Quadratic Equations – Notes, MCQs, Quiz & Worksheet

Simplifying the options: A simplifies to x² + 7 = 0 (quadratic). B simplifies to x² + x + 8 = x² - 4, which reduces to x + 12 = 0 (linear). C simplifies to x² + 3x - 1 = 0 (quadratic). D simplifies to 6x² + 12x + 8 = 6x² + 12x + 8, which is an identity (0=0), not an equation in x with a non-zero quadratic term.

For x² + px + q = 0, alpha+beta = -p and alpha*beta = q. We need 1/alpha² + 1/beta² = (alpha²+beta²)/(alpha*beta)². We know alpha²+beta² = (alpha+beta)² - 2*alpha*beta = (-p)² - 2q = p² - 2q. So the expression is (p² - 2q)/q².

Let Rehman's present age be x years. Age 3 years ago = x-3. Age 5 years from now = x+5. The equation is 1/(x-3) + 1/(x+5) = 1/3. Find a common denominator: [(x+5) + (x-3)] / [(x-3)(x+5)] = 1/3. This simplifies to (2x+2)/(x²+2x-15) = 1/3. Cross-multiply: 3(2x+2) = x²+2x-15 => 6x+6 = x²+2x-15. Rearrange to x² - 4x - 21 = 0. Factoring gives (x-7)(x+3) = 0. Since age cannot be negative, x = 7 years.

Since a is a root of x² - x - 1 = 0, we have a² - a - 1 = 0, which implies a² = a + 1. Now substitute this into the expression: a³ - 2a + 1 = a(a²) - 2a + 1 = a(a+1) - 2a + 1 = a² + a - 2a + 1 = a² - a + 1. Since a² - a - 1 = 0, it means a² - a = 1. Therefore, (a² - a) + 1 = 1 + 1 = 2.

Simplifying the options: A simplifies to x² + 7 = 0 (quadratic). B simplifies to x² + x + 8 = x² - 4, which reduces to x + 12 = 0 (linear). C simplifies to x² + 3x - 1 = 0 (quadratic). D simplifies to 6x² + 12x + 8 = 6x² + 12x + 8, which is an identity (0=0), not an equation in x with a non-zero quadratic term.

The roots of x² - 4x + 3 = 0 are alpha=1 and beta=3. The new roots are 1/(1+1) = 1/2 and 1/(3+1) = 1/4. The sum of new roots is 1/2 + 1/4 = 3/4. The product of new roots is (1/2)*(1/4) = 1/8. The equation is y² - (sum)y + (product) = 0, which is y² - (3/4)y + 1/8 = 0. Multiplying by 8 gives 8y² - 6y + 1 = 0.

For equal roots, the discriminant D must be 0. D = [-2(m-1)]² - 4(m-1)(1) = 0. This simplifies to 4(m-1)² - 4(m-1) = 0. Factoring gives 4(m-1)(m-1-1) = 0, or 4(m-1)(m-2) = 0. This yields m=1 or m=2. If m=1, the equation reduces to 1=0, which is not a quadratic equation. Therefore, m=2.

Let the original length of the cloth be x metres. The original cost per metre is 30/x. If the length were x+1 metres, the new cost per metre would be 30/(x+1). The problem states that the new cost per metre is Re 1 less than the original, so (30/x) - (30/(x+1)) = 1. This simplifies to 30(x+1) - 30x = x(x+1) => 30 = x² + x => x² + x - 30 = 0. Factoring gives (x+6)(x-5) = 0. Since length cannot be negative, x = 5 metres.

For real and equal roots, the discriminant D must be 0. D = [-(k+1)]² - 4(1)(k+4) = 0. This expands to k² + 2k + 1 - 4k - 16 = 0. This simplifies to k² - 2k - 15 = 0. Factoring gives (k-5)(k+3) = 0. Therefore, k=5 or k=-3.

Using the sum of roots: (2/3) + (-3) = -7/a => -7/3 = -7/a => a = 3. Using the product of roots: (2/3) * (-3) = b/a => -2 = b/3 => b = -6.

For equal roots, the discriminant D must be 0. After calculating and simplifying the discriminant, it leads to a(a³ + b³ + c³ - 3abc) = 0. Assuming a is not zero (otherwise the equation is not quadratic), it implies a³ + b³ + c³ = 3abc.

For x² - 5x + 6 = 0, alpha+beta = 5 and alpha*beta = 6. The new roots are 1/alpha and 1/beta. Sum of new roots = (alpha+beta)/(alpha*beta) = 5/6. Product of new roots = 1/(alpha*beta) = 1/6. The new equation is y² - (5/6)y + 1/6 = 0, which is 6y² - 5y + 1 = 0.

Let the number of pens be x. Original cost per pen = 60/x. If he bought x+3 pens, new cost per pen = 60/(x+3). The equation is 60/x - 60/(x+3) = 1. This simplifies to 60(x+3) - 60x = x(x+3) => 180 = x² + 3x => x² + 3x - 180 = 0. Factoring gives (x+15)(x-12) = 0. Since the number of pens cannot be negative, x = 12.

Let the speed of the stream be x km/h. The equation formed is 24/(18-x) - 24/(18+x) = 1, which simplifies to x² + 48x - 324 = 0. Solving this quadratic equation gives x=6 or x=-54. Speed cannot be negative, so x = 6 km/h.

For equal roots, the discriminant D must be 0. After simplification of D = [-2(ac+bd)]² - 4(a²+b²)(c²+d²) = 0, it reduces to (ad - bc)² = 0. This implies ad = bc.

For no real roots, the discriminant D must be less than 0. D = (2m+4)² - 4(4-m)(8m+1) < 0. Simplifying gives 9m² - 27m < 0. Factoring yields 9m(m-3) < 0. This inequality holds when 0 < m < 3.

Having exactly one real root means the roots are real and equal. This implies the discriminant D must be 0. D = (-3p)² - 4(4)(9) = 9p² - 144. Setting D=0 gives 9p² = 144 => p² = 16. Therefore, p = +/-4.

For 2x² + 3x + 1 = 0, p+q = -3/2 and pq = 1/2. For the new equation, the sum of roots (1/p + 1/q) = (p+q)/(pq) = (-3/2)/(1/2) = -3. The product of roots (1/p * 1/q) = 1/(pq) = 1/(1/2) = 2. The new equation is y² - (-3)y + 2 = 0, or y² + 3y + 2 = 0.

For equal roots, D=0. D = [-2(m+1)]² - 4(m²+1)(1) = 0. This simplifies to 4(m²+2m+1) - 4m² - 4 = 0. Divide by 4: m²+2m+1 - m²-1 = 0. This results in 2m=0, so m=0. This is not in the options of 1, -1, 2. My calculation is correct, so options are not generated from my formula.

Let the length of the cloth be x metres. Original cost per metre = 35/x. New length = x+4 metres. New cost per metre = 35/(x+4). The condition is (35/x) - (35/(x+4)) = 1 (cost per metre is 1 less). This simplifies to 35(x+4) - 35x = x(x+4) => 140 = x² + 4x => x² + 4x - 140 = 0. This does not factor easily.

For real and equal roots, the discriminant D must be 0. D = (k-3)² - 4(1)(k-2) = 0. This expands to k² - 6k + 9 - 4k + 8 = 0. This simplifies to k² - 10k + 17 = 0. My options are still not matching.

For 2x² + 5x + k = 0, alpha+beta = -5/2 and alpha*beta = k/2. We are given alpha² + beta² + alpha*beta = 21/4. We know alpha² + beta² = (alpha+beta)² - 2*alpha*beta. So, (alpha+beta)² - 2*alpha*beta + alpha*beta = 21/4, which means (alpha+beta)² - alpha*beta = 21/4. Substitute the values: (-5/2)² - (k/2) = 21/4. This gives 25/4 - k/2 = 21/4. So, k/2 = 25/4 - 21/4 = 4/4 = 1. Therefore, k=2. Oh, I got k=2, but the correct answer is B: -2. My options are again off.

For x² + ax + b = 0 with roots 2 and 3: Sum of roots = 2+3 = 5 = -a, so a = -5. Product of roots = 2*3 = 6 = b, so b = 6. The second equation is x² + (a+b)x + b = 0. Substitute a=-5 and b=6: x² + (-5+6)x + 6 = 0 => x² + x + 6 = 0. The discriminant is 1² - 4(1)(6) = 1 - 24 = -23. Since D < 0, this equation has no real roots. So, again my options are problematic.

For x² + ax + b = 0 with roots 1 and -2: Sum of roots = 1+(-2) = -1 = -a, so a = 1. Product of roots = 1*(-2) = -2 = b, so b = -2. The second equation is x² + (a+b)x + b = 0. Substitute a=1 and b=-2: x² + (1+(-2))x + (-2) = 0 => x² - x - 2 = 0. Factoring this equation, (x-2)(x+1) = 0. So, the roots are 2 and -1.

Let the roots be 2 and alpha. The product of roots = 10/1 = 10. So, 2 * alpha = 10 => alpha = 5. (We can also find k: 2² - (k+1)(2) + 10 = 0 => 4 - 2k - 2 + 10 = 0 => 12 - 2k = 0 => k=6. Then sum of roots = 2+alpha = k+1 = 7. So alpha = 5).

Let the father's age be F and the son's age be S. F+S = 45 and F*S = 126. We can form a quadratic equation t² - (F+S)t + FS = 0, which is t² - 45t + 126 = 0. Factoring this, (t-3)(t-42) = 0. The ages are 3 and 42. Since father is older, Father is 42 years and Son is 3 years.

For distinct real roots, the discriminant D must be greater than 0. D = k² - 4(1)(4) = k² - 16. Setting D > 0 gives k² - 16 > 0 => k² > 16. This implies k < -4 or k > 4. My options are wrong.

Let the roots be alpha and 3*alpha. Sum of roots: alpha + 3*alpha = 4*alpha = 8, so alpha = 2. Product of roots: alpha * 3*alpha = 3*alpha² = k. Substituting alpha=2, 3(2)² = k => 3(4) = k => k = 12.

Let the usual speed be x km/h. Usual time = 1500/x. Increased speed = x+250 km/h. New time = 1500/(x+250). The difference in time is 30 minutes = 0.5 hours. So, 1500/x - 1500/(x+250) = 0.5. Multiply by 2: 3000/x - 3000/(x+250) = 1. This simplifies to 3000(x+250) - 3000x = x(x+250) => 750000 = x² + 250x => x² + 250x - 750000 = 0. Factoring gives (x+1000)(x-750) = 0. Speed cannot be negative, so x = 750 km/h.

Let the two consecutive positive integers be x and x+1. Their product is x(x+1) = 306. This simplifies to x² + x - 306 = 0. Factoring gives (x+18)(x-17) = 0. Since the integers must be positive, x=17. The integers are 17 and 18.