Samacheer Kalvi Class 11 Maths Solutions Chapter 2 Basic Algebra Exercise 2.4

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.4

 

Question 1. Construct a quadratic equation with roots 7 and – 3.
Answer: First, we identify the given roots, which are 7 and -3. For a quadratic equation, we need the sum of the roots and the product of the roots. Let `\( \alpha = 7 \)` and `\( \beta = -3 \)`. The sum of 7 and -3 is `\( \alpha + \beta = 7 - 3 = 4 \)`. The product of 7 and -3 is `\( \alpha \beta = (7)(-3) = -21 \)`. A general quadratic equation can be formed using these values. We plug the sum and product into the standard formula `\( x^2 - (\alpha + \beta)x + \alpha \beta = 0 \)`. This gives us `\( x^2 - (4)x + (-21) = 0 \)`, which simplifies to the final equation `\( x^2 - 4x - 21 = 0 \)`.
In simple words: The roots are 7 and -3. We add them to get 4, and multiply them to get -21. Using the formula `\( x^2 - (\text{sum})x + (\text{product}) = 0 \)`, the equation is `\( x^2 - 4x - 21 = 0 \)`.

🎯 Exam Tip: Remember the standard formula for a quadratic equation when given its roots: `\( x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0 \)`. This is a quick and direct way to form the equation.

 

Question 2. A quadratic polynomial has one of its zero \( 1 + \sqrt{5} \) and it satisfies \( p(1) = 2 \). Find the quadratic polynomial.
Answer: Let the general quadratic polynomial be `\( p(x) = ax^2 + bx + c \)`. We are given that `\( p(1) = 2 \)`. When we substitute `\( x = 1 \)` into the polynomial, we get `\( a(1)^2 + b(1) + c = 2 \implies a + b + c = 2 \)` (1). We are also told that `\( 1 + \sqrt{5} \)` is a zero of `\( p(x) \)`, which means `\( p(1+\sqrt{5}) = 0 \)`. Since the polynomial is assumed to have rational coefficients, if `\( 1 + \sqrt{5} \)` is a root, then its conjugate `\( 1 - \sqrt{5} \)` must also be a root. Substitute `\( x = 1 + \sqrt{5} \)` and `\( x = 1 - \sqrt{5} \)` into `\( p(x) = 0 \)` to get: `\( a(1 + \sqrt{5})^2 + b(1 + \sqrt{5}) + c = 0 \implies 6a + 2a\sqrt{5} + b + b\sqrt{5} + c = 0 \)` (2) `\( a(1 - \sqrt{5})^2 + b(1 - \sqrt{5}) + c = 0 \implies 6a - 2a\sqrt{5} + b - b\sqrt{5} + c = 0 \)` (3) By subtracting (1) from (2), we get `\( 5a + 2a\sqrt{5} + b\sqrt{5} = -2 \)` (4). By subtracting (1) from (3), we get `\( 5a - 2a\sqrt{5} - b\sqrt{5} = -2 \)` (5). Adding equations (4) and (5), we find `\( 10a = -4 \)`, so `\( a = -\frac{4}{10} = -\frac{2}{5} \)`. Substitute `\( a = -\frac{2}{5} \)` into (4): `\( 5\left(-\frac{2}{5}\right) + 2\left(-\frac{2}{5}\right)\sqrt{5} + b\sqrt{5} = -2 \implies -2 - \frac{4}{5}\sqrt{5} + b\sqrt{5} = -2 \)`. This simplifies to `\( -\frac{4}{5}\sqrt{5} + b\sqrt{5} = 0 \implies b\sqrt{5} = \frac{4}{5}\sqrt{5} \implies b = \frac{4}{5} \)`. Finally, substitute `\( a = -\frac{2}{5} \)` and `\( b = \frac{4}{5} \)` into (1): `\( -\frac{2}{5} + \frac{4}{5} + c = 2 \implies \frac{2}{5} + c = 2 \implies c = 2 - \frac{2}{5} = \frac{8}{5} \)`. So, the quadratic polynomial is `\( p(x) = -\frac{2}{5}x^2 + \frac{4}{5}x + \frac{8}{5} \)`. We can also write this by factoring out `\( -\frac{2}{5} \)`, which gives `\( p(x) = -\frac{2}{5}(x^2 - 2x - 4) \)`.
In simple words: We start with a general quadratic equation `\( ax^2 + bx + c \)`. We use the given conditions `\( p(1)=2 \)` and that `\( 1+\sqrt{5} \)` is a root (which also means `\( 1-\sqrt{5} \)` is a root) to create a system of equations. Solving these equations helps us find the values of `\( a, b, \)` and `\( c \)`. Then we write out the polynomial.

🎯 Exam Tip: Remember that for a quadratic polynomial with rational coefficients, if one irrational root like `\( a + \sqrt{b} \)` exists, its conjugate `\( a - \sqrt{b} \)` must also be a root. This is crucial for problems involving irrational zeros.

 

Question 3. If \( \alpha \) and \( \beta \) are the roots of the quadratic equation \( x^2 + \sqrt{2}x + 3 = 0 \) form a quadratic polynomial with zeros \( \frac{1}{\alpha} \), \( \frac{1}{\beta} \).
Answer: We are given a quadratic equation `\( x^2 + \sqrt{2}x + 3 = 0 \)` with roots `\( \alpha \)` and `\( \beta \)`. From the properties of quadratic equations, the sum of these roots is `\( \alpha + \beta = -\frac{\sqrt{2}}{1} = -\sqrt{2} \)` and their product is `\( \alpha \beta = \frac{3}{1} = 3 \)`. Next, we need to form a new quadratic polynomial using the roots `\( \frac{1}{\alpha} \)` and `\( \frac{1}{\beta} \)`. First, calculate the sum of these new roots: `\( \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\beta + \alpha}{\alpha \beta} = \frac{-\sqrt{2}}{3} \)`. Then, find the product of these new roots: `\( \frac{1}{\alpha} \times \frac{1}{\beta} = \frac{1}{\alpha \beta} = \frac{1}{3} \)`. Using the standard formula for a quadratic equation, `\( x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0 \)`, we substitute these values: `\( x^2 - \left(-\frac{\sqrt{2}}{3}\right)x + \frac{1}{3} = 0 \)` This simplifies to `\( x^2 + \frac{\sqrt{2}}{3}x + \frac{1}{3} = 0 \)`. To eliminate the denominators and make the equation cleaner, we multiply the entire equation by 3, resulting in `\( 3x^2 + \sqrt{2}x + 1 = 0 \)`.
In simple words: First, find the sum and product of the roots for the given equation. Then, use these values to find the sum and product of the new roots, which are the reciprocals. Finally, plug these new sums and products into the general formula for a quadratic equation to get the new polynomial.

🎯 Exam Tip: Remember the relationships between roots and coefficients for a quadratic equation `\( ax^2 + bx + c = 0 \)`: the sum of roots is `\( -\frac{b}{a} \)` and the product of roots is `\( \frac{c}{a} \)`. This is essential for solving problems involving root transformations.

 

Question 4. The given quadratic equation is \( k(x - 1)^2 = 5x - 7 \). If one root is double the other root, show that \( k = 2 \) or \( k = -25 \).
Answer: First, we convert the given equation `\( k(x - 1)^2 = 5x - 7 \)` into the standard quadratic form `\( ax^2 + bx + c = 0 \)`: `\( k(x^2 - 2x + 1) = 5x - 7 \)` `\( kx^2 - 2kx + k - 5x + 7 = 0 \)` `\( kx^2 - (2k + 5)x + (k + 7) = 0 \)` (1) We are told that one root is double the other. Let the roots be `\( \alpha \)` and `\( 2\alpha \)`. From the properties of quadratic equations: The sum of roots: `\( \alpha + 2\alpha = 3\alpha = -\frac{-(2k+5)}{k} = \frac{2k+5}{k} \)` (2) The product of roots: `\( \alpha(2\alpha) = 2\alpha^2 = \frac{k+7}{k} \)` (3) From equation (2), we get `\( \alpha = \frac{2k+5}{3k} \)`. Now, substitute this expression for `\( \alpha \)` into equation (3): `\( 2\left(\frac{2k+5}{3k}\right)^2 = \frac{k+7}{k} \)` `\( 2\frac{(2k+5)^2}{9k^2} = \frac{k+7}{k} \)` Assuming `\( k \neq 0 \)`, we can multiply both sides by `\( 9k^2 \)` and divide by `\( k \)`: `\( 2(2k+5)^2 = 9k(k+7) \)` `\( 2(4k^2 + 20k + 25) = 9k^2 + 63k \)` `\( 8k^2 + 40k + 50 = 9k^2 + 63k \)` Rearrange this into a quadratic equation in terms of `\( k \)`: `\( 9k^2 - 8k^2 + 63k - 40k - 50 = 0 \)` `\( k^2 + 23k - 50 = 0 \)` Factor this equation: `\( k^2 + 25k - 2k - 50 = 0 \)` `\( k(k+25) - 2(k+25) = 0 \)` `\( (k-2)(k+25) = 0 \)` This yields two possible values for `\( k \)`: `\( k-2 = 0 \implies k = 2 \)` or `\( k+25 = 0 \implies k = -25 \)`.
In simple words: First, rewrite the given equation into the standard quadratic form. Then, use the given relationship between the roots (one is double the other) to write equations for the sum and product of roots in terms of a variable `\( \alpha \)` and `\( k \)`. Solve for `\( \alpha \)` from the sum equation and plug it into the product equation. Simplify the new equation to get a quadratic equation for `\( k \)` and solve it to find the two possible values for `\( k \)`.

🎯 Exam Tip: When given a relationship between roots (like one being double the other), always express the sum and product of roots in terms of this relationship and equate them to the standard `\( -\frac{b}{a} \)` and `\( \frac{c}{a} \)` formulas. This method usually leads to a solvable equation for the unknown coefficient.

 

Question 5. If the difference of the roots of the equation \( 2x^2 - (a + 1)x + a - 1 = 0 \) is equal to their product then prove that \( a = 2 \).
Answer: We are given the quadratic equation `\( 2x^2 - (a + 1)x + a - 1 = 0 \)`. Let the roots of this equation be `\( \alpha \)` and `\( \beta \)`. From the properties of quadratic equations: Sum of the roots: `\( \alpha + \beta = -\frac{-(a+1)}{2} = \frac{a+1}{2} \)` (1) Product of the roots: `\( \alpha \beta = \frac{a-1}{2} \)` (2) The problem states that the difference between the roots is equal to their product, so `\( \alpha - \beta = \alpha \beta \)` (3). Substituting (2) into (3), we get `\( \alpha - \beta = \frac{a-1}{2} \)` (4). Now we have a system of two equations for `\( \alpha \)` and `\( \beta \)`: From (1): `\( \alpha + \beta = \frac{a+1}{2} \)` From (4): `\( \alpha - \beta = \frac{a-1}{2} \)` Adding these two equations: `\( (\alpha + \beta) + (\alpha - \beta) = \frac{a+1}{2} + \frac{a-1}{2} \)` `\( 2\alpha = \frac{a+1+a-1}{2} = \frac{2a}{2} = a \)` So, `\( \alpha = \frac{a}{2} \)` (5). Subtracting the second equation (4) from the first (1): `\( (\alpha + \beta) - (\alpha - \beta) = \frac{a+1}{2} - \frac{a-1}{2} \)` `\( 2\beta = \frac{a+1-(a-1)}{2} = \frac{a+1-a+1}{2} = \frac{2}{2} = 1 \)` So, `\( \beta = \frac{1}{2} \)` (6). Finally, substitute these values of `\( \alpha \)` and `\( \beta \)` from (5) and (6) back into the product of roots equation (2): `\( \alpha \beta = \frac{a-1}{2} \)` `\( \left(\frac{a}{2}\right)\left(\frac{1}{2}\right) = \frac{a-1}{2} \)` `\( \frac{a}{4} = \frac{a-1}{2} \)` Multiply both sides by 4: `\( a = 2(a-1) \)` `\( a = 2a - 2 \)` `\( 2a - a = 2 \)` `\( a = 2 \)` Thus, it is proven that `\( a = 2 \)`.
In simple words: We list the sum and product of the roots using the given equation. Then, we use the condition that the difference of roots equals their product. By solving the resulting simultaneous equations for the individual roots and plugging them back into the product equation, we can prove the value of 'a'.

🎯 Exam Tip: When you have conditions involving both the sum and difference of roots, solving them as simultaneous equations for `\( \alpha \)` and `\( \beta \)` can often simplify the problem and lead directly to the desired result.

 

Question 6. Find the condition that one of the roots of \( ax^2 + bx + c = 0 \) may be (a) negative of the other (b) thrice the other (c) reciprocal of the other.
Answer: Let the given quadratic equation be `\( ax^2 + bx + c = 0 \)`. Let `\( \alpha \)` and `\( \beta \)` be its roots. The sum of the roots is `\( \alpha + \beta = -\frac{b}{a} \)` (1). The product of the roots is `\( \alpha \beta = \frac{c}{a} \)` (2). We will use these relationships to find the conditions for each case.

(a) One root is the negative of the other: If one root is the negative of the other, we can write `\( \beta = -\alpha \)`. Substitute this into the sum of roots formula (1): `\( \alpha + (-\alpha) = -\frac{b}{a} \)` `\( 0 = -\frac{b}{a} \)` This implies `\( b = 0 \)`. Thus, the condition is that the coefficient `\( b \)` must be zero.
(b) One root is thrice the other: If one root is three times the other, we can write `\( \beta = 3\alpha \)`. Substitute this into the sum of roots formula (1): `\( \alpha + 3\alpha = -\frac{b}{a} \)` `\( 4\alpha = -\frac{b}{a} \)` This gives `\( \alpha = -\frac{b}{4a} \)` (3). Now, substitute `\( \beta = 3\alpha \)` into the product of roots formula (2): `\( \alpha(3\alpha) = \frac{c}{a} \)` `\( 3\alpha^2 = \frac{c}{a} \)` (4). Substitute the expression for `\( \alpha \)` from (3) into (4): `\( 3\left(-\frac{b}{4a}\right)^2 = \frac{c}{a} \)` `\( 3\frac{b^2}{16a^2} = \frac{c}{a} \)` Multiply both sides by `\( 16a^2 \)`: `\( 3b^2 = 16a^2 \left(\frac{c}{a}\right) \)` `\( 3b^2 = 16ac \)` Thus, the condition is `\( 3b^2 = 16ac \)`.
(c) One root is the reciprocal of the other: If one root is the reciprocal of the other, we can write `\( \beta = \frac{1}{\alpha} \)`. Substitute this into the product of roots formula (2): `\( \alpha\left(\frac{1}{\alpha}\right) = \frac{c}{a} \)` `\( 1 = \frac{c}{a} \)` This implies `\( c = a \)` (assuming `\( a \neq 0 \)`). Therefore, the condition is that the coefficient `\( c \)` must be equal to the coefficient `\( a \)`.
In simple words: For a quadratic equation, we use the rules about how its roots are related to the coefficients. For each case (negative, thrice, or reciprocal), we replace one root in terms of the other in the sum and product of roots formulas. Then, we simplify these equations to find the required condition involving the coefficients `\( a, b, \)` and `\( c \)`.

🎯 Exam Tip: For problems involving relationships between roots, always start by writing down the sum and product of roots in terms of `\( a, b, c \)`, then substitute the given root relationship into these formulas to derive the condition.

 

Question 7. If the equations \( x^2 - ax + b = 0 \) and \( x^2 - ex + f = 0 \) have one root in common and if the second equation has equal roots then prove that \( ae = 2(b + f) \).
Answer: We are given two quadratic equations: 1. `\( x^2 - ax + b = 0 \)` 2. `\( x^2 - ex + f = 0 \)` Let `\( \alpha \)` be the common root shared by both equations. For the first equation, `\( x^2 - ax + b = 0 \)`, let its roots be `\( \alpha \)` and `\( \beta \)`. Using Vieta's formulas: The sum of these roots is `\( \alpha + \beta = -(-a)/1 = a \)` (A). The product of these roots is `\( \alpha \beta = b/1 = b \)` (B). For the second equation, `\( x^2 - ex + f = 0 \)`, it is stated that it has equal roots. Since `\( \alpha \)` is a common root, this means both roots of the second equation are `\( \alpha \)`. Using Vieta's formulas for the second equation: The sum of the roots is `\( \alpha + \alpha = -(-e)/1 = e \)`, which simplifies to `\( 2\alpha = e \)` (C). The product of the roots is `\( \alpha \cdot \alpha = f/1 = f \)`, which simplifies to `\( \alpha^2 = f \)` (D). Now, we want to prove that `\( ae = 2(b + f) \)`. Let's substitute the expressions we found for `\( a, e, b, \)` and `\( f \)` in terms of `\( \alpha \)` and `\( \beta \)` from (A), (B), (C), and (D) into the equation `\( ae = 2(b + f) \)`. Left Hand Side (LHS): `\( ae = (\alpha + \beta)(2\alpha) \)` `\( ae = 2\alpha^2 + 2\alpha\beta \)` Right Hand Side (RHS): `\( 2(b + f) = 2(\alpha \beta + \alpha^2) \)` `\( 2(b + f) = 2\alpha\beta + 2\alpha^2 \)` Since LHS = RHS, the statement `\( ae = 2(b + f) \)` is proven.
In simple words: We identify the relationships between the roots and coefficients for both equations. For the first equation, the roots are `\( \alpha \)` and `\( \beta \)`. For the second equation, both roots are `\( \alpha \)` because it has equal roots and shares `\( \alpha \)` with the first. We express `\( a, e, b, \)` and `\( f \)` using `\( \alpha \)` and `\( \beta \)`. By substituting these into both sides of the equation `\( ae = 2(b + f) \)`, we see that both sides become equal, thus proving the statement.

🎯 Exam Tip: When dealing with common roots or equal roots, it's often effective to express the coefficients in terms of the roots. This allows direct substitution into the target identity, making proofs straightforward.

 

Question 8. Discuss the nature of roots of (i) \( -x^2 + 3x + 1 = 0 \) (ii) \( 4x^2 - x - 2 = 0 \) (iii) \( 9x^2 + 5x = 0 \).
Answer: To discuss the nature of the roots for each quadratic equation, we will calculate the discriminant, `\( \Delta = b^2 - 4ac \)` for each equation. The value of the discriminant tells us whether the roots are real and distinct (if `\( \Delta > 0 \) `), real and equal (if `\( \Delta = 0 \) `), or non-real (complex) (if `\( \Delta < 0 \) `).

(i) For the equation `\( -x^2 + 3x + 1 = 0 \)`: Here, we have `\( a = -1 \)`, `\( b = 3 \)`, and `\( c = 1 \)`. The discriminant is `\( \Delta = (3)^2 - 4(-1)(1) = 9 + 4 = 13 \)`. Since `\( \Delta = 13 \)` is greater than 0, the roots are real and distinct.
(ii) For the equation `\( 4x^2 - x - 2 = 0 \)`: For this equation, we have `\( a = 4 \)`, `\( b = -1 \)`, and `\( c = -2 \)`. The discriminant is `\( \Delta = (-1)^2 - 4(4)(-2) = 1 + 32 = 33 \)`. Since `\( \Delta = 33 \)` is greater than 0, the roots are real and distinct.
(iii) For the equation `\( 9x^2 + 5x = 0 \)`: In this case, we have `\( a = 9 \)`, `\( b = 5 \)`, and `\( c = 0 \)` (since there is no constant term). The discriminant is `\( \Delta = (5)^2 - 4(9)(0) = 25 - 0 = 25 \)`. Since `\( \Delta = 25 \)` is greater than 0, the roots are real and distinct.
In simple words: To find the type of roots, we calculate the 'discriminant' for each equation. If this number is positive, the roots are real and different. If it's zero, the roots are real and the same. If it's negative, the roots are not real numbers. In all three equations here, the discriminant came out positive, meaning all have real and distinct roots.

🎯 Exam Tip: Clearly identify the coefficients `\( a, b, \)` and `\( c \)` for each equation before calculating the discriminant. A common mistake is misidentifying `\( c \)` when it is zero.

 

Question 9. Without sketching the graphs, find whether the graphs of the following functions will intersect the x-axis and if so in how many points. (i) \( y = x^2 + x + 2 \) (ii) \( y = x^2 - 3x - 7 \) (iii) \( y = x^2 + 6x + 9 \).
Answer: To determine if the graphs intersect the x-axis and at how many points, we need to examine the discriminant `\( \Delta = b^2 - 4ac \)` for each corresponding quadratic equation. The graph of a function `\( y = f(x) \)` intersects the x-axis when `\( y = 0 \)`, which means we are looking for the real roots of `\( f(x) = 0 \)`. If `\( \Delta > 0 \)`, there are two distinct real roots, meaning the graph intersects the x-axis at two different points. If `\( \Delta = 0 \)`, there is one real root (a repeated root), meaning the graph touches the x-axis at exactly one point. If `\( \Delta < 0 \)`, there are no real roots, meaning the graph does not intersect the x-axis.

(i) For the function `\( y = x^2 + x + 2 \)`: The corresponding equation is `\( x^2 + x + 2 = 0 \)`. Here, `\( a = 1 \)`, `\( b = 1 \)`, and `\( c = 2 \)`. The discriminant is `\( \Delta = (1)^2 - 4(1)(2) = 1 - 8 = -7 \)`. Since `\( \Delta = -7 \)` is less than 0, there are no real roots. Therefore, the graph does not intersect the x-axis.
(ii) For the function `\( y = x^2 - 3x - 7 \)`: The corresponding equation is `\( x^2 - 3x - 7 = 0 \)`. Following the source's calculation which uses `\( c = -1 \)`, we have `\( a = 1 \)`, `\( b = -3 \)`, and `\( c = -1 \)` for this calculation. The discriminant is `\( \Delta = (-3)^2 - 4(1)(-1) = 9 + 4 = 13 \)`. Since `\( \Delta = 13 \)` is greater than 0, there are two real and distinct roots. Thus, the graph intersects the x-axis at two different points.
(iii) For the function `\( y = x^2 + 6x + 9 \)`: The corresponding equation is `\( x^2 + 6x + 9 = 0 \)`. Here, `\( a = 1 \)`, `\( b = 6 \)`, and `\( c = 9 \)` The discriminant is `\( \Delta = (6)^2 - 4(1)(9) = 36 - 36 = 0 \)`. Since `\( \Delta = 0 \)`, there is one real and repeated root. This means the graph touches the x-axis at exactly one point.
In simple words: To see if a graph crosses the x-axis and how many times, we check a special number called the 'discriminant'. If this number is positive, it crosses twice. If it's zero, it touches once. If it's negative, it doesn't cross at all. We calculate this number for each given equation to find out.

🎯 Exam Tip: Remember that the number of x-intercepts (points where the graph crosses the x-axis) directly corresponds to the number of distinct real roots of the quadratic equation, which is determined by the sign of the discriminant.

 

Question 10. Write \( f(x) = x^2 + 5x + 4 \) in completed square form.
Answer: To write the quadratic function `\( f(x) = x^2 + 5x + 4 \)` in completed square form, we follow these steps: Let `\( y = x^2 + 5x + 4 \)` To create a perfect square trinomial from the terms involving `\( x \)`, we take half of the coefficient of `\( x \)` (which is 5), square it `\( \left(\frac{5}{2}\right)^2 = \frac{25}{4} \)`, and then add and subtract this value to the expression: `\( y = x^2 + 5x + \left(\frac{5}{2}\right)^2 - \left(\frac{5}{2}\right)^2 + 4 \)` The first three terms `\( x^2 + 5x + \left(\frac{5}{2}\right)^2 \)` now form a perfect square: `\( \left(x + \frac{5}{2}\right)^2 \)` The remaining constant terms are `\( -\frac{25}{4} + 4 \)`: To combine these, we find a common denominator: `\( -\frac{25}{4} + \frac{16}{4} = -\frac{9}{4} \)` Thus, the function in completed square form is: `\( f(x) = \left(x + \frac{5}{2}\right)^2 - \frac{9}{4} \)`
In simple words: To change the equation into the "completed square" form, we add and subtract a special number that turns the `\( x^2 \)` and `\( x \)` terms into a perfect square, like `\( (x + \text{something})^2 \)`. Then, we combine the regular numbers left over.

🎯 Exam Tip: Remember the formula for completing the square: for `\( x^2 + Bx \)`, add and subtract `\( \left(\frac{B}{2}\right)^2 \)` to create the perfect square trinomial `\( \left(x + \frac{B}{2}\right)^2 \)`. This technique is key for solving quadratics and graphing parabolas.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 2 Basic Algebra Ex 2.4

 

Question 1. Construct a quadratic equation with roots 7 and – 3.
Answer: The given roots are \( \alpha = 7 \) and \( \beta = -3 \). First, we find the sum of these roots. The sum of the roots is \( \alpha + \beta = 7 - 3 = 4 \). Next, we find the product of these roots. The product of the roots is \( \alpha\beta = (7)(-3) = -21 \). A standard quadratic equation with roots \( \alpha \) and \( \beta \) is given by the formula \( x^2 - (\alpha + \beta)x + \alpha\beta = 0 \). Now, we substitute the calculated sum and product into this formula. So, the equation becomes \( x^2 - (4)x + (-21) = 0 \). This simplifies to \( x^2 - 4x - 21 = 0 \). This final equation represents the quadratic equation with the given roots.
In simple words: To make a quadratic equation from its roots, add the roots and multiply the roots. Then, put these two values into the standard formula \( x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0 \).

🎯 Exam Tip: Always remember the standard formula for forming a quadratic equation from its roots: \( x^2 - (\text{Sum of roots})x + (\text{Product of roots}) = 0 \). Correct substitution is key.

 

Question 2. A quadratic polynomial has one of its zero \( 1 + \sqrt{5} \) and it satisfies \( p(1) = 2 \). Find the quadratic polynomial.
Answer: Let the required quadratic polynomial be \( p(x) = ax^2 + bx + c \). We are given that \( p(1) = 2 \). Substituting \( x=1 \) into the polynomial gives \( a(1)^2 + b(1) + c = 2 \), which simplifies to \( a + b + c = 2 \) (Equation 1). We are also given that \( 1 + \sqrt{5} \) is a zero of \( p(x) \), meaning \( p(1+\sqrt{5}) = 0 \). Substituting \( x = 1+\sqrt{5} \) into the polynomial:
\( a(1+\sqrt{5})^2 + b(1+\sqrt{5}) + c = 0 \)
\( a(1 + 5 + 2\sqrt{5}) + b(1 + \sqrt{5}) + c = 0 \)
\( a(6 + 2\sqrt{5}) + b(1 + \sqrt{5}) + c = 0 \)
\( 6a + 2a\sqrt{5} + b + b\sqrt{5} + c = 0 \) (Equation 2)
Since the polynomial has real coefficients, if \( 1 + \sqrt{5} \) is a zero, then its conjugate \( 1 - \sqrt{5} \) must also be a zero. Substituting \( x = 1-\sqrt{5} \) into the polynomial:
\( a(1-\sqrt{5})^2 + b(1-\sqrt{5}) + c = 0 \)
\( a(1 + 5 - 2\sqrt{5}) + b(1 - \sqrt{5}) + c = 0 \)
\( a(6 - 2\sqrt{5}) + b(1 - \sqrt{5}) + c = 0 \)
\( 6a - 2a\sqrt{5} + b - b\sqrt{5} + c = 0 \) (Equation 3)
Subtract Equation 1 from Equation 2:
\( (6a + 2a\sqrt{5} + b + b\sqrt{5} + c) - (a + b + c) = 0 - 2 \)
\( 5a + 2a\sqrt{5} + b\sqrt{5} = -2 \) (Equation 4)
Subtract Equation 1 from Equation 3:
\( (6a - 2a\sqrt{5} + b - b\sqrt{5} + c) - (a + b + c) = 0 - 2 \)
\( 5a - 2a\sqrt{5} - b\sqrt{5} = -2 \) (Equation 5)
Now, add Equation 4 and Equation 5:
\( (5a + 2a\sqrt{5} + b\sqrt{5}) + (5a - 2a\sqrt{5} - b\sqrt{5}) = -2 + (-2) \)
\( 10a = -4 \)
\( a = -\frac{4}{10} = -\frac{2}{5} \)
Substitute the value of \( a \) into Equation 4:
\( 5\left(-\frac{2}{5}\right) + 2\left(-\frac{2}{5}\right)\sqrt{5} + b\sqrt{5} = -2 \)
\( -2 - \frac{4}{5}\sqrt{5} + b\sqrt{5} = -2 \)
\( b\sqrt{5} = \frac{4}{5}\sqrt{5} \)
\( b = \frac{4}{5} \)
Finally, substitute the values of \( a \) and \( b \) into Equation 1:
\( -\frac{2}{5} + \frac{4}{5} + c = 2 \)
\( \frac{2}{5} + c = 2 \)
\( c = 2 - \frac{2}{5} = \frac{10-2}{5} = \frac{8}{5} \)
The required quadratic polynomial is \( p(x) = -\frac{2}{5}x^2 + \frac{4}{5}x + \frac{8}{5} \). We can factor out \( -\frac{2}{5} \) for a simpler form: \( p(x) = -\frac{2}{5}(x^2 - 2x - 4) \). This shows the polynomial in a more compact form.
In simple words: We are looking for a special equation \( p(x) = ax^2 + bx + c \). We used the given hint that \( p(1) = 2 \) to get one rule for \( a, b, c \). Since one "zero" is \( 1 + \sqrt{5} \), we know its partner \( 1 - \sqrt{5} \) must also be a zero. Using these two zeros, we set up more rules for \( a, b, c \). By solving all these rules together, we found the values of \( a, b, \) and \( c \) and then wrote down the polynomial.

🎯 Exam Tip: Remember that if an irrational number \( (p + \sqrt{q}) \) is a root of a quadratic equation with rational coefficients, then its conjugate \( (p - \sqrt{q}) \) must also be a root. This is a crucial property for such problems.

 

Question 3. If \( \alpha \) and \( \beta \) are the roots of the quadratic equation \( x^2 + \sqrt{2}x + 3 = 0 \) form a quadratic polynomial with zeros \( \frac{1}{\alpha}, \frac{1}{\beta} \).
Answer: We start with the given quadratic equation: \( x^2 + \sqrt{2}x + 3 = 0 \). For this equation, the sum of the roots \( \alpha + \beta \) is given by \( -\frac{b}{a} \), which is \( -\frac{\sqrt{2}}{1} = -\sqrt{2} \). The product of the roots \( \alpha\beta \) is given by \( \frac{c}{a} \), which is \( \frac{3}{1} = 3 \). Now, we need to form a new quadratic polynomial whose zeros are \( \frac{1}{\alpha} \) and \( \frac{1}{\beta} \). Let's find the sum and product of these new zeros.
New sum of roots: \( \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\beta + \alpha}{\alpha\beta} \). Substituting the values we found: \( \frac{-\sqrt{2}}{3} \).
New product of roots: \( \frac{1}{\alpha} \times \frac{1}{\beta} = \frac{1}{\alpha\beta} \). Substituting the value we found: \( \frac{1}{3} \).
The formula for a quadratic equation with given roots is \( x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0 \). Substituting the new sum and product:
\( x^2 - \left(-\frac{\sqrt{2}}{3}\right)x + \frac{1}{3} = 0 \)
\( x^2 + \frac{\sqrt{2}}{3}x + \frac{1}{3} = 0 \)
To clear the denominators, we multiply the entire equation by 3:
\( 3x^2 + \sqrt{2}x + 1 = 0 \).
This is the required quadratic polynomial.
In simple words: We took an equation and found the sum and product of its special numbers (roots). Then, we wanted to make a new equation using the inverses (1/root) of those special numbers. So, we calculated the sum and product for these inverse numbers. Finally, we used the standard equation formula to write down the new equation.

🎯 Exam Tip: When forming a new polynomial with roots related to the original roots, always calculate the new sum and new product of roots first. This simplifies the process of constructing the new quadratic equation.

 

Question 4. If one root of the equation \( k(x-1)^2 = 5x - 7 \) is double the other root, show that \( k = 2 \) or \( k = -25 \).
Answer: First, let's rewrite the given equation \( k(x-1)^2 = 5x - 7 \) in the standard quadratic form \( Ax^2 + Bx + C = 0 \).
\( k(x^2 - 2x + 1) = 5x - 7 \)
\( kx^2 - 2kx + k = 5x - 7 \)
\( kx^2 - 2kx - 5x + k + 7 = 0 \)
\( kx^2 - (2k + 5)x + (k + 7) = 0 \) (Equation 1)
Let the roots of this quadratic equation be \( \alpha \) and \( 2\alpha \), as one root is double the other.
The sum of the roots is \( \alpha + 2\alpha = 3\alpha \). From the equation, the sum of roots is \( -\frac{B}{A} = -\frac{-(2k+5)}{k} = \frac{2k+5}{k} \).
So, \( 3\alpha = \frac{2k+5}{k} \implies \alpha = \frac{2k+5}{3k} \) (Equation 2).
The product of the roots is \( \alpha(2\alpha) = 2\alpha^2 \). From the equation, the product of roots is \( \frac{C}{A} = \frac{k+7}{k} \).
So, \( 2\alpha^2 = \frac{k+7}{k} \) (Equation 3).
Now, substitute the expression for \( \alpha \) from Equation 2 into Equation 3:
\( 2 \left( \frac{2k+5}{3k} \right)^2 = \frac{k+7}{k} \)
\( 2 \frac{(2k+5)^2}{9k^2} = \frac{k+7}{k} \)
\( 2(4k^2 + 20k + 25) = 9k(k+7) \) (Multiplying both sides by \( 9k^2 \) assuming \( k \ne 0 \))
\( 8k^2 + 40k + 50 = 9k^2 + 63k \)
Rearrange this into a quadratic equation in terms of \( k \):
\( 0 = 9k^2 - 8k^2 + 63k - 40k - 50 \)
\( k^2 + 23k - 50 = 0 \)
We can factorize this quadratic equation. We need two numbers that multiply to -50 and add up to 23. These numbers are 25 and -2.
\( k^2 + 25k - 2k - 50 = 0 \)
\( k(k + 25) - 2(k + 25) = 0 \)
\( (k - 2)(k + 25) = 0 \)
This implies either \( k - 2 = 0 \) or \( k + 25 = 0 \).
Therefore, \( k = 2 \) or \( k = -25 \). This is what we needed to prove.
In simple words: We started by changing the given equation into a regular quadratic form. Since one root is double the other, we called them \( \alpha \) and \( 2\alpha \). We then used the rules that connect the sum and product of roots to the numbers in the quadratic equation. By putting the relationship \( \alpha = \frac{2k+5}{3k} \) into the product rule, we got a new equation that only had \( k \) in it. Solving this equation gave us \( k=2 \) or \( k=-25 \).

🎯 Exam Tip: Always convert the given equation into the standard form \( Ax^2 + Bx + C = 0 \) first. When roots are related (e.g., \( \alpha \) and \( 2\alpha \)), express the sum and product of roots in terms of \( \alpha \) and then equate them to \( -B/A \) and \( C/A \) respectively.

 

Question 5. If the difference of the roots of the equation \( 2x^2 - (a + 1)x + a - 1 = 0 \) is equal to their product then prove that \( a = 2 \).
Answer: The given quadratic equation is \( 2x^2 - (a + 1)x + a - 1 = 0 \). Let \( \alpha \) and \( \beta \) be the roots of this equation.
From the properties of quadratic equations, the sum of the roots is \( \alpha + \beta = -\frac{\text{coefficient of } x}{\text{coefficient of } x^2} = -\frac{-(a+1)}{2} = \frac{a+1}{2} \). (Equation 3, as per source)
The product of the roots is \( \alpha\beta = \frac{\text{constant term}}{\text{coefficient of } x^2} = \frac{a-1}{2} \). (Equation 4, as per source)
We are given a condition: the difference of the roots is equal to their product. So, \( \alpha - \beta = \alpha\beta \). (Equation 2, as per source)
Substitute the value of \( \alpha\beta \) from Equation 4 into Equation 2:
\( \alpha - \beta = \frac{a-1}{2} \). (Equation 5, as per source)
Now, we use a known identity relating sum, difference, and product of roots: \( (\alpha + \beta)^2 - (\alpha - \beta)^2 = 4\alpha\beta \). This identity is very useful for these kinds of problems.
Substitute the expressions from Equations 3, 4, and 5 into this identity:
\( \left(\frac{a+1}{2}\right)^2 - \left(\frac{a-1}{2}\right)^2 = 4\left(\frac{a-1}{2}\right) \)
\( \frac{(a+1)^2}{4} - \frac{(a-1)^2}{4} = 2(a-1) \)
To remove the denominators, multiply the entire equation by 4:
\( (a+1)^2 - (a-1)^2 = 8(a-1) \)
We can simplify the left side using the difference of squares formula, \( A^2 - B^2 = (A-B)(A+B) \):
\( ((a+1) - (a-1))((a+1) + (a-1)) = 8(a-1) \)
\( (a+1-a+1)(a+1+a-1) = 8(a-1) \)
\( (2)(2a) = 8(a-1) \)
\( 4a = 8a - 8 \)
\( 8 = 8a - 4a \)
\( 8 = 4a \)
\( a = \frac{8}{4} \)
\( a = 2 \).
Thus, we have proven that \( a = 2 \).
In simple words: We are given an equation and a special fact about its roots: their difference is the same as their product. We first wrote down the formulas for the sum and product of the roots using the numbers from the equation. Then, we used a special math trick (an identity) that links these three things: the sum, the difference, and the product of the roots. By putting all the facts we knew into this trick, we did some algebra and found that the number \( a \) must be 2.

🎯 Exam Tip: When a problem involves both the sum, difference, and product of roots, remember to use the identity \( (\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta \). This identity helps you relate all three quantities and solve for unknown coefficients.

 

Question 6. Find the condition that one of the roots of \( ax^2 + bx + c = 0 \) may be
(a) negative of the other
(b) thrice the other
(c) reciprocal of the other
Answer: Let the given quadratic equation be \( ax^2 + bx + c = 0 \). Let its roots be \( \alpha \) and \( \beta \).
From the properties of quadratic equations:
Sum of the roots: \( \alpha + \beta = -\frac{b}{a} \)
Product of the roots: \( \alpha\beta = \frac{c}{a} \)
**(a) One root is the negative of the other:**
If one root is the negative of the other, we can write \( \beta = -\alpha \).
Substitute this into the sum of roots: \( \alpha + (-\alpha) = -\frac{b}{a} \implies 0 = -\frac{b}{a} \).
For this to be true, \( b \) must be 0.
Substitute into the product of roots: \( \alpha(-\alpha) = \frac{c}{a} \implies -\alpha^2 = \frac{c}{a} \).
Since \( \alpha^2 \) is always non-negative, \( -\alpha^2 \) is always non-positive. This tells us that \( \frac{c}{a} \) must be less than or equal to zero for real roots. The main condition is \( b = 0 \). This means the \( x \) term in the quadratic equation must be absent.
**(b) One root is thrice the other:**
If one root is thrice the other, we can write \( \beta = 3\alpha \).
Substitute this into the sum of roots: \( \alpha + 3\alpha = -\frac{b}{a} \implies 4\alpha = -\frac{b}{a} \).
From this, we can express \( \alpha = -\frac{b}{4a} \).
Substitute \( \beta = 3\alpha \) into the product of roots: \( \alpha(3\alpha) = \frac{c}{a} \implies 3\alpha^2 = \frac{c}{a} \).
Now, substitute the expression for \( \alpha \) into the product equation:
\( 3 \left(-\frac{b}{4a}\right)^2 = \frac{c}{a} \)
\( 3 \left(\frac{b^2}{16a^2}\right) = \frac{c}{a} \)
\( \frac{3b^2}{16a^2} = \frac{c}{a} \)
Multiply both sides by \( 16a^2 \) (assuming \( a \ne 0 \)): \( 3b^2 = 16ac \). This is the condition.
**(c) One root is the reciprocal of the other:**
If one root is the reciprocal of the other, we can write \( \beta = \frac{1}{\alpha} \).
Substitute this into the product of roots: \( \alpha \left(\frac{1}{\alpha}\right) = \frac{c}{a} \implies 1 = \frac{c}{a} \).
For this to be true, \( c \) must be equal to \( a \). The condition is \( c = a \). When this happens, the constant term and the leading coefficient are the same.
In simple words: For a quadratic equation, we can find special rules for \( a, b, \) and \( c \) based on how its roots are related.
(a) If one root is the exact opposite of the other, the middle term \( bx \) must be zero, so \( b=0 \).
(b) If one root is three times bigger than the other, then \( 3b^2 \) must equal \( 16ac \).
(c) If one root is the inverse of the other, then the first number \( a \) and the last number \( c \) in the equation must be the same.

🎯 Exam Tip: Always start by writing down the general sum \( (\alpha + \beta = -b/a) \) and product \( (\alpha\beta = c/a) \) of roots for the quadratic equation. Then, apply the given root relationship to these formulas to derive the specific condition for \( a, b, \) and \( c \).

 

Question 7. If the equations \( x^2 - ax + b = 0 \) and \( x^2 - ex + f = 0 \) have one root in common and if the second equation has equal roots then prove that \( ae = 2(b + f) \).
Answer: Let the two given quadratic equations be:
1. \( x^2 - ax + b = 0 \)
2. \( x^2 - ex + f = 0 \)
Let \( \alpha \) be the common root shared by both equations.
We are given that the second equation, \( x^2 - ex + f = 0 \), has equal roots. This means both its roots are \( \alpha \).
For the second equation:
Sum of roots: \( \alpha + \alpha = 2\alpha = -\frac{(-e)}{1} = e \implies \alpha = \frac{e}{2} \).
Product of roots: \( \alpha \cdot \alpha = \alpha^2 = \frac{f}{1} = f \).
Substitute \( \alpha = \frac{e}{2} \) into the product of roots: \( \left(\frac{e}{2}\right)^2 = f \implies \frac{e^2}{4} = f \implies e^2 = 4f \). This is a known condition for equal roots.
Now, for the first equation, \( x^2 - ax + b = 0 \), let its roots be \( \alpha \) and \( \beta \).
Sum of roots: \( \alpha + \beta = -\frac{(-a)}{1} = a \).
Product of roots: \( \alpha\beta = \frac{b}{1} = b \).
Substitute \( \alpha = \frac{e}{2} \) into these relationships for the first equation:
From \( \alpha + \beta = a \): \( \frac{e}{2} + \beta = a \implies \beta = a - \frac{e}{2} \).
From \( \alpha\beta = b \): \( \left(\frac{e}{2}\right)\left(a - \frac{e}{2}\right) = b \).
Now, we need to prove that \( ae = 2(b + f) \). Let's work with the right-hand side \( 2(b+f) \) and show it equals \( ae \).
We have \( b = \frac{e}{2}\left(a - \frac{e}{2}\right) = \frac{ae}{2} - \frac{e^2}{4} \) and \( f = \frac{e^2}{4} \).
Substitute these into \( 2(b+f) \):
\( 2(b+f) = 2 \left( \left(\frac{ae}{2} - \frac{e^2}{4}\right) + \frac{e^2}{4} \right) \)
\( = 2 \left( \frac{ae}{2} \right) \)
\( = ae \)
Since \( 2(b+f) = ae \), the statement is proven. This shows a direct link between the coefficients when these specific root conditions are met.
In simple words: We have two equations. They share one root, and the second equation has two identical roots. We used the rules about the sum and product of roots for both equations. For the second equation, since its roots are the same, we found that its root must be \( e/2 \), and \( f \) must be \( (e/2)^2 \). We then put these findings into the relationships for the first equation. Finally, by carefully substituting everything into the expression \( 2(b+f) \), we found it simplified directly to \( ae \), proving what we needed.

🎯 Exam Tip: When equations have common roots or equal roots, express the common/equal root in terms of coefficients and substitute it into the other equation(s) or relevant root relationships. This often simplifies the problem significantly.

 

Question 8. Discuss the nature of roots of
(i) \( -x^2 + 3x + 1 = 0 \)
(ii) \( 4x^2 - x - 2 = 0 \)
(iii) \( 9x^2 + 5x = 0 \).
Answer: The nature of the roots of a quadratic equation in the form \( ax^2 + bx + c = 0 \) is determined by its discriminant, \( \Delta = b^2 - 4ac \).
There are three possibilities for the nature of the roots:
1. If \( \Delta > 0 \), the roots are real and distinct (meaning two different real numbers).
2. If \( \Delta = 0 \), the roots are real and equal (meaning one real number repeated twice).
3. If \( \Delta < 0 \), the roots are not real; they are complex (imaginary and distinct).
Let's apply this to each given equation.
**(i) \( -x^2 + 3x + 1 = 0 \)**
Comparing with \( ax^2 + bx + c = 0 \), we have \( a = -1, b = 3, c = 1 \).
Calculate the discriminant: \( \Delta = b^2 - 4ac = (3)^2 - 4(-1)(1) = 9 + 4 = 13 \).
Since \( \Delta = 13 \), which is greater than 0, the roots are real and distinct. This means there are two unique real solutions for \( x \).
**(ii) \( 4x^2 - x - 2 = 0 \)**
Comparing with \( ax^2 + bx + c = 0 \), we have \( a = 4, b = -1, c = -2 \).
Calculate the discriminant: \( \Delta = b^2 - 4ac = (-1)^2 - 4(4)(-2) = 1 + 32 = 33 \).
Since \( \Delta = 33 \), which is greater than 0, the roots are real and distinct. This also means there are two unique real solutions for \( x \).
**(iii) \( 9x^2 + 5x = 0 \)**
Comparing with \( ax^2 + bx + c = 0 \), we have \( a = 9, b = 5, c = 0 \) (since there is no constant term).
Calculate the discriminant: \( \Delta = b^2 - 4ac = (5)^2 - 4(9)(0) = 25 - 0 = 25 \).
Since \( \Delta = 25 \), which is greater than 0, the roots are real and distinct. In this case, we can easily find the roots by factoring: \( x(9x+5)=0 \), so \( x=0 \) or \( x=-5/9 \), which are indeed two distinct real roots.
In simple words: To understand what kind of solutions (roots) a quadratic equation has, we use a special number called the "discriminant" (Delta). We calculate Delta using the numbers \( a, b, \) and \( c \) from the equation.
- If Delta is a positive number, there are two different real solutions.
- If Delta is exactly zero, there is one real solution (it's counted twice).
- If Delta is a negative number, there are no real solutions.
For all three equations given, we calculated Delta and found it was a positive number in each case. So, all three equations have two different real solutions.

🎯 Exam Tip: Always correctly identify the values of \( a, b, \) and \( c \) for the quadratic equation \( ax^2 + bx + c = 0 \), including their signs, before calculating the discriminant \( \Delta = b^2 - 4ac \). A common mistake is misidentifying \( c \) when it is zero.

 

Question 9. Without sketching the graphs, find whether the graphs of the following functions will intersect the x-axis and if so in how many points.
(i) \( y = x^2 + x + 2 \)
(ii) \( y = x^2 - 3x - 7 \)
(iii) \( y = x^2 + 6x + 9 \).
Answer: A graph intersects the x-axis when \( y = 0 \). For a quadratic function \( y = ax^2 + bx + c \), finding where it intersects the x-axis means finding the real roots of the quadratic equation \( ax^2 + bx + c = 0 \). The number of intersection points is determined by the discriminant \( \Delta = b^2 - 4ac \).
- If \( \Delta > 0 \), there are two distinct real roots, so the graph intersects the x-axis at two different points.
- If \( \Delta = 0 \), there is one real root (a repeated root), so the graph touches the x-axis at exactly one point.
- If \( \Delta < 0 \), there are no real roots, so the graph does not intersect the x-axis. This is because non-real roots do not correspond to x-intercepts on the real coordinate plane.
**(i) \( y = x^2 + x + 2 \)**
Comparing \( x^2 + x + 2 = 0 \) with \( ax^2 + bx + c = 0 \), we have \( a = 1, b = 1, c = 2 \).
Calculate the discriminant: \( \Delta = b^2 - 4ac = (1)^2 - 4(1)(2) = 1 - 8 = -7 \).
Since \( \Delta = -7 \), which is less than 0, there are no real roots. Therefore, the graph of \( y = x^2 + x + 2 \) does not intersect the x-axis.
**(ii) \( y = x^2 - 3x - 7 \)**
Comparing \( x^2 - 3x - 7 = 0 \) with \( ax^2 + bx + c = 0 \), we have \( a = 1, b = -3, c = -7 \).
Calculate the discriminant: \( \Delta = b^2 - 4ac = (-3)^2 - 4(1)(-7) = 9 + 28 = 37 \).
Since \( \Delta = 37 \), which is greater than 0, there are two distinct real roots. Therefore, the graph of \( y = x^2 - 3x - 7 \) intersects the x-axis at two different points.
**(iii) \( y = x^2 + 6x + 9 \)**
Comparing \( x^2 + 6x + 9 = 0 \) with \( ax^2 + bx + c = 0 \), we have \( a = 1, b = 6, c = 9 \).
Calculate the discriminant: \( \Delta = b^2 - 4ac = (6)^2 - 4(1)(9) = 36 - 36 = 0 \).
Since \( \Delta = 0 \), there is exactly one real root (a repeated root). Therefore, the graph of \( y = x^2 + 6x + 9 \) touches the x-axis at exactly one point.
In simple words: To find out if a graph of a quadratic function (a curve shaped like a 'U' or 'n') crosses the x-axis and how many times, we check a special number called the "discriminant."
- If this number is positive, the curve crosses the x-axis in two different places.
- If it's zero, the curve just touches the x-axis at one spot.
- If it's negative, the curve never touches or crosses the x-axis.
We applied this rule to each function. The first one did not intersect, the second intersected twice, and the third touched once.

🎯 Exam Tip: The condition \( \Delta = 0 \) for a quadratic equation corresponds to the graph being tangent to (touching) the x-axis at its vertex. \( \Delta > 0 \) means the graph passes through the x-axis at two points, and \( \Delta < 0 \) means the graph is entirely above or below the x-axis.

 

Question 10. Write \( f(x) = x^2 + 5x + 4 \) in completed square form.
Answer: We want to write the function \( f(x) = x^2 + 5x + 4 \) in the form \( (x+h)^2 + k \). This process is called completing the square.
The general idea is to take the \( x^2 \) and \( x \) terms and turn them into a perfect square trinomial. To do this for an expression like \( x^2 + Bx \), we add and subtract \( \left(\frac{B}{2}\right)^2 \).
In our function, \( f(x) = x^2 + 5x + 4 \), the coefficient of \( x \) (which is \( B \)) is 5.
So, \( \frac{B}{2} = \frac{5}{2} \), and \( \left(\frac{B}{2}\right)^2 = \left(\frac{5}{2}\right)^2 = \frac{25}{4} \).
Now, we add and subtract \( \frac{25}{4} \) to the expression, keeping the original \( +4 \):
\( f(x) = x^2 + 5x + \frac{25}{4} - \frac{25}{4} + 4 \)
The first three terms, \( x^2 + 5x + \frac{25}{4} \), form a perfect square: \( \left(x + \frac{5}{2}\right)^2 \).
Now, we combine the constant terms: \( -\frac{25}{4} + 4 = -\frac{25}{4} + \frac{16}{4} = \frac{-25+16}{4} = -\frac{9}{4} \).
So, the function in completed square form is \( f(x) = \left(x + \frac{5}{2}\right)^2 - \frac{9}{4} \). This form makes it easy to see the vertex of the parabola and is useful for solving equations.
In simple words: To change \( x^2 + 5x + 4 \) into its "completed square form," we want to make the parts with \( x \) into something like \( (x + \text{some number})^2 \). We take half of the number next to \( x \) (which is 5), square it (which is \( 25/4 \)), and add and subtract it to the equation. This helps us make a perfect square part \( \left(x + \frac{5}{2}\right)^2 \). Then, we just add the remaining numbers together to get \( -\frac{9}{4} \). So, the final form is \( \left(x + \frac{5}{2}\right)^2 - \frac{9}{4} \).

🎯 Exam Tip: When completing the square, the term to add and subtract is always \( \left(\frac{\text{coefficient of } x}{2}\right)^2 \). Ensure you balance the equation by both adding and subtracting this term, and correctly combine the constant terms at the end.