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Algebra - Quadratic Equations and Inequalities

Grade 12IGCSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Standard Form: ax2+bx+c=0ax^2 + bx + c = 0, where a≠0a \neq 0.

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Methods of Solving: Factorization, Completing the Square, and the Quadratic Formula.

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The Discriminant (Δ=b2−4ac\Delta = b^2 - 4ac): Determines the nature of the roots (Real and distinct if Δ>0\Delta > 0, Real and equal if Δ=0\Delta = 0, No real roots if Δ<0\Delta < 0).

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Relationships between Roots: For roots α\alpha and β\beta, sum α+β=−b/a\alpha + \beta = -b/a and product αβ=c/a\alpha\beta = c/a.

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Quadratic Inequalities: Solved by finding critical values (roots) and testing intervals or using the graph of the parabola.

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Hidden Quadratics: Equations like ax4+bx2+c=0ax^4 + bx^2 + c = 0 that can be transformed into quadratics using substitution (e.g., let u=x2u = x^2).

📐Formulae

x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} (Quadratic Formula)

Δ=b2−4ac\Delta = b^2 - 4ac (Discriminant)

a(x−h)2+k=0a(x - h)^2 + k = 0 (Vertex Form, where (h,k)(h, k) is the vertex)

α+β=−ba\alpha + \beta = -\frac{b}{a} (Sum of Roots)

αβ=ca\alpha\beta = \frac{c}{a} (Product of Roots)

💡Examples

Problem 1:

Solve the inequality x2−5x+6>0x^2 - 5x + 6 > 0.

Solution:

  1. Factorize the quadratic: (x−2)(x−3)>0(x - 2)(x - 3) > 0. 2. Identify critical values: x=2x = 2 and x=3x = 3. 3. Test intervals or use a sketch: For x<2x < 2, the expression is positive. For 2<x<32 < x < 3, it is negative. For x>3x > 3, it is positive. 4. Solution: x<2x < 2 or x>3x > 3.

Explanation:

To solve a quadratic inequality, we first find the roots of the corresponding equation. These roots divide the number line into intervals. Since the coefficient of x2x^2 is positive, the parabola opens upwards, meaning the expression is positive outside the roots.

Problem 2:

Find the range of values of kk for which the equation x2+kx+4=0x^2 + kx + 4 = 0 has two distinct real roots.

Solution:

  1. For distinct real roots, Δ>0\Delta > 0. 2. Substitute values: k2−4(1)(4)>0k^2 - 4(1)(4) > 0. 3. k2−16>0k^2 - 16 > 0. 4. (k−4)(k+4)>0(k - 4)(k + 4) > 0. 5. Range: k<−4k < -4 or k>4k > 4.

Explanation:

The nature of the roots is determined by the discriminant. 'Two distinct real roots' strictly requires the discriminant to be greater than zero. We then solve the resulting quadratic inequality in terms of kk.

Problem 3:

Express 2x2−8x+52x^2 - 8x + 5 in the form a(x−h)2+ka(x - h)^2 + k.

Solution:

  1. Factor out 2 from the first two terms: 2(x2−4x)+52(x^2 - 4x) + 5. 2. Complete the square inside: 2[(x−2)2−4]+52[(x - 2)^2 - 4] + 5. 3. Expand: 2(x−2)2−8+52(x - 2)^2 - 8 + 5. 4. Result: 2(x−2)2−32(x - 2)^2 - 3.

Explanation:

Completing the square is a process used to find the vertex of a parabola and to solve quadratic equations that are not easily factorable. Here, a=2,h=2,k=−3a=2, h=2, k=-3.