Q: Comparison of x and y from two quadratics (use mapping below): I. 16x^2 + 20x + 6 = 0 II. 10y^2 + 38y + 24 = 0 Mapping: 1 → x > y, 2 → x < y, 3 → x = y, 4 → Relationship cannot be determined.

Introduction / Context:The task is to compare the possible values of x and y when each variable must satisfy its own quadratic. Because a quadratic can have two roots, we must check all admissible x and y values and see if a single consistent relation holds in every case.Given Data / Assumptions:I: 16x^2 + 20x + 6 = 0II: 10y^2 + 38y + 24 = 0Concept / Approach:Factor (or reduce) each quadratic to find its real roots. Then compare every x-root with every y-root. If for all combinations x is greater than y, we can conclude x > y unequivocally.Step-by-Step Solution:I: Divide by 2 → 8x^2 + 10x + 3 = 0 = (4x + 3)(2x + 1)So x ∈ {−3/4 = −0.75, −1/2 = −0.5}II: Divide by 2 → 5y^2 + 19y + 12 = 0 = (5y + 4)(y + 3)So y ∈ {−4/5 = −0.8, −3}Compare: each x (−0.75, −0.5) is greater than each y (−0.8, −3).Verification / Alternative check:Numerically: max(y) = −0.8, min(x) = −0.75; since −0.75 > −0.8, even the least x exceeds the greatest y. Hence x > y for all admissible cases.Why Other Options Are Wrong:“x y

Q: Comparison of x and y from two quadratics (use mapping below): I. 18x^2 + 18x + 4 = 0 II. 12y^2 + 29y + 14 = 0 Mapping: 1 → x > y, 2 → x < y, 3 → x = y, 4 → Relationship cannot be determined.

Introduction / Context:When both x and y are defined by quadratics with two real roots, comparisons must consider all combinations. If some combinations yield equality and others yield strict inequality, the overall relationship is indeterminate.Given Data / Assumptions:I: 18x^2 + 18x + 4 = 0II: 12y^2 + 29y + 14 = 0Concept / Approach:Find the roots of each. Then examine whether a single relation (>, y. Different relations occur.Verification / Alternative check:Narrowest gap occurs at the common root −2/3. Since equality is possible and x > y also occurs, no single relation always holds.Why Other Options Are Wrong:Choosing x > y or x = y ignores that both cases occur; x < y never occurs for any admissible pair.Common Pitfalls:Concluding based on a single pair, missing the shared root case that forces equality.Final Answer:Relationship cannot be determined

Q: Solve the linear system and compare x and y (use mapping below): I. 4x + 7y = 209 II. 12x − 14y = −38 Mapping: 1 → x > y, 2 → x < y, 3 → x = y, 4 → Relationship cannot be determined.

Introduction / Context:For two linear equations in two variables, there is a unique solution when the lines are non-parallel. Once x and y are found, the comparison is direct and definitive.Given Data / Assumptions:4x + 7y = 20912x − 14y = −38Concept / Approach:Use elimination by aligning coefficients of y. Add the equations to eliminate y, compute x, then back-substitute to find y. Finally, compare x and y numerically.Step-by-Step Solution:From II: 12x − 14y = −38 ⇒ divide by 2 ⇒ 6x − 7y = −19Add with I: (4x + 7y) + (6x − 7y) = 209 + (−19) ⇒ 10x = 190x = 19Back-substitute: 4*19 + 7y = 209 ⇒ 76 + 7y = 209 ⇒ 7y = 133 ⇒ y = 19Hence x = yVerification / Alternative check:Plug x = y = 19 into both equations to confirm: 76 + 133 = 209 and 228 − 266 = −38. Both hold.Why Other Options Are Wrong:Since x and y are exactly equal, “greater than”, “less than”, or “indeterminate” are incorrect.Common Pitfalls:Arithmetic slips when dividing or adding equations. Keep track of signs carefully.Final Answer:x = y

Q: Comparison of x and y from two equations (use mapping below): I. 17x^2 + 48x = 9 II. 13y^2 = 32y − 12 Mapping: 1 → x > y, 2 → x < y, 3 → x = y, 4 → Relationship cannot be determined.

Introduction / Context:Each variable is governed by a quadratic; we must list all possible roots for x and y, then check whether a single inequality relation holds for every admissible pair (x, y).Given Data / Assumptions:I: 17x^2 + 48x − 9 = 0II: 13y^2 − 32y + 12 = 0Concept / Approach:Compute discriminants to find exact roots. Then compare the sets: if every x is less than every y, conclude x < y conclusively.Step-by-Step Solution:I: D = 48^2 − 4*17*(−9) = 2304 + 612 = 2916 = 54^2x = [−48 ± 54]/(2*17) ⇒ x ∈ {3/17 ≈ 0.1765, −3}II: D = (−32)^2 − 4*13*12 = 1024 − 624 = 400 = 20^2y = [32 ± 20]/(2*13) ⇒ y ∈ {2, 6/13 ≈ 0.4615}Compare every pair: −3 y” or “x = y” never occurs with these root sets; “indeterminate” is wrong because the inequality is consistent.Common Pitfalls:Missing one root or miscomputing the discriminant can flip the conclusion; compute precisely.Final Answer:x

Q: Comparison of x and y from two equations (use mapping below): I. 8x^2 + 6x = 5 II. 12y^2 − 22y + 8 = 0 Mapping: 1 → x > y, 2 → x < y, 3 → x = y, 4 → Relationship cannot be determined.

Introduction / Context:When multiple roots exist for both variables, the relation between x and y may vary across admissible pairs. We conclude “indeterminate” if both equality and strict inequality occur among the valid combinations.Given Data / Assumptions:I: 8x^2 + 6x − 5 = 0II: 12y^2 − 22y + 8 = 0Concept / Approach:Find roots of both equations, then compare. A single consistent relation is needed to declare x > y, x < y, or x = y; otherwise choose “Relationship cannot be determined.”Step-by-Step Solution:I: D = 6^2 + 4*8*5 = 36 + 160 = 196 = 14^2x = [−6 ± 14]/16 ⇒ x ∈ {0.5, −1.25}II: Divide by 2 ⇒ 6y^2 − 11y + 4 = 0, D = 121 − 96 = 25 = 5^2y = [11 ± 5]/12 ⇒ y ∈ {4/3 ≈ 1.333, 1/2 = 0.5}Comparisons: x = 0.5 with y = 0.5 gives equality; x = −1.25 with y = 0.5 gives x y ignores actual values; choosing x = y ignores the other x-root; choosing x < y ignores the equality case.Common Pitfalls:Overlooking repeated roots or failing to consider every possible pair when deciding the relation.Final Answer:Relationship cannot be determined

Q: Read the following passage and answer the given questions. Two algebraic expressions are given as (I) $6x^2 + 29x + P$ and (II) $3x + 1$. (I) is divided by (II) leaving $(2x + 9)$ as quotient and 19 as remainder. What is the value of largest root …

Concept & Quadratic Equation Roots: To find the roots of a quadratic equation, we can use factorization by splitting the middle term. ax^2 + bx + c = 0 Step-by-Step Solution: • From the previous problem, we found that P = 28. • Substitute P into the equation: 6x^2 + 29x + 28 = 0. • To factorize, we need two numbers that multiply to a × c (6 × 28 = 168) and add to b (29). • These numbers are 21 and 8 (21 + 8 = 29 and 21 × 8 = 168). • Split the middle term: 6x^2 + 21x + 8x + 28 = 0. • Factor by grouping: 3x(2x + 7) + 4(2x + 7) = 0. • This gives (3x + 4)(2x + 7) = 0. • Setting each factor to zero yields the roots: x = -(4/3) and x = -(7/2). • Convert to mixed fractions/decimals to compare: x = -1(1/3) and x = -3.5. • Since -1.33 is greater than -3.5, the largest root is -1(1/3). Exam Strategy & Shortcut: Once you have the roots as -(21/6) and -(8/6), you can immediately see that -(8/6) (which simplifies to -(4/3)) is a smaller negative number and thus the larger value. There's no need to convert fully to decimals if you recognize relative fractional sizes. Common Pitfall: Choosing -(7/2) (or -3.5) instead of -1.33 because 3.5 is a larger magnitude. Remember that on the negative side of the number line, numbers closer to zero have a larger mathematical value. Final Answer: Therefore, the correct answer is -1(1/3).

Q: Directions: Solve the quadratic and answer the following questions. A : $(x-2)^2 = (-3x^2) + 2^2 + 25x - P$ B: $\left(10y^2 - 3^2y + \frac{2}{3}\right) \times 3 + 10y = 0$ One root of equation A is $5$. $\frac{7P}{P - 24} \times 0.2P - 91$ is eq…

Concept & Algebraic Substitution: To find the value of the given expression, we must first determine the value of the unknown constant P. We can find P by utilizing the fact that x=5 is a root of equation A. ax^2 + bx + c = 0 Step-by-Step Solution: 1. Simplify Equation A: (x-2)^2 = -3x^2 + 2^2 + 25x - P x^2 - 4x + 4 = -3x^2 + 4 + 25x - P Bring all terms to one side: 4x^2 - 29x + P = 0 2. Find the value of P: Since x = 5 is a root, substitute x = 5 into the simplified equation: 4(5)^2 - 29(5) + P = 0 100 - 145 + P = 0 -45 + P = 0 \implies P = 45 3. Evaluate the target expression: Substitute P = 45 into (7P/P - 24) × 0.2P - 91: = (7(45)/45 - 24) × 0.2(45) - 91 = (315/21) × 9 - 91 = 15 × 9 - 91 = 135 - 91 = 44 4. Check the options by substituting P = 45: (a) P + 2 = 47 (b) 2P + 2 = 92 (c) 2P - 1 = 89 (d) P - 1 = 44 The evaluated expression matches option (d). Exam Strategy & Shortcut: Whenever a root is given for an equation with a single unknown variable, substituting the root directly is the fastest way to isolate the variable. Simplify expressions early to prevent arithmetic errors with large numbers. Common Pitfall: A common mistake is failing to simplify equation A correctly before substitution, particularly forgetting to move the -3x^2 over to make it 4x^2, which will result in an incorrect P value. Final Answer: Therefore, the correct answer is P - 1.

Question 1

Exactly one real root (repeated) condition: Determine all real values of p such that x^2 + 5px + 16 = 0 has exactly one real root (i.e., equal roots).

Accepted Answer

Introduction / Context:“Exactly one real root” for a quadratic means the graph touches the x-axis at a single point: the discriminant must be zero. We find all p that make Δ = 0 for x^2 + 5px + 16 = 0.Given Data / Assumptions:a = 1, b = 5p, c = 16.p is real.Concept / Approach:Set Δ = b^2 − 4ac = 0 and solve for p. This yields the exact parameter values where the quadratic has a repeated real root.Step-by-Step Solution:Δ = (5p)^2 − 4*1*16 = 25p^2 − 64.Set Δ = 0 ⇒ 25p^2 − 64 = 0 ⇒ 25p^2 = 64 ⇒ p^2 = 64/25.Therefore p = ± 8/5.Verification / Alternative check:For p = 8/5, the quadratic is x^2 + 8x + 16 = 0 ⇒ (x + 4)^2 = 0. For p = −8/5, x^2 − 8x + 16 = 0 ⇒ (x − 4)^2 = 0. Both have a single (repeated) real root.Why Other Options Are Wrong:−8/5 < p < 8/5: Gives Δ < 0 (no real roots).p ≤ −8/5 or p ≥ 8/5: Includes Δ > 0 as well as Δ = 0; not “exactly one.”No real value of p: False; two real values exist.Common Pitfalls:Confusing “at least one real root” (Δ ≥ 0) with “exactly one real root” (Δ = 0). Here the equality is essential.Final Answer:p = 8/5 or p = −8/5

Question 2

Quadratic roots identity (Vieta's relations): If α and β are the roots of ax^2 + bx + c = 0 (a ≠ 0), find the value of (α/β) + (β/α) in terms of a, b, c.

Accepted Answer

Introduction / Context:This problem checks your comfort with Vieta's relations for quadratics and your ability to manipulate symmetric expressions in the roots. Instead of solving for the roots explicitly, we will express (α/β) + (β/α) directly using a, b, and c.Given Data / Assumptions:Quadratic: ax^2 + bx + c = 0 with a ≠ 0.Roots: α and β (possibly real or complex).We require (α/β) + (β/α).Concept / Approach:By Vieta's relations, α + β = −b/a and αβ = c/a. Note that (α/β) + (β/α) = (α^2 + β^2)/(αβ). Also, α^2 + β^2 = (α + β)^2 − 2αβ. Substitute Vieta's expressions and simplify carefully to avoid sign errors.Step-by-Step Solution:α + β = −b/aαβ = c/aα^2 + β^2 = (α + β)^2 − 2αβ = (b^2/a^2) − 2(c/a)(α/β) + (β/α) = (α^2 + β^2)/(αβ) = [ (b^2/a^2) − 2(c/a) ] / (c/a)= (b^2/a − 2c)/c = (b^2 − 2ac)/(ac)Verification / Alternative check:Plugging a simple example (e.g., a = 1, b = −3, c = 1 with roots 1 and 2) confirms the formula numerically equals (1/2 + 2/1) = 2.5 and (b^2 − 2ac)/(ac) = (9 − 2)/1 = 7; wait, with a = 1, b = −3, c = 2 (roots 1 and 2), (α/β + β/α) = 1/2 + 2 = 2.5 and (b^2 − 2ac)/(ac) = (9 − 4)/(2) = 2.5—consistent.Why Other Options Are Wrong:Dividing by 2ac or adding 2ac in the numerator changes the value; these arise from algebra slips in the transformation from α^2 + β^2 to (α/β + β/α).Common Pitfalls:Forgetting that α^2 + β^2 = (α + β)^2 − 2αβ, or mishandling complex fractions while clearing denominators.Final Answer:(b^2 - 2ac) / (ac)

Question 3

Construct the quadratic equation whose roots are √2 and 2√2. Express the equation with integer constant term and a rational middle coefficient if possible.

Accepted Answer

Introduction / Context:Building a quadratic from its roots is a classic application of Vieta's relations. If r1 and r2 are the roots, then the monic quadratic is x^2 − (r1 + r2)x + r1 r2 = 0. The task uses irrational roots √2 and 2√2, but the product becomes an integer, simplifying the constant term.Given Data / Assumptions:Roots: √2 and 2√2.Monic quadratic form: x^2 − (sum)x + product = 0.Concept / Approach:Compute the sum and product of the given roots, and substitute into the monic quadratic template. Double-check arithmetic with radicals to avoid mistakes.Step-by-Step Solution:Sum = √2 + 2√2 = 3√2Product = √2 * 2√2 = 2*(√2*√2) = 2*2 = 4Quadratic: x^2 − (3√2)x + 4 = 0Verification / Alternative check:Substitute x = √2: LHS = 2 − 3√2*√2 + 4 = 2 − 6 + 4 = 0. Substitute x = 2√2: LHS = 8 − 3√2*2√2 + 4 = 8 − 12 + 4 = 0. Both roots satisfy the equation.Why Other Options Are Wrong:Sign errors on the middle term or constant produce equations that do not vanish at both roots; testing either root quickly reveals nonzero residuals.Common Pitfalls:Miscomputing the product of radicals (√2 * 2√2 = 4, not 2√4). Keep the rule √2*√2 = 2 in mind.Final Answer:x^2 − 3√2 x + 4 = 0

Question 4

Comparison of x and y from two quadratics (use mapping below): I. 16x^2 + 20x + 6 = 0 II. 10y^2 + 38y + 24 = 0  Mapping: 1 → x > y,  2 → x < y,  3 → x = y,  4 → Relationship cannot be determined.

Accepted Answer

Introduction / Context:The task is to compare the possible values of x and y when each variable must satisfy its own quadratic. Because a quadratic can have two roots, we must check all admissible x and y values and see if a single consistent relation holds in every case.Given Data / Assumptions:I: 16x^2 + 20x + 6 = 0II: 10y^2 + 38y + 24 = 0Concept / Approach:Factor (or reduce) each quadratic to find its real roots. Then compare every x-root with every y-root. If for all combinations x is greater than y, we can conclude x > y unequivocally.Step-by-Step Solution:I: Divide by 2 → 8x^2 + 10x + 3 = 0 = (4x + 3)(2x + 1)So x ∈ {−3/4 = −0.75, −1/2 = −0.5}II: Divide by 2 → 5y^2 + 19y + 12 = 0 = (5y + 4)(y + 3)So y ∈ {−4/5 = −0.8, −3}Compare: each x (−0.75, −0.5) is greater than each y (−0.8, −3).Verification / Alternative check:Numerically: max(y) = −0.8, min(x) = −0.75; since −0.75 > −0.8, even the least x exceeds the greatest y. Hence x > y for all admissible cases.Why Other Options Are Wrong:“x < y” and “x = y” are contradicted by direct comparison; “Relationship cannot be determined” is invalid because the relationship is consistent across all root pairs.Common Pitfalls:Stopping after one comparison. You must ensure the relation holds for every root combination.Final Answer:x > y

Question 5

Comparison of x and y from two quadratics (use mapping below): I. 18x^2 + 18x + 4 = 0 II. 12y^2 + 29y + 14 = 0  Mapping: 1 → x > y,  2 → x < y,  3 → x = y,  4 → Relationship cannot be determined.

Accepted Answer

Introduction / Context:When both x and y are defined by quadratics with two real roots, comparisons must consider all combinations. If some combinations yield equality and others yield strict inequality, the overall relationship is indeterminate.Given Data / Assumptions:I: 18x^2 + 18x + 4 = 0II: 12y^2 + 29y + 14 = 0Concept / Approach:Find the roots of each. Then examine whether a single relation (>, <, or =) holds for every admissible pair (x, y). If not, choose “Relationship cannot be determined.”Step-by-Step Solution:I: Divide by 2 → 9x^2 + 9x + 2 = 0 = (3x + 1)(3x + 2)x ∈ {−1/3, −2/3}II: Discriminant = 29^2 − 4*12*14 = 169 ⇒ √D = 13y = (−29 ± 13)/24 ⇒ y ∈ {−2/3, −7/4}Comparisons: x = −2/3 and y = −2/3 give equality; x = −1/3 compared with y = −2/3 gives x > y. Different relations occur.Verification / Alternative check:Narrowest gap occurs at the common root −2/3. Since equality is possible and x > y also occurs, no single relation always holds.Why Other Options Are Wrong:Choosing x > y or x = y ignores that both cases occur; x < y never occurs for any admissible pair.Common Pitfalls:Concluding based on a single pair, missing the shared root case that forces equality.Final Answer:Relationship cannot be determined

Question 6

Solve the linear system and compare x and y (use mapping below): I. 4x + 7y = 209 II. 12x − 14y = −38  Mapping: 1 → x > y,  2 → x < y,  3 → x = y,  4 → Relationship cannot be determined.

Accepted Answer

Introduction / Context:For two linear equations in two variables, there is a unique solution when the lines are non-parallel. Once x and y are found, the comparison is direct and definitive.Given Data / Assumptions:4x + 7y = 20912x − 14y = −38Concept / Approach:Use elimination by aligning coefficients of y. Add the equations to eliminate y, compute x, then back-substitute to find y. Finally, compare x and y numerically.Step-by-Step Solution:From II: 12x − 14y = −38 ⇒ divide by 2 ⇒ 6x − 7y = −19Add with I: (4x + 7y) + (6x − 7y) = 209 + (−19) ⇒ 10x = 190x = 19Back-substitute: 4*19 + 7y = 209 ⇒ 76 + 7y = 209 ⇒ 7y = 133 ⇒ y = 19Hence x = yVerification / Alternative check:Plug x = y = 19 into both equations to confirm: 76 + 133 = 209 and 228 − 266 = −38. Both hold.Why Other Options Are Wrong:Since x and y are exactly equal, “greater than”, “less than”, or “indeterminate” are incorrect.Common Pitfalls:Arithmetic slips when dividing or adding equations. Keep track of signs carefully.Final Answer:x = y

Question 7

Comparison of x and y from two equations (use mapping below): I. 17x^2 + 48x = 9 II. 13y^2 = 32y − 12  Mapping: 1 → x > y,  2 → x < y,  3 → x = y,  4 → Relationship cannot be determined.

Accepted Answer

Introduction / Context:Each variable is governed by a quadratic; we must list all possible roots for x and y, then check whether a single inequality relation holds for every admissible pair (x, y).Given Data / Assumptions:I: 17x^2 + 48x − 9 = 0II: 13y^2 − 32y + 12 = 0Concept / Approach:Compute discriminants to find exact roots. Then compare the sets: if every x is less than every y, conclude x < y conclusively.Step-by-Step Solution:I: D = 48^2 − 4*17*(−9) = 2304 + 612 = 2916 = 54^2x = [−48 ± 54]/(2*17) ⇒ x ∈ {3/17 ≈ 0.1765, −3}II: D = (−32)^2 − 4*13*12 = 1024 − 624 = 400 = 20^2y = [32 ± 20]/(2*13) ⇒ y ∈ {2, 6/13 ≈ 0.4615}Compare every pair: −3 < 0.4615 and −3 < 2; 0.1765 < 0.4615 and 0.1765 < 2.Verification / Alternative check:Note the largest x is about 0.1765 while the smallest y is about 0.4615, ensuring x < y for all combinations.Why Other Options Are Wrong:“x > y” or “x = y” never occurs with these root sets; “indeterminate” is wrong because the inequality is consistent.Common Pitfalls:Missing one root or miscomputing the discriminant can flip the conclusion; compute precisely.Final Answer:x

Question 8

Comparison of x and y from two equations (use mapping below): I. 8x^2 + 6x = 5 II. 12y^2 − 22y + 8 = 0  Mapping: 1 → x > y,  2 → x < y,  3 → x = y,  4 → Relationship cannot be determined.

Accepted Answer

Introduction / Context:When multiple roots exist for both variables, the relation between x and y may vary across admissible pairs. We conclude “indeterminate” if both equality and strict inequality occur among the valid combinations.Given Data / Assumptions:I: 8x^2 + 6x − 5 = 0II: 12y^2 − 22y + 8 = 0Concept / Approach:Find roots of both equations, then compare. A single consistent relation is needed to declare x > y, x < y, or x = y; otherwise choose “Relationship cannot be determined.”Step-by-Step Solution:I: D = 6^2 + 4*8*5 = 36 + 160 = 196 = 14^2x = [−6 ± 14]/16 ⇒ x ∈ {0.5, −1.25}II: Divide by 2 ⇒ 6y^2 − 11y + 4 = 0, D = 121 − 96 = 25 = 5^2y = [11 ± 5]/12 ⇒ y ∈ {4/3 ≈ 1.333, 1/2 = 0.5}Comparisons: x = 0.5 with y = 0.5 gives equality; x = −1.25 with y = 0.5 gives x < y.Verification / Alternative check:Since both equality and strict inequality (x < y) occur, no single relation always holds across all valid pairs.Why Other Options Are Wrong:Choosing x > y ignores actual values; choosing x = y ignores the other x-root; choosing x < y ignores the equality case.Common Pitfalls:Overlooking repeated roots or failing to consider every possible pair when deciding the relation.Final Answer:Relationship cannot be determined

Question 9

Read the following passage and answer the given questions.  Two algebraic expressions are given as (I) $6x^2 + 29x + P$ and (II) $3x + 1$. (I) is divided by (II) leaving $(2x + 9)$ as quotient and 19 as remainder.  What is the value of largest root …

Accepted Answer

Concept & Quadratic Equation Roots: To find the roots of a quadratic equation, we can use factorization by splitting the middle term. ax^2 + bx + c = 0 Step-by-Step Solution: • From the previous problem, we found that P = 28. • Substitute P into the equation: 6x^2 + 29x + 28 = 0. • To factorize, we need two numbers that multiply to a × c (6 × 28 = 168) and add to b (29). • These numbers are 21 and 8 (21 + 8 = 29 and 21 × 8 = 168). • Split the middle term: 6x^2 + 21x + 8x + 28 = 0. • Factor by grouping: 3x(2x + 7) + 4(2x + 7) = 0. • This gives (3x + 4)(2x + 7) = 0. • Setting each factor to zero yields the roots: x = -(4/3) and x = -(7/2). • Convert to mixed fractions/decimals to compare: x = -1(1/3) and x = -3.5. • Since -1.33 is greater than -3.5, the largest root is -1(1/3). Exam Strategy & Shortcut: Once you have the roots as -(21/6) and -(8/6), you can immediately see that -(8/6) (which simplifies to -(4/3)) is a smaller negative number and thus the larger value. There's no need to convert fully to decimals if you recognize relative fractional sizes. Common Pitfall: Choosing -(7/2) (or -3.5) instead of -1.33 because 3.5 is a larger magnitude. Remember that on the negative side of the number line, numbers closer to zero have a larger mathematical value. Final Answer: Therefore, the correct answer is -1(1/3).

Question 10

Directions: Solve the quadratic and answer the following questions.  A : $(x-2)^2 = (-3x^2) + 2^2 + 25x - P$  B: $\left(10y^2 - 3^2y + \frac{2}{3}ight) 	imes 3 + 10y = 0$  One root of equation A is $5$.  $\frac{7P}{P - 24} 	imes 0.2P - 91$ is eq…

Accepted Answer

Concept & Algebraic Substitution: To find the value of the given expression, we must first determine the value of the unknown constant P. We can find P by utilizing the fact that x=5 is a root of equation A. ax^2 + bx + c = 0 Step-by-Step Solution: 1. Simplify Equation A: (x-2)^2 = -3x^2 + 2^2 + 25x - P x^2 - 4x + 4 = -3x^2 + 4 + 25x - P Bring all terms to one side: 4x^2 - 29x + P = 0 2. Find the value of P: Since x = 5 is a root, substitute x = 5 into the simplified equation: 4(5)^2 - 29(5) + P = 0 100 - 145 + P = 0 -45 + P = 0 \implies P = 45 3. Evaluate the target expression: Substitute P = 45 into (7P/P - 24) × 0.2P - 91: = (7(45)/45 - 24) × 0.2(45) - 91 = (315/21) × 9 - 91 = 15 × 9 - 91 = 135 - 91 = 44 4. Check the options by substituting P = 45: (a) P + 2 = 47 (b) 2P + 2 = 92 (c) 2P - 1 = 89 (d) P - 1 = 44 The evaluated expression matches option (d). Exam Strategy & Shortcut: Whenever a root is given for an equation with a single unknown variable, substituting the root directly is the fastest way to isolate the variable. Simplify expressions early to prevent arithmetic errors with large numbers. Common Pitfall: A common mistake is failing to simplify equation A correctly before substitution, particularly forgetting to move the -3x^2 over to make it 4x^2, which will result in an incorrect P value. Final Answer: Therefore, the correct answer is P - 1.

Question 11

Directions: Solve the quadratic and answer the following questions.  A : $(x-2)^2 = (-3x^2) + 2^2 + 25x - P$  B: $\left(10y^2 - 3^2y + \frac{2}{3}ight) 	imes 3 + 10y = 0$  One root of equation A is $5$.  Find the product of smallest root of equat…

Accepted Answer

Concept & Quadratic Roots: To solve this, we need to find all the roots of Equation A and identify the smallest one, then multiply it by the smallest two-digit prime number. Product of roots = (c/a) Step-by-Step Solution: 1. Find Equation A and its roots: From expanding and rearranging Equation A, we previously found: 4x^2 - 29x + P = 0 Substituting x=5, we found P=45. The complete quadratic equation is: 4x^2 - 29x + 45 = 0 2. Solve for the second root: We can factorize the quadratic: 4x^2 - 20x - 9x + 45 = 0 4x(x - 5) - 9(x - 5) = 0 (4x - 9)(x - 5) = 0 The roots are x = 5 and x = (9/4). The smallest root of Equation A is (9/4). 3. Identify the smallest two-digit prime number: The prime numbers starting from single digits are 2, 3, 5, 7, 11, 13 \dots The smallest two-digit prime number is 11. 4. Calculate the final product: Product = (9/4) × 11 = (99/4) Convert (99/4) into a mixed fraction: 99 \div 4 = 24 with a remainder of 3. So, (99/4) = 24(3/4). Exam Strategy & Shortcut: Instead of factorizing, use the sum or product of roots. Product of roots = (c/a) = (45/4). Since one root is 5, the other is (45/4) \div 5 = (9/4). This is significantly faster! Common Pitfall: A frequent error is misidentifying the smallest two-digit prime number as 10 (which is composite) or 13 (missing 11), throwing off the final calculation. Final Answer: Therefore, the correct answer is 24(3/4).

Question 12

Directions: Solve the quadratic and answer the following questions.  A : $(x-2)^2 = (-3x^2) + 2^2 + 25x - P$  B: $\left(10y^2 - 3^2y + \frac{2}{3}ight) 	imes 3 + 10y = 0$  One root of equation A is $5$.  Which of the following are the smallest ro…

Accepted Answer

Concept & Factorization: We need to determine the roots for both Equation A and Equation B. After finding all roots, we simply compare them to select the smallest root from A and the largest root from B. x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} Step-by-Step Solution: 1. Find roots of Equation A: As derived previously, Equation A simplifies to 4x^2 - 29x + 45 = 0. The roots are x = 5 and x = (9/4) (or 2.25). The smallest root of Equation A is (9/4). 2. Simplify Equation B: (10y^2 - 3^2y + (2/3)) × 3 + 10y = 0 Expand the equation by multiplying by 3: 30y^2 - 27y + 2 + 10y = 0 30y^2 - 17y + 2 = 0 3. Find roots of Equation B: Factorize by splitting the middle term. We need two numbers that multiply to (30 × 2 = 60) and add to -17. These are -12 and -5. 30y^2 - 12y - 5y + 2 = 0 6y(5y - 2) - 1(5y - 2) = 0 (6y - 1)(5y - 2) = 0 The roots are y = (1/6) and y = (2/5). 4. Identify the largest root of Equation B: Compare (1/6) (approx 0.166) and (2/5) (which is 0.4). The largest root of Equation B is (2/5). 5. Combine the results: Smallest root of A = (9/4). Largest root of B = (2/5). Respectively, the values are (9/4), (2/5). Exam Strategy & Shortcut: For Equation B, once you reach 30y^2 - 17y + 2 = 0, you can find the roots using the shortcut of changing the signs of the split middle terms and dividing by a: The split terms are -12, -5. Change signs: 12, 5. Divide by 30: (12/30), (5/30) \implies (2/5), (1/6). Common Pitfall: Be careful when comparing fractions. Students sometimes assume that a larger denominator implies a larger number, mistakenly choosing (1/6) over (2/5). Always convert to decimals or cross-multiply to verify. Final Answer: Therefore, the correct answer is (9/4), (2/5).

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