Samacheer Kalvi Class 12 Maths Solutions Chapter 3 Theory of Equations Exercise 3.4

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 3 Theory of Equations Ex 3.4

Question 1. Solve:
(i) \( (x - 5) (x - 7) (x + 6)(x + 4) = 504 \)
(ii) \( (x - 4)(x - 2)(x - 7)(x + 1) = 16 \)
Answer:
(i) We need to solve the equation \( (x - 5) (x - 7) (x + 6)(x + 4) = 504 \).
First, rearrange the terms to group them smartly:
\( (x - 5) (x + 4) (x - 7) (x + 6) = 504 \)
Now, multiply the first two terms and the last two terms:
\( (x^2 + 4x - 5x - 20) (x^2 + 6x - 7x - 42) = 504 \)
\( (x^2 - x - 20) (x^2 - x - 42) = 504 \)
Let \( y = x^2 - x \) to simplify the equation. This makes it easier to handle.
\( (y - 20) (y - 42) = 504 \)
Expand this quadratic equation:
\( y^2 - 42y - 20y + 840 = 504 \)
\( y^2 - 62y + 840 = 504 \)
Subtract 504 from both sides to set the equation to zero:
\( y^2 - 62y + 840 - 504 = 0 \)
\( y^2 - 62y + 336 = 0 \)
Now, factor this quadratic equation for y:
\( (y - 56) (y - 6) = 0 \)
This gives two possible values for y:
\( y - 56 = 0 \implies y = 56 \)
Or
\( y - 6 = 0 \implies y = 6 \)
Next, substitute back \( x^2 - x \) for y and solve for x.
Case 1: \( x^2 - x = 56 \)
\( x^2 - x - 56 = 0 \)
Factor the quadratic equation:
\( (x - 8) (x + 7) = 0 \)
This gives two solutions for x:
\( x - 8 = 0 \implies x = 8 \)
Or
\( x + 7 = 0 \implies x = -7 \)
Case 2: \( x^2 - x = 6 \)
\( x^2 - x - 6 = 0 \)
Factor the quadratic equation:
\( (x - 3) (x + 2) = 0 \)
This gives two more solutions for x:
\( x - 3 = 0 \implies x = 3 \)
Or
\( x + 2 = 0 \implies x = -2 \)
So, the roots of the equation are \( 8, -7, 3, -2 \). Finding common terms like \( x^2 - x \) is key to simplifying these complex equations.

(ii) We need to solve the equation \( (x - 4)(x - 2)(x - 7)(x + 1) = 16 \).
First, rearrange the terms to group them smartly:
\( (x - 4) (x + 1) (x - 2) (x - 7) = 16 \)
Multiply the first two terms and the last two terms:
\( (x^2 + x - 4x - 4) (x^2 - 7x - 2x + 14) = 16 \)
\( (x^2 - 3x - 4) (x^2 - 9x + 14) = 16 \) (Correction from source: the original PDF's grouping of terms in this part seems to have a copy-paste error and doesn't lead to \( x^2 - 6x \). The PDF's next step uses `(x^2 - 6x + 8) (x^2 - 6x - 7) = 16`, implying a different grouping: \( (x - 4)(x - 2) \) and \( (x - 7)(x + 1) \). I will follow the grouping that leads to the \( x^2 - 6x \) substitution shown in the solution, which is \( (x-4)(x-2) \) and \( (x-7)(x+1) \).)
Let's follow the PDF's implied grouping for `x^2 - 6x` to match the solution steps:
Group terms: \( (x - 4) (x - 2) (x - 7) (x + 1) = 16 \)
Multiply \( (x-4)(x-2) \) and \( (x-7)(x+1) \):
\( (x^2 - 2x - 4x + 8) (x^2 + x - 7x - 7) = 16 \)
\( (x^2 - 6x + 8) (x^2 - 6x - 7) = 16 \)
Let \( y = x^2 - 6x \) to simplify the equation.
\( (y + 8) (y - 7) = 16 \)
Expand this quadratic equation:
\( y^2 - 7y + 8y - 56 = 16 \)
\( y^2 + y - 56 = 16 \)
Subtract 16 from both sides:
\( y^2 + y - 56 - 16 = 0 \)
\( y^2 + y - 72 = 0 \)
Factor the quadratic equation for y:
\( (y + 9) (y - 8) = 0 \)
This gives two possible values for y:
\( y + 9 = 0 \implies y = -9 \)
Or
\( y - 8 = 0 \implies y = 8 \)
Substitute back \( x^2 - 6x \) for y and solve for x.
Case 1: \( x^2 - 6x = -9 \)
\( x^2 - 6x + 9 = 0 \)
This is a perfect square trinomial:
\( (x - 3)^2 = 0 \)
This gives a repeated root for x:
\( x = 3, 3 \)
Case 2: \( x^2 - 6x = 8 \)
\( x^2 - 6x - 8 = 0 \)
This quadratic equation cannot be factored easily, so use the quadratic formula \( x = \frac { -b \pm \sqrt { b^2 - 4ac } }{ 2a } \). Here, \( a=1, b=-6, c=-8 \).
\( x = \frac { -(-6) \pm \sqrt { (-6)^2 - 4(1)(-8) } }{ 2(1) } \)
\( x = \frac { 6 \pm \sqrt { 36 + 32 } }{ 2 } \)
\( x = \frac { 6 \pm \sqrt { 68 } }{ 2 } \)
Simplify the square root: \( \sqrt{68} = \sqrt{4 \times 17} = 2\sqrt{17} \).
\( x = \frac { 6 \pm 2\sqrt { 17 } }{ 2 } \)
Divide by 2:
\( x = 3 \pm \sqrt { 17 } \)
So, the roots of the equation are \( 3, 3, 3 + \sqrt{17}, 3 - \sqrt{17} \). Recognizing patterns for substitution helps simplify complex polynomial equations into easier quadratic forms.
In simple words: For equations like these, group terms to find a common expression you can replace with a single letter (like 'y'). Solve for 'y', then put the original expression back in and solve for 'x'. This makes a big problem into two smaller, easier ones.

🎯 Exam Tip: When faced with products of four linear factors, always try to group them such that the quadratic terms formed have a common \( x^2 + bx \) part. This allows for a simplifying substitution.

Question 2. Solve:
\( (2x - 1)(x + 3)(x - 2)(2x + 3) + 20 = 0 \)
Answer:
We need to solve the equation \( (2x - 1)(x + 3)(x - 2)(2x + 3) + 20 = 0 \).
First, rearrange the terms to group them smartly. We'll group them to find a common quadratic expression:
\( (x + 3) (x - 2) (2x - 1) (2x + 3) + 20 = 0 \)
Multiply the first two factors and the last two factors:
\( (x^2 + 3x - 2x - 6) (4x^2 + 6x - 2x - 3) + 20 = 0 \)
Simplify both resulting quadratic expressions:
\( (x^2 + x - 6) (4x^2 + 4x - 3) + 20 = 0 \)
Now, we want to make a substitution. Notice that the term \( x^2 + x \) appears in the first quadratic. In the second, \( 4x^2 + 4x \) is present. We can factor out 4 from the second term to get \( 4(x^2 + x) \).
\( (x^2 + x - 6) [4(x^2 + x) - 3] + 20 = 0 \)
Let \( y = x^2 + x \) to simplify the equation.
\( (y - 6) (4y - 3) + 20 = 0 \)
Expand this equation:
\( 4y^2 - 3y - 24y + 18 + 20 = 0 \)
Combine like terms:
\( 4y^2 - 27y + 38 = 0 \)
Now, factor this quadratic equation for y. We need two numbers that multiply to \( 4 \times 38 = 152 \) and add up to -27. These numbers are -19 and -8.
\( 4y^2 - 19y - 8y + 38 = 0 \)
Factor by grouping:
\( y(4y - 19) - 2(4y - 19) = 0 \)
\( (4y - 19) (y - 2) = 0 \)
This gives two possible values for y:
\( 4y - 19 = 0 \implies 4y = 19 \implies y = \frac { 19 }{ 4 } \)
Or
\( y - 2 = 0 \implies y = 2 \)
Next, substitute back \( x^2 + x \) for y and solve for x.
Case 1: \( x^2 + x = 2 \)
\( x^2 + x - 2 = 0 \)
Factor the quadratic equation:
\( (x + 2) (x - 1) = 0 \)
This gives two solutions for x:
\( x + 2 = 0 \implies x = -2 \)
Or
\( x - 1 = 0 \implies x = 1 \)
Case 2: \( x^2 + x = \frac { 19 }{ 4 } \)
Multiply by 4 to clear the fraction:
\( 4(x^2 + x) = 19 \)
\( 4x^2 + 4x = 19 \)
\( 4x^2 + 4x - 19 = 0 \)
This quadratic equation cannot be factored easily, so we use the quadratic formula \( x = \frac { -b \pm \sqrt { b^2 - 4ac } }{ 2a } \). Here, \( a=4, b=4, c=-19 \).
\( x = \frac { -4 \pm \sqrt { 4^2 - 4(4)(-19) } }{ 2(4) } \)
\( x = \frac { -4 \pm \sqrt { 16 + 304 } }{ 8 } \)
\( x = \frac { -4 \pm \sqrt { 320 } }{ 8 } \)
Simplify the square root: \( \sqrt{320} = \sqrt{64 \times 5} = 8\sqrt{5} \).
\( x = \frac { -4 \pm 8\sqrt { 5 } }{ 8 } \)
Factor out 4 from the numerator and simplify:
\( x = \frac { 4(-1 \pm 2\sqrt { 5 }) }{ 8 } \)
\( x = \frac { -1 \pm 2\sqrt { 5 } }{ 2 } \)
So, the roots of the equation are \( 1, -2, \frac { -1 + 2\sqrt { 5 } }{ 2 }, \frac { -1 - 2\sqrt { 5 } }{ 2 } \). Complex equations can often be simplified by recognizing common algebraic patterns for substitution.
In simple words: Group the factors carefully so that a common expression like \( x^2 + x \) appears. Replace this common part with 'y' to make the equation simpler. Solve for 'y', then put the original expression back to find all the 'x' values.

🎯 Exam Tip: When you have four linear factors, try pairing them up so that the resulting quadratic expressions have a common middle term, like \( ax^2 + bx \). This substitution technique drastically simplifies the problem.