Quadratic Expressions and Equations

14 Grade 10 Core Mathematics questions on Quadratic Expressions and Equations, with worked marking schemes. 14 questions are available on this topic in full.

1. Question 1

   3 marks

   Factorise completely the expression $4x^2-25y^2$. Hence evaluate $4\times 51^2-25\times 20^2$ without using a calculator.

   (3 marks)

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2. Question 2

   1 mark

   The area of a rectangular vegetable plot is ($x^2+5x+6$) square metres and its length is (x + 3) metres. State an expression for its width.

   A. (x + 1) m

   B. (x − 2) m

   C. (x + 6) m

   D. (x + 2) m

   (1 mark)

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3. Question 3

   1 mark

   Form the quadratic equation whose roots are x = −3 and x = 4.

   A. $x^2-x-12=0$

   B. $x^2-7x+12=0$

   C. $x^2+x-12=0$

   D. $x^2+7x+12=0$

   (1 mark)

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4. Question 4

   1 mark

   Factorise completely the expression $9x^2-49y^2$.

   A. ($3x-7y$)($3x+7y$)

   B. $(3x+7y{)}^2$

   C. $(3x-7y{)}^2$

   D. ($9x-49y$)($9x+49y$)

   (1 mark)

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5. Question 5

   1 mark

   The expression $x^2+kx+25$ is a perfect square and k is positive. Identify the value of k.

   A. 50

   B. 5

   C. 10

   D. 25

   (1 mark)

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6. Question 6

   3 marks

   Solve the equation $x^2+2x-15=0$ by factorisation.

   (3 marks)

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7. Question 7

   1 mark

   A learner is asked to solve the equation $x^2=9x$. Which of the following gives the complete solution?

   A. x = 9 only

   B. x = 3 only

   C. x = 0 or x = 9

   D. x = −9 or x = 9

   (1 mark)

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8. Question 8

   1 mark

   Expand $(2x-5{)}^2$.

   A. $4x^2-20x+25$

   B. $2x^2-20x+25$

   C. $4x^2-10x+25$

   D. $4x^2+25$

   (1 mark)

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9. Question 9

   4 marks

   A tomato nursery bed on a farm in Kirinyaga is rectangular. Its length is 4 m longer than its width and it covers an area of 96 m². Form a quadratic equation for the width and use it to determine the length of the nursery bed.

   (4 marks)

   Marking scheme

   Award 1 mark: width taken as w metres and length as (w + 4) metres, giving w(w + 4) = 96. Award 1 mark: expanded and rearranged to $w^2+4w-96=0$. Award 1 mark: factorised as (w + 12)(w − 8) = 0. Award 1 mark: w = 8 with w = −12 rejected because a width cannot be negative, hence length = 12 m. Note: the fourth mark requires the negative root to be rejected and the length (not the width) stated.

10. Question 10

    1 mark

    State the roots of the equation ($2x-3$)(x + 4) = 0.

    A. x = −1.5 or x = 4

    B. $x=\frac{2}{3}$ or x = −4

    C. x = 1.5 or x = −4

    D. x = 3 or x = −4

    (1 mark)

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11. Question 11

    2 marks

    Factorise completely the expression $3x^2-12x$.

    (2 marks)

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12. Question 12

    2 marks

    Distinguish between a quadratic expression and a quadratic equation.

    (2 marks)

    Marking scheme

    Award 1 mark: a quadratic expression, such as $x^2+3x-4$, has no equals sign; it can only be simplified, factorised or evaluated for a given value of x. Award 1 mark: a quadratic equation, such as $x^2+3x-4=0$, states that the expression equals a value, so it can be solved and has roots (solutions). Note: an answer that merely repeats "one has = and one does not" without the consequence (evaluated versus solved) earns 1 mark only.

13. Question 13

    4 marks

    Solve the equation $2x^2-5x-3=0$ by factorisation.

    (4 marks)

    Marking scheme

    Award 1 mark: two numbers with product $2\times (-3)$ = −6 and sum −5 found, i.e. −6 and +1. Award 1 mark: middle term split, $2x^2-6x+x-3=0$. Award 1 mark: grouped and factorised, 2x(x − 3) + 1(x − 3) = 0 giving (x − 3)$(2x+1)=0$. Award 1 mark: x = 3 or x = −0.5 (accept −$\frac{1}{2}$); both roots needed. Note: allow follow-through on the final mark if the factors are consistent with an earlier slip in splitting the middle term.

14. Question 14

    6 marks

    A trader at a market in Kisii spent KSh 1 200 on sukuma wiki seedlings. Had each seedling cost KSh 2 less, she would have bought 20 more seedlings for the same money. Form a quadratic equation and use it to determine the price she paid for one seedling.

    (6 marks)

    Marking scheme

    Award 1 mark: price of one seedling taken as KSh p, so number bought = 1200/p. Award 1 mark: at the lower price the number bought = 1200/(p − 2). Award 1 mark: relationship formed, 1200/(p − 2) = 1200/p + 20 (or the difference of the two equals 20). Award 1 mark: multiplied through by p(p − 2) and simplified to $2400=20p^2-40p$, hence $p^2-2p-120=0$. Award 1 mark: factorised as (p − 12)(p + 10) = 0. Award 1 mark: p = 12, rejecting p = −10 as a price cannot be negative, so one seedling cost KSh 12. Note: a correct check ($1200÷12=100$ and $1200÷10=120$, which is 20 more) may replace the rejection statement for the final mark.

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