) and \(x=4\).

Because the coefficient of \(x^2\) is positive, the parabola opens upwards.

An upward-opening parabola is below the x-axis between the roots and above the x-axis outside them.

So the solution is \(-2 < x < 4\).

Factorise:

\(x^2+2x-15=(x+5)(x-3)\).

So the roots are \(x=-5\) and \(x=3\).

Because the coefficient of \(x^2\) is positive, the parabola opens upwards.

An upward-opening parabola is below the x-axis between the roots and above the x-axis outside them.

So the solution is \(x \le -5\) or \(x \ge 3\).

Factorise:

\(4x^2-4x-3=(2x+1)(2x-3)\).

So the roots are \(x=-\dfrac12\) and \(x=\dfrac32\).

Because the coefficient of \(x^2\) is positive, the parabola opens upwards.

An upward-opening parabola is below the x-axis between the roots and above the x-axis outside them.

So the solution is \(x < -\dfrac{1}{2}\) or \(x > \dfrac{3}{2}\).

Factorise:

\(6x^2+6x-12=6(x+2)(x-1)\).

So the roots are \(x=-2\) and \(x=1\).

Because the coefficient of \(x^2\) is positive, the parabola opens upwards.

An upward-opening parabola is below the x-axis between the roots and above the x-axis outside them.

So the solution is \(-2 \le x \le 1\).

Factorise:

\(x^2-9=(x-3)(x+3)\).

So the roots are \(x=-3\) and \(x=3\).

Because the coefficient of \(x^2\) is positive, the parabola opens upwards.

An upward-opening parabola is below the x-axis between the roots and above the x-axis outside them.

So the solution is \(x < -3\) or \(x > 3\).

Rearrange:

\(2x^2-18=2(x^2-9)=2(x-3)(x+3)\).

So the roots are \(x=-3\) and \(x=3\).

Because the coefficient of \(x^2\) is positive, the parabola opens upwards.

An upward-opening parabola is below the x-axis between the roots and above the x-axis outside them.

So the solution is \(-3 \le x \le 3\).

Factorise:

\(x^2+6x+5=(x+5)(x+1)\).

So the roots are \(x=-5\) and \(x=-1\).

Because the coefficient of \(x^2\) is positive, the parabola opens upwards.

An upward-opening parabola is below the x-axis between the roots and above the x-axis outside them.

So the solution is \(x \le -5\) or \(x \ge -1\).

Factorise:

\(x^2-8x+15=(x-3)(x-5)\).

So the roots are \(x=3\) and \(x=5\).

Because the coefficient of \(x^2\) is positive, the parabola opens upwards.

An upward-opening parabola is below the x-axis between the roots and above the x-axis outside them.

So the solution is \(3 < x < 5\).

Factorise:

\(3x^2-27=3(x^2-9)=3(x-3)(x+3)\).

So the roots are \(x=-3\) and \(x=3\).

Because the coefficient of \(x^2\) is positive, the parabola opens upwards.

An upward-opening parabola is below the x-axis between the roots and above the x-axis outside them.

So the solution is \(x < -3\) or \(x > 3\).

Factorise:

\(4x^2+4x-8=4(x+2)(x-1)\).

So the roots are \(x=-2\) and \(x=1\).

Because the coefficient of \(x^2\) is positive, the parabola opens upwards.

An upward-opening parabola is below the x-axis between the roots and above the x-axis outside them.

So the solution is \(-2 \le x \le 1\).

Factorise:

\(2x^2-14x+24=2(x-3)(x-4)\).

So the roots are \(x=3\) and \(x=4\).

Because the coefficient of \(x^2\) is positive, the parabola opens upwards.

An upward-opening parabola is below the x-axis between the roots and above the x-axis outside them.

So the solution is \(x \le 3\) or \(x \ge 4\).

Factorise:

\(5x^2+5x-30=5(x+3)(x-2)\).

So the roots are \(x=-3\) and \(x=2\).

Because the coefficient of \(x^2\) is positive, the parabola opens upwards.

An upward-opening parabola is below the x-axis between the roots and above the x-axis outside them.

So the solution is \(-3 < x < 2\).

Find the roots:

\(-x^2+4x+5=0\iff x^2-4x-5=0\).

\((x-5)(x+1)=0\), so the roots are \(x=-1\) and \(x=5\).

Because the coefficient of \(x^2\) is negative, the parabola opens downwards.

A downward-opening parabola is above the x-axis between the roots and below the x-axis outside them.

So the solution is \(-1 < x < 5\).

Find the roots:

\(-x^2+x+6=0\iff x^2-x-6=0\).

\((x-3)(x+2)=0\), so the roots are \(x=-2\) and \(x=3\).

Because the coefficient of \(x^2\) is negative, the parabola opens downwards.

A downward-opening parabola is above the x-axis between the roots and below the x-axis outside them.

So the solution is \(x \le -2\) or \(x \ge 3\).

Find the roots:

\(-2x^2+8x-6=0\iff 2x^2-8x+6=0\).

\(2(x-1)(x-3)=0\), so the roots are \(x=1\) and \(x=3\).

Because the coefficient of \(x^2\) is negative, the parabola opens downwards.

A downward-opening parabola is above the x-axis between the roots and below the x-axis outside them.

So the solution is \(x < 1\) or \(x > 3\).

Find the roots:

\(-3x^2-3x+18=0\iff 3x^2+3x-18=0\).

\(3(x+3)(x-2)=0\), so the roots are \(x=-3\) and \(x=2\).

Because the coefficient of \(x^2\) is negative, the parabola opens downwards.

A downward-opening parabola is above the x-axis between the roots and below the x-axis outside them.

So the solution is \(-3 \le x \le 2\).

Find the roots:

\(-5x^2+5x+30=0\iff x^2-x-6=0\).

\((x-3)(x+2)=0\), so the roots are \(x=-2\) and \(x=3\).

Because the coefficient of \(x^2\) is negative, the parabola opens downwards.

A downward-opening parabola is above the x-axis between the roots and below the x-axis outside them.

So the solution is \(x < -2\) or \(x > 3\).

Find the roots:

\(-2x^2-10x-12=0\iff x^2+5x+6=0\).

\((x+2)(x+3)=0\), so the roots are \(x=-3\) and \(x=-2\).

Because the coefficient of \(x^2\) is negative, the parabola opens downwards.

A downward-opening parabola is above the x-axis between the roots and below the x-axis outside them.

So the solution is \(-3 \le x \le -2\).

Find the roots:

\(-x^2+2x+8=0\iff x^2-2x-8=0\).

\((x-4)(x+2)=0\), so the roots are \(x=-2\) and \(x=4\).

Because the coefficient of \(x^2\) is negative, the parabola opens downwards.

A downward-opening parabola is above the x-axis between the roots and below the x-axis outside them.

So the solution is \(-2 < x < 4\).

Find the roots:

\(-x^2-2x+15=0\iff x^2+2x-15=0\).

\((x+5)(x-3)=0\), so the roots are \(x=-5\) and \(x=3\).

Because the coefficient of \(x^2\) is negative, the parabola opens downwards.

A downward-opening parabola is above the x-axis between the roots and below the x-axis outside them.

So the solution is \(x \le -5\) or \(x \ge 3\).

Find the roots:

\(-4x^2+4x+3=0\iff 4x^2-4x-3=0\).

\((2x+1)(2x-3)=0\), so the roots are \(x=-\dfrac12\) and \(x=\dfrac32\).

Because the coefficient of \(x^2\) is negative, the parabola opens downwards.

A downward-opening parabola is above the x-axis between the roots and below the x-axis outside them.

So the solution is \(-\dfrac{1}{2} < x < \dfrac{3}{2}\).

Find the roots:

\(-6x^2-6x+12=0\iff x^2+x-2=0\).

\((x+2)(x-1)=0\), so the roots are \(x=-2\) and \(x=1\).

Because the coefficient of \(x^2\) is negative, the parabola opens downwards.

A downward-opening parabola is above the x-axis between the roots and below the x-axis outside them.

So the solution is \(x < -2\) or \(x > 1\).

Find the roots:

\(-x^2+9=0\iff x^2-9=0\).

\((x-3)(x+3)=0\), so the roots are \(x=-3\) and \(x=3\).

Because the coefficient of \(x^2\) is negative, the parabola opens downwards.

A downward-opening parabola is above the x-axis between the roots and below the x-axis outside them.

So the solution is \(-3 \le x \le 3\).

Find the roots:

\(-2x^2+18=0\iff x^2-9=0\).

\((x-3)(x+3)=0\), so the roots are \(x=-3\) and \(x=3\).

Because the coefficient of \(x^2\) is negative, the parabola opens downwards.

A downward-opening parabola is above the x-axis between the roots and below the x-axis outside them.

So the solution is \(x < -3\) or \(x > 3\).

Factorise:

\(x^2-4x+4=(x-2)^2\).

The graph touches the x-axis at \(x=2\) and does not go below it.

Because the coefficient of \(x^2\) is positive, the parabola opens upwards.

So the graph is never below the x-axis.

Therefore, the solution is No real solution.

Factorise:

\(x^2-6x+9=(x-3)^2\).

The graph touches the x-axis at \(x=3\) and is above the x-axis elsewhere.

Because the coefficient of \(x^2\) is positive, the parabola opens upwards.

So the graph is zero only at \(x=3\).

Therefore, the solution is \(x=3\).

Factorise:

\(2x^2+8x+8=2(x+2)^2\).

The graph touches the x-axis at \(x=-2\) and is above the x-axis everywhere else.

Because the coefficient of \(x^2\) is positive, the parabola opens upwards.

So the graph is positive for all \(x\) except at \(x=-2\).

Therefore, the solution is \(x\ne -2\).

Factorise:

\(3x^2-12x+12=3(x-2)^2\).

The graph touches the x-axis at \(x=2\) and is never below the x-axis.

Because the coefficient of \(x^2\) is positive, the parabola opens upwards.

So the graph is on or above the x-axis for every real value of \(x\).

Therefore, the solution is \(x\in\mathbb{R}\).

Factorise:

\(-x^2+2x-1=-(x-1)^2\).

The graph touches the x-axis at \(x=1\) and is below the x-axis elsewhere.

Because the coefficient of \(x^2\) is negative, the parabola opens downwards.

So the graph is never above the x-axis.

Therefore, the solution is No real solution.

Factorise:

\(-x^2+10x-25=-(x-5)^2\).

The graph touches the x-axis at \(x=5\) and is below the x-axis elsewhere.

Because the coefficient of \(x^2\) is negative, the parabola opens downwards.

So the graph is zero only at \(x=5\).

Therefore, the solution is \(x=5\).

Factorise:

\(-2x^2-8x-8=-2(x+2)^2\).

The graph touches the x-axis at \(x=-2\) and is below the x-axis everywhere else.

Because the coefficient of \(x^2\) is negative, the parabola opens downwards.

So the graph is negative for all \(x\) except at \(x=-2\).

Therefore, the solution is \(x\ne -2\).

Factorise:

\(-3x^2+18x-27=-3(x-3)^2\).

The graph touches the x-axis at \(x=3\) and is never above the x-axis.

Because the coefficient of \(x^2\) is negative, the parabola opens downwards.

So the graph is on or below the x-axis for every real value of \(x\).

Therefore, the solution is \(x\in\mathbb{R}\).

For \(y=x^2+4x+5\), the discriminant is \(b^2-4ac=4^2-4(1)(5)=16-20=-4\).

Since the discriminant is negative and the coefficient of \(x^2\) is positive, the graph never meets the x-axis and stays entirely above it.

Therefore, the inequality is true for every real value of \(x\).

For \(y=2x^2-4x+5\), the discriminant is \((-4)^2-4(2)(5)=16-40=-24\).

Since the discriminant is negative and the coefficient of \(x^2\) is positive, the graph is entirely above the x-axis.

Therefore, the inequality is never true for any real value of \(x\).

For \(y=3x^2+6x+4\), the discriminant is \(6^2-4(3)(4)=36-48=-12\).

Since the discriminant is negative and the coefficient of \(x^2\) is positive, the graph is entirely above the x-axis.

Therefore, the inequality is true for every real value of \(x\).

For \(y=x^2-2x+2\), the discriminant is \((-2)^2-4(1)(2)=4-8=-4\).

Since the discriminant is negative and the coefficient of \(x^2\) is positive, the graph stays entirely above the x-axis.

Therefore, the inequality is never true for any real value of \(x\).

For \(y=-x^2+4x-5\), the discriminant is \(4^2-4(-1)(-5)=16-20=-4\).

Since the discriminant is negative and the coefficient of \(x^2\) is negative, the graph stays entirely below the x-axis.

Therefore, the inequality is true for every real value of \(x\).

For \(y=-2x^2+8x-10\), the discriminant is \(8^2-4(-2)(-10)=64-80=-16\).

Since the discriminant is negative and the coefficient of \(x^2\) is negative, the graph is entirely below the x-axis.

Therefore, the inequality is never true for any real value of \(x\).

For \(y=-3x^2-6x-4\), the discriminant is \((-6)^2-4(-3)(-4)=36-48=-12\).

Since the discriminant is negative and the coefficient of \(x^2\) is negative, the graph stays entirely below the x-axis.

Therefore, the inequality is true for every real value of \(x\).

For \(y=-x^2+2x-2\), the discriminant is \(2^2-4(-1)(-2)=4-8=-4\).

Since the discriminant is negative and the coefficient of \(x^2\) is negative, the graph stays entirely below the x-axis.

Therefore, the inequality is never true for any real value of \(x\).

Solve \(x^2+4x+2=0\):

\(x=\dfrac{-4\pm\sqrt{16-8}}{2}=\dfrac{-4\pm\sqrt8}{2}=-2\pm\sqrt2\).

So the roots are \(x=-2-\sqrt2\) and \(x=-2+\sqrt2\).

Because the coefficient of \(x^2\) is positive, the parabola opens upwards.

An upward-opening parabola is below the x-axis between the roots and above the x-axis outside them.

So the solution is \(-2-\sqrt2 < x < -2+\sqrt2\).

Solve \(2x^2-8x-3=0\):

\(x=\dfrac{8\pm\sqrt{64+24}}{4}=\dfrac{8\pm\sqrt{88}}{4}=2\pm\dfrac{\sqrt{22}}{2}\).

So the roots are \(x=2-\dfrac{\sqrt{22}}{2}\) and \(x=2+\dfrac{\sqrt{22}}{2}\).

Because the coefficient of \(x^2\) is positive, the parabola opens upwards.

An upward-opening parabola is below the x-axis between the roots and above the x-axis outside them.

So the solution is \(x < 2-\dfrac{\sqrt{22}}{2}\) or \(x > 2+\dfrac{\sqrt{22}}{2}\).

Maths4U original, AI-assisted. Teacher editorial review pending. One point for the final answer; written reasoning is not assessed here.

Maths4U original, AI-assisted. Teacher editorial review pending. One point for the final answer; written reasoning is not assessed here.

Maths4U original, AI-assisted. Teacher editorial review pending. One point for the final answer; written reasoning is not assessed here.

Maths4U original, AI-assisted. Teacher editorial review pending. One point for the final answer; written reasoning is not assessed here.