Question

1. $x^2 + x < -10$
2. $3x^2 - x - 40 < x^2 + 2x - 2$
3. $x^2 + 3x < 12$
4. $5x^2 - x < 7x - 2$

Explanation:

Let's solve each inequality one by one.

Problem 5: \(x^2 + x < -10\)

Step 1: Rewrite the inequality

Rewrite the inequality as \(x^2 + x + 10 < 0\).

Step 2: Analyze the quadratic function

Consider the quadratic function \(y = x^2 + x + 10\). The discriminant of this quadratic is \(\Delta = b^2 - 4ac\), where \(a = 1\), \(b = 1\), and \(c = 10\). So \(\Delta = 1^2 - 4\times1\times10 = 1 - 40 = -39\). Since \(\Delta < 0\) and \(a = 1>0\), the parabola \(y = x^2 + x + 10\) is always above the \(x\)-axis. So there are no real solutions for \(x^2 + x + 10 < 0\).

Step 1: Rewrite the inequality

Rewrite the inequality as \(x^2 + 3x - 12 < 0\).

Step 2: Find the roots of the quadratic equation

For the quadratic equation \(x^2 + 3x - 12 = 0\), we use the quadratic formula \(x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = 3\), and \(c=- 12\). The discriminant \(\Delta=b^2 - 4ac=3^2-4\times1\times(- 12)=9 + 48 = 57\). Then the roots are \(x_1=\frac{-3+\sqrt{57}}{2}\) and \(x_2=\frac{-3-\sqrt{57}}{2}\).

Step 3: Determine the solution of the inequality

Since the coefficient of \(x^2\) ( \(a = 1>0\)) in the quadratic function \(y=x^2 + 3x - 12\) is positive, the parabola opens upwards. The inequality \(x^2 + 3x - 12<0\) holds for the values of \(x\) between the two roots. So the solution is \(\frac{-3-\sqrt{57}}{2}

Step 1: Rewrite the inequality

Rewrite the inequality as \(5x^2-8x + 2<0\).

Step 2: Find the roots of the quadratic equation

For the quadratic equation \(5x^2-8x + 2 = 0\), using the quadratic formula \(x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}\), where \(a = 5\), \(b=-8\), and \(c = 2\). The discriminant \(\Delta=b^2-4ac=(-8)^2-4\times5\times2=64 - 40 = 24\). Then the roots are \(x_1=\frac{8+\sqrt{24}}{10}=\frac{8 + 2\sqrt{6}}{10}=\frac{4+\sqrt{6}}{5}\) and \(x_2=\frac{8-\sqrt{24}}{10}=\frac{8-2\sqrt{6}}{10}=\frac{4 - \sqrt{6}}{5}\).

Step 3: Determine the solution of the inequality

Since the coefficient of \(x^2\) ( \(a = 5>0\)) in the quadratic function \(y = 5x^2-8x + 2\) is positive, the parabola opens upwards. The inequality \(5x^2-8x + 2<0\) holds for the values of \(x\) between the two roots. So the solution is \(\frac{4-\sqrt{6}}{5}

Answer:

No real solutions.

Problem 7: \(x^2 + 3x < 12\)