ICSE Class 8 Maths Algebra Chapter 11 Quadratic Equations

Quadratic Equations

A quadratic equation in x is an equation in which the highest power of x is 2. The standard form of a quadratic equation in x is \(ax^2 + bx + c = 0\), where a, b and c are constants and \(a \ne 0\).

Examples \(3x^2 + 5x = 2\), \(-6x^2 = 7x + 9\) and \(3x^2 - 25 = 0\) are quadratic equations in x.

Solution Of A Quadratic Equation

In general, a quadratic equation has two solutions, which are also called roots. The two solutions may be equal.

Let the roots of the equation \(ax^2 + bx + c = 0\) be \(x = \alpha\) and \(x = \beta\). Then each of the values of x must satisfy the equation, that is, \(a\alpha^2 + b\alpha + c = 0\) and \(a\beta^2 + b\beta + c = 0\).

Solving A Quadratic Equation

To solve a quadratic equation, we first factorize it and then use the principle of zero product, which states: If the product of two expressions (or numbers) is zero then at least one of the expressions must be zero. We can express this symbolically as follows.

If \(ab = 0\) then \(a = 0\) or \(b = 0\) or both \(a = 0\) and \(b = 0\).

Example If \((x - 1)(x - 2) = 0\) then by the principle of zero product, either \(x - 1 = 0\) or \(x - 2 = 0\).

If \(x - 1 = 0\) then \(x = 1\). If \(x - 2 = 0\) then \(x = 2\). Therefore \(x = 1\) or 2.

Now we can write the steps for solving a quadratic equation.

Steps 1. Write the equation in the standard form (that is, in the form \(ax^2 + bx + c = 0\)).

2. Factorize the quadratic expression on the LHS.

3. Set each factor equal to zero.

4. Solve each of the resulting linear equations.

Solved Examples

Example 1 Solve the equation \(x^2 + 4x + 3 = 0\).

Solution To begin with, let us factorize the LHS of the given equation with 0 on RHS.

\(x^2 + 4x + 3 = 0\) or \(x^2 + x + 3x + 3 = 0\)

or \(x(x + 1) + 3(x + 1) = 0\) or \((x + 1)(x + 3) = 0\).

By the zero-product principle, either \(x + 1 = 0\), which gives \(x = -1\) or \(x + 3 = 0\), which gives \(x = -3\).

Therefore, the roots (or solutions) of the equation are -1 and -3.

Verification

Let us substitute the two values of x in the given equation one by one.

When \(x = -1\), \(x^2 + 4x + 3 = (-1)^2 + 4 \times (-1) + 3 = 1 - 4 + 3 = 0\), which is true.

When \(x = -3\), \(x^2 + 4x + 3 = (-3)^2 + 4 \times (-3) + 3 = 9 - 12 + 3 = 0\), which is also true.

Hence, the solutions \(x = -1\) and \(x = -3\) are correct.

Example 2 Solve the quadratic equation \(8m^2 - 10m - 7 = 0\).

Solution Given, \(8m^2 - 10m - 7 = 0\) or \(8m^2 - 14m + 4m - 7 = 0\) or \(2m(4m - 7) + 1(4m - 7) = 0\) or \((4m - 7)(2m + 1) = 0\).

By the zero-product principle,

either \(4m - 7 = 0\), i.e., \(4m = 7\), i.e., \(m = \frac{7}{4}\)

or \(2m + 1 = 0\), i.e., \(2m = -1\), i.e., \(m = \frac{-1}{2}\).

Hence, the solutions = \(\frac{7}{4}\) and \(\frac{-1}{2}\).

Example 3 Solve the equation \(4x^2 - 12x + 9 = 0\).

Solution Given, \(4x^2 - 12x + 9 = 0\) or \(4x^2 - 6x - 6x + 9 = 0\) or \(2x(2x - 3) - 3(2x - 3) = 0\) or \((2x - 3)(2x - 3) = 0\).

By the zero-product rule, either \(2x - 3 = 0\) or \(2x - 3 = 0\).

Therefore either \(x = \frac{3}{2}\) or \(x = \frac{3}{2}\).

Hence, the solutions are \(\frac{3}{2}\) and \(\frac{3}{2}\), which are equal.

Example 4 Solve \(4p^2 - 81 = 0\).

Solution Given, \(4p^2 - 81 = 0\) or \((2p)^2 - (9)^2 = 0\) or \((2p + 9)(2p - 9) = 0\).

This means either \(2p + 9 = 0\) or \(2p - 9 = 0\).

Therefore, either \(p = -\frac{9}{2}\) or \(p = \frac{9}{2}\).

Hence, the roots of the equation = \(\frac{-9}{2}, \frac{9}{2}\).

Example 5 Solve the equation \((x - 3)(x + 5) = 9\).

Solution Here, \((x - 3)(x + 5) = 9\) or \(x^2 + (5 - 3)x - 5 \times 3 = 9\)

(Since \((x + a)(x - b) = x^2 + (a - b)x - ab\))

or \(x^2 + 2x - 15 = 9\) or \(x^2 + 2x - 15 - 9 = 0\) or \(x^2 + 2x - 24 = 0\) or \(x^2 + 6x - 4x - 24 = 0\) or \(x(x + 6) - 4(x + 6) = 0\) or \((x + 6)(x - 4) = 0\).

This means either \(x + 6 = 0\), that is, \(x = -6\) or \(x - 4 = 0\), that is, \(x = 4\).

Hence, \(x = -6\) or \(x = 4\). So, the solutions of the equation are -6 and 4.

Example 6 Solve the equation \(x - \frac{40}{x} = 3, x \ne 0\).

Solution Multiplying both sides of the given equation by x,

\(x\left(x - \frac{40}{x}\right) = 3x\) or \(x^2 - 40 = 3x\) or \(x^2 - 3x - 40 = 0\) or \(x^2 - 8x + 5x - 40 = 0\) or \(x(x - 8) + 5(x - 8) = 0\) or \((x - 8)(x + 5) = 0\).

Hence, either \(x - 8 = 0\), which gives \(x = 8\) or \(x + 5 = 0\), which gives \(x = -5\).

Therefore, \(x = 8\) or \(x = -5\). So, the solutions are \(x = 8, -5\).

Example 7 Solve the equation \(\frac{3}{x - 2} + \frac{8}{x + 3} = 2\).

Solution Multiplying both sides of the given equation by \((x - 2)(x + 3)\), the LCM of the denominators,

\((x - 2)(x + 3)\left[\frac{3}{x - 2} + \frac{8}{x + 3}\right] = 2(x - 2)(x + 3)\)

or \(3(x + 3) + 8(x - 2) = 2(x^2 + (3 - 2)x - 2 \times 3)\)

or \(3x + 9 + 8x - 16 = 2(x^2 + x - 6)\) or \(2x^2 + 2x - 12 = 11x - 7\) or \(2x^2 - 9x - 5 = 0\) or \(2x^2 - 10x + x - 5 = 0\) or \(2x(x - 5) + 1(x - 5) = 0\) or \((x - 5)(2x + 1) = 0\).

Hence, either \(x - 5 = 0\), which gives \(x = 5\) or \(2x + 1 = 0\), which gives \(x = -\frac{1}{2}\).

Therefore, \(x = 5\) or \(x = -\frac{1}{2}\). Hence, the solutions are \(x = 5, -\frac{1}{2}\).

Example 8 Solve the equation \(\frac{x + 3}{x + 2} = \frac{3x - 7}{2x - 3}\).

Solution Given, \(\frac{x + 3}{x + 2} = \frac{3x - 7}{2x - 3}\) or \((x + 3)(2x - 3) = (x + 2)(3x - 7)\)

or \(x(2x - 3) + 3(2x - 3) = x(3x - 7) + 2(3x - 7)\)

or \(2x^2 - 3x + 6x - 9 = 3x^2 - 7x + 6x - 14\) or \(2x^2 + 3x - 9 = 3x^2 - x - 14\) or \(3x^2 - 2x^2 - x - 3x - 14 + 9 = 0\) or \(x^2 - 4x - 5 = 0\) or \(x^2 - 5x + x - 5 = 0\) or \(x(x - 5) + 1(x - 5) = 0\) or \((x - 5)(x + 1) = 0\).

Hence, either \(x - 5 = 0\), which gives \(x = 5\) or \(x + 1 = 0\), which gives \(x = -1\).

Therefore, the solutions of the equation are \(x = 5, -1\).

Teacher's Note

Quadratic equations appear frequently in real-world scenarios such as calculating projectile motion, determining profit and loss in business, or finding dimensions of structures - whenever a quantity depends on the square of another variable.

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