Q: What will be the factors of (x^2-13x+40=0)?

((x-5)(x-8)) (5+8=13) and (5\times8=40), so ((x-5)(x-8)) is correct. In exams, take both negative signs for a negative middle term.

Q: Which is the correct factorised form of the quadratic equation $2x^2-7x+5=0$?

$(2x-5)(x-1)=0$ The correct answer is A. Expanding $(2x-5)(x-1)$ gives $2x^2-2x-5x+5=2x^2-7x+5$, which matches the quadratic expression in the equation. Therefore, its factorised form is $(2x-5)(x-1)=0$.

Q: What are the roots of the equation $2x^2-7x+5=0$?

$x=1,\frac{5}{2}$ The factorisation is $2x^2-7x+5=(2x-5)(x-1)$. Therefore, $(2x-5)(x-1)=0$ gives $x=\frac{5}{2}$ or $x=1$. The values in option D do not satisfy the original equation. As an exam tip, substitute the obtained roots back into the equation to verify them.

Q: What values of \$x\$ are obtained by solving the equation \$x+3)^2=16\$?

\$x=1,-7\$ Taking the square root gives \$x+3=\\pm4\$. Thus, \$x+3=4\$ gives \$x=1\$, while \$x+3=-4\$ gives \$x=-7\$. Therefore, option A is correct. The values in option B do not satisfy the original equation. In an exam, remember to consider both the positive and negative square-root cases.

Question 1

What are the roots of (9x^2-16=0)?

Accepted Answer

(x=\frac{4}{3},-\frac{4}{3}) ((3x-4)(3x+4)=0), so (x=\frac{4}{3}) or (x=-\frac{4}{3}). In exams, solve linear factors carefully.

Question 2

Which identity directly applies to (x^2-100=0)?

Accepted Answer

(a^2-b^2=(a-b)(a+b)) The direct answer is option A: a²−b²=(a−b)(a+b). Rewrite 100 as 10². Then x²−100=x²−10², which has the form a²−b² with a=x and b=10. Therefore the identity factors it as (x−10)(x+10), and the equation becomes (x−10)(x+10)=0. Option A applies exactly because the two square terms are separated by subtraction. Option B is the identity for the square of a sum and would require a middle term +2ab. Option C is for the square of a difference and would contain a middle term −2ab. Option D is false in general: a²+b² is not equal to (a+b)², because the latter also contains 2ab. For example, x²−100=0 gives x=±10 after factoring. Exam cue: two perfect squares with a minus sign indicate difference of squares.

Question 3

Which are the roots of (x^2+8x+12=0)?

Accepted Answer

(x=-2,-6) (x^2+8x+12=(x+2)(x+6)), so the roots are (-2) and (-6). In exams, a positive middle term can give negative roots.

Question 4

Why is (x=0) a root of (x^2-5x=0)?

Accepted Answer

Because (x(x-5)=0) (x^2-5x=x(x-5)), so zero product rule gives (x=0). In exams, do not lose this root by dividing by the variable.

Question 5

What will be the factors of (x^2-13x+40=0)?

Accepted Answer

((x-5)(x-8)) (5+8=13) and (5	imes8=40), so ((x-5)(x-8)) is correct. In exams, take both negative signs for a negative middle term.

Question 6

What are the roots of (x^2-13x+40=0)?

Accepted Answer

(x=5,8) ((x-5)(x-8)=0), so (x=5) and (x=8). In exams, roots are obtained by taking opposite signs of factors.

Question 7

What is the correct splitting of the middle term in (2x^2-7x+5=0)?

Accepted Answer

(2x^2-2x-5x+5=0) ((-2)+(-5)=-7) and ((-2)(-5)=10=ac), so (-2x-5x) is correct. In exams, checking (ac) is important while splitting the middle term.

Question 8

Which is the correct factorised form of the quadratic equation $2x^2-7x+5=0$?

Accepted Answer

$(2x-5)(x-1)=0$ The correct answer is A. Expanding $(2x-5)(x-1)$ gives $2x^2-2x-5x+5=2x^2-7x+5$, which matches the quadratic expression in the equation. Therefore, its factorised form is $(2x-5)(x-1)=0$.

Question 9

What are the roots of the equation $2x^2-7x+5=0$?

Accepted Answer

$x=1,\frac{5}{2}$ The factorisation is $2x^2-7x+5=(2x-5)(x-1)$. Therefore, $(2x-5)(x-1)=0$ gives $x=\frac{5}{2}$ or $x=1$. The values in option D do not satisfy the original equation. As an exam tip, substitute the obtained roots back into the equation to verify them.

Question 10

What values of $x$ are obtained by solving the equation $x+3)^2=16$?

Accepted Answer

$x=1,-7$ Taking the square root gives $x+3=\pm4$. Thus, $x+3=4$ gives $x=1$, while $x+3=-4$ gives $x=-7$. Therefore, option A is correct. The values in option B do not satisfy the original equation. In an exam, remember to consider both the positive and negative square-root cases.

Question 11

When solving the equation $x^2+16x+11=0$ by completing the square, which number must be added to both sides to make $x^2+16x$ a perfect square?

Accepted Answer

64 The coefficient of $x$ is 16. Its half is 8, and the square of 8 is $8^2=64$. Thus, adding 64 to both sides gives $x^2+16x+64=(x+8)^2$. The number 8 is only half the coefficient, not the number to be added. Exam tip: For $x^2+bx$, add $\left(\frac{b}{2}\right)^2$ to complete the square.

Question 12

The expression $x^2+16x+64$ is equal to which of the following?

Accepted Answer

$(x+8)^2$ $(x+8)^2=x^2+2\cdot8x+8^2=x^2+16x+64$, so option A is correct. Option B would produce the middle term \-16x\), not the given $+16x$. In an exam, identify a perfect square by taking the square root of the constant term and checking whether twice that number gives the middle-term coefficient.

Question 13

What are the correct factors of (x^2+5x-14=0)?

Accepted Answer

((x+7)(x-2)) (7+(-2)=5) and (7	imes(-2)=-14), so ((x+7)(x-2)) is correct. In exams, keep one sign positive and one negative.

Question 14

What are the roots of (x^2+5x-14=0)?

Accepted Answer

(x=2,-7) ((x+7)(x-2)=0), so (x=-7) and (x=2). In exams, the sign changes while finding roots from factors.

Question 15

To solve the equation $8x^2=72$, which equation is obtained first after simplifying both sides?

Accepted Answer

$x^2=9$ Dividing both sides of $8x^2=72$ by 8 gives $x^2=9$. Taking square roots then gives $x=\pm3$. Option B is incorrect because 72 has not been divided by 8. In an exam, first remove the coefficient of $x^2$ by performing the same operation on both sides.

Question 16

What are the solutions of the equation $8x^2=72$?

Accepted Answer

$x=\pm3$ Dividing both sides by 8 gives $x^2=9$. Therefore, $x=\pm\sqrt{9}=\pm3$, so both $x=3$ and $x=-3$ are solutions. Writing only $x=3$ or only $x=-3$ is incomplete. Exam tip: remember to include both positive and negative values when taking the square root of a positive number.

Question 17

What is the correct denominator in the quadratic formula?

Accepted Answer

(2a) In (x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}), the denominator is (2a). In exams, forgetting (2a) is a common mistake.

Question 18

If $x^2-6x+k$ is in the perfect-square form $(x-3)^2$, what is the value of $k$?

Accepted Answer

9 Expanding the perfect square gives $(x-3)^2=x^2-6x+9$. Comparing this with $x^2-6x+k$, the constant term is $k=9$. Therefore, option A is correct. Exam tip: in $(x-a)^2=x^2-2ax+a^2$, the constant term is $a^2$.

Question 19

In which method is (b^2-4ac) found first?

Accepted Answer

Quadratic formula method In the quadratic formula, (b^2-4ac) is used as the discriminant. In exams, identify (a), (b), and (c) before using the formula method.

Question 20

In which perfect square form can (4x^2+4x+1=0) be written?

Accepted Answer

((2x+1)^2=0) (4x^2+4x+1=(2x+1)^2), so it is a perfect square equation. In exams, recognize ((a+b)^2) to solve quickly.

Question 21

Using factorisation method, what will be the roots of (x^2-15x+56=0)?

Accepted Answer

(x=7,8) The direct answer is option A: x=7 and x=8. Factorisation means writing the quadratic as a product of simpler factors. We need two numbers whose product is 56 and whose sum is 15, because the middle term is −15x. The numbers are 7 and 8. Thus x²−15x+56=(x−7)(x−8). Setting the product equal to zero gives (x−7)(x−8)=0, so either x−7=0 or x−8=0. Hence x=7 or x=8. Option A is correct. Option B, −7 and −8, would arise from factors (x+7)(x+8), whose middle term is +15x, not −15x. Option C, 4 and 14, has product 56 but sum 18, so it cannot produce the required middle term. Option D, 1 and 56, has product 56 but sum 57, so it also fails. Check both conditions—product 56 and sum 15—to avoid guessing.

Question 22

What is the correct factorised form of (x^2+13x+42=0)?

Accepted Answer

((x+6)(x+7)=0) (6+7=13) and (6	imes7=42), so the correct factors are ((x+6)(x+7)). In exams, match the signs carefully.

Question 23

Which is the simplest method to solve (x^2-49=0)?

Accepted Answer

Difference of squares method (x^2-49=x^2-7^2), so the difference of squares method is fastest. In exams, recognizing (a^2-b^2) is useful.

Question 24

Using the square-root method, what are the solutions of the equation $x^2=144$?

Accepted Answer

$x=\pm12$ Taking square roots on both sides of $x^2=144$ gives $x=\pm\sqrt{144}=\pm12$. Thus, the two solutions are $x=12$ and $x=-12$. Options B and C give only one solution each, while option D uses an incorrect value for $\sqrt{144}$. Exam tip: when solving $x^2=a$ for positive $a$, remember both roots, $x=\pm\sqrt{a}$.

Question 25

Using zero product rule, what are the roots of ((x-9)(x+2)=0)?

Accepted Answer

(x=9,-2) ((x-9)=0) or ((x+2)=0), so (x=9) or (x=-2). In exams, set each factor equal to zero separately.

Methods of Solving Quadratic Equations

Practice questions

01 What are the roots of (9x^2-16=0)?

02 Which identity directly applies to (x^2-100=0)?

03 Which are the roots of (x^2+8x+12=0)?

04 Why is (x=0) a root of (x^2-5x=0)?

05 What will be the factors of (x^2-13x+40=0)?

06 What are the roots of (x^2-13x+40=0)?

07 What is the correct splitting of the middle term in (2x^2-7x+5=0)?

08 Which is the correct factorised form of the quadratic equation \(2x^2-7x+5=0\)?

09 What are the roots of the equation \(2x^2-7x+5=0\)?

10 What values of \\(x\\) are obtained by solving the equation \\(x+3)^2=16\\)?

11 When solving the equation \(x^2+16x+11=0\) by completing the square, which number must be added to both sides to make \(x^2+16x\) a perfect square?

12 The expression \\(x^2+16x+64\\) is equal to which of the following?

13 What are the correct factors of (x^2+5x-14=0)?

14 What are the roots of (x^2+5x-14=0)?

15 To solve the equation \(8x^2=72\), which equation is obtained first after simplifying both sides?

16 What are the solutions of the equation \(8x^2=72\)?

17 What is the correct denominator in the quadratic formula?

18 If \(x^2-6x+k\) is in the perfect-square form \((x-3)^2\), what is the value of \(k\)?

19 In which method is (b^2-4ac) found first?

20 In which perfect square form can (4x^2+4x+1=0) be written?

21 Using factorisation method, what will be the roots of (x^2-15x+56=0)?

22 What is the correct factorised form of (x^2+13x+42=0)?

23 Which is the simplest method to solve (x^2-49=0)?

24 Using the square-root method, what are the solutions of the equation \(x^2=144\)?

25 Using zero product rule, what are the roots of ((x-9)(x+2)=0)?

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