Q: In the given equation $x^2+8x=0$, what does $x^2+8x+16$ become when a perfect square is formed by completing the square?

$(x+4)^2$ In $x^2+8x+16$, the coefficient of $x$ is 8. Taking half of 8 gives 4, and adding its square, 16, gives $x^2+8x+16=(x+4)^2$. Therefore, A is correct. In $(x+8)^2$, the middle term would be $16x$, so B is incorrect. Exam tip: To complete the square in $x^2+bx$, add $\left(\frac{b}{2}\right)^2$.

Q: What are the values of $x$ when the equation $x^2=49$ is solved by the square root method?

$x=\pm 7$ Using the square root method, $x^2=49$ gives $x=\pm\sqrt{49}$. Hence, $x=7$ or $x=-7$, so $x=\pm7$. Writing only the positive root (option B) is a common mistake because a positive number has two square roots. Exam tip: For $x^2=a$, always write $x=\pm\sqrt{a}$.

Q: Which of the following is the factorised form of the equation $4x^2-1=0$?

$(2x-1)(2x+1)=0$ Here, $4x^2-1=(2x)^2-1^2$. Using the difference-of-squares identity $a^2-b^2=(a-b)(a+b)$, it becomes $(2x-1)(2x+1)$, so option A is correct. Option C expands to $4x^2-4x+1$, not $4x^2-1$. In an exam, first check whether the expression matches the difference-of-squares pattern.

Q: What will be the factors of (x^2-8x+15=0)?

((x-3)(x-5)) (3+5=8) and (3\times5=15), so with signs we get ((x-3)(x-5)). In exams, for a negative middle term, look for both negative factors.

Question 1

In the given equation $x^2+8x=0$, what does $x^2+8x+16$ become when a perfect square is formed by completing the square?

Accepted Answer

$(x+4)^2$ In $x^2+8x+16$, the coefficient of $x$ is 8. Taking half of 8 gives 4, and adding its square, 16, gives $x^2+8x+16=(x+4)^2$. Therefore, A is correct. In $(x+8)^2$, the middle term would be $16x$, so B is incorrect. Exam tip: To complete the square in $x^2+bx$, add $\left(\frac{b}{2}\right)^2$.

Question 2

What are the values of $x$ when the equation $x^2=49$ is solved by the square root method?

Accepted Answer

$x=\pm 7$ Using the square root method, $x^2=49$ gives $x=\pm\sqrt{49}$. Hence, $x=7$ or $x=-7$, so $x=\pm7$. Writing only the positive root (option B) is a common mistake because a positive number has two square roots. Exam tip: For $x^2=a$, always write $x=\pm\sqrt{a}$.

Question 3

Which of the following is the factorised form of the equation $4x^2-1=0$?

Accepted Answer

$(2x-1)(2x+1)=0$ Here, $4x^2-1=(2x)^2-1^2$. Using the difference-of-squares identity $a^2-b^2=(a-b)(a+b)$, it becomes $(2x-1)(2x+1)$, so option A is correct. Option C expands to $4x^2-4x+1$, not $4x^2-1$. In an exam, first check whether the expression matches the difference-of-squares pattern.

Question 4

What are the roots of (4x^2-1=0)?

Accepted Answer

(x=\frac{1}{2},-\frac{1}{2}) ((2x-1)(2x+1)=0), so (x=\frac{1}{2}) or (x=-\frac{1}{2}). In exams, solve each linear factor carefully.

Question 5

Which rule directly applies to x² − 1 = 0?

Accepted Answer

a² − b² = (a − b)(a + b) The expression x² − 1 is a difference of two perfect squares because 1 = 1². The identity for this pattern is a² − b² = (a − b)(a + b). Taking a = x and b = 1 gives x² − 1² = (x − 1)(x + 1). Hence the equation can be rewritten as (x − 1)(x + 1) = 0, leading to x = 1 or x = −1 by the zero-product property. Therefore option A is correct. Option B is the identity for the square of a sum and would require a middle term 2ab. Option C is not a valid identity in general, and option D is not the zero-product rule; from ab = 0, the correct conclusion is a = 0 or b = 0.

Question 6

Which are the correct roots for (x^2+6x+8=0)?

Accepted Answer

(x=-2,-4) (x^2+6x+8=(x+2)(x+4)), so (x=-2) and (x=-4). In exams, a positive middle term can give negative roots.

Question 7

Why is (x=0) a root of (x^2-3x=0)?

Accepted Answer

Because (x(x-3)=0) (x^2-3x=x(x-3)), so zero product rule gives (x=0). In exams, do not lose (x=0) by dividing by (x).

Question 8

If x² − 3x = 0 is divided by x to write x − 3 = 0, which root is missed?

Accepted Answer

x = 0 First factor the equation without dividing by a variable: x² − 3x = x(x − 3) = 0. By the zero-product property, either x = 0 or x − 3 = 0, which gives x = 3. If we divide the original equation by x, we implicitly assume x is nonzero, so the solution x = 0 is excluded before the solving process is complete. The reduced equation x − 3 = 0 therefore retains only x = 3 and misses x = 0. Thus option A is correct. The safe procedure is to factor first and then set each factor equal to zero. Options B is a valid remaining root, while −3 and 1 do not satisfy the original equation.

Question 9

What will be the factors of (x^2-8x+15=0)?

Accepted Answer

((x-3)(x-5)) (3+5=8) and (3	imes5=15), so with signs we get ((x-3)(x-5)). In exams, for a negative middle term, look for both negative factors.

Question 10

What is the correct splitting of the middle term in (2x^2-5x+3=0)?

Accepted Answer

(2x^2-2x-3x+3=0) ((-2)+(-3)=-5) and ((-2)(-3)=6=ac), so the correct split is (-2x-3x). In exams, the product of the two numbers must be (ac).

Question 11

Which is the factorised form of the quadratic equation $2x^2-5x+3=0$?

Accepted Answer

$(2x-3)(x-1)=0$ Multiplying $(2x-3)(x-1)$ gives $2x^2-2x-3x+3=2x^2-5x+3$, so the correct factorised form is $(2x-3)(x-1)=0$. In option B, the coefficient of the middle term becomes $1$ instead of $-5$, while options C and D do not produce the required leading and constant terms. In an exam, multiply the factors back to verify the middle and constant terms.

Question 12

Which equation can be solved directly by square root method?

Accepted Answer

((x-2)^2=9) In ((x-2)^2=9), square root can be taken directly. In exams, recognize the form ((	ext{expression})^2=k).

Question 13

Which values of $x$ are obtained by solving the equation $x-2$^2=9?

Accepted Answer

$x=5$ या $x=-1$ Taking the square root gives $x-2=\pm3$. Thus, $x-2=3$ gives $x=5$, while $x-2=-3$ gives $x=-1$. Option D is incorrect because the second value should be $x=-1$, not $x=1$. In an exam, remember to consider both cases represented by $\pm$.

Question 14

When solving the equation $x^2-10x+7=0$ by completing the square method, which number should be added to make $x^2-10x$ a perfect square?

Accepted Answer

25 The coefficient of $x$ is $-10$, whose half is $-5$. Squaring it gives $(-5)^2=25$, so 25 must be added: $x^2-10x+25=(x-5)^2$. Exam tip: in the completing-square method, add the square of half the coefficient of $x$.

Question 15

The expression $x^2-10x+25$ is equal to which of the following?

Accepted Answer

$(x-5)^2$ Using the perfect-square identity $(x-a)^2=x^2-2ax+a^2$ with $a=5$, we get $(x-5)^2=x^2-10x+25$. Hence, option A is correct. Option B would produce a middle term of $+10x$, while option D represents a difference of squares. In an exam, identify $25=5^2$ and verify the middle term as $-2\times5x=-10x$.

Question 16

What are the correct factors for (x^2+x-6=0)?

Accepted Answer

((x+3)(x-2)) (3+(-2)=1) and (3	imes(-2)=-6), so ((x+3)(x-2)) is correct. In exams, match signs with the product.

Question 17

What are the roots of (x^2+x-6=0)?

Accepted Answer

(x=2,-3) ((x+3)(x-2)=0), so (x=-3) and (x=2). In exams, the sign changes while finding roots from factors.

Question 18

What is the correct statement about the real roots of (x^2+9=0)?

Accepted Answer

No real roots (x^2=-9) gives no solution in real numbers. In exams, remember that (x^2) cannot be negative for real (x).

Question 19

What is the correct simplified form to solve (5x^2=20)?

Accepted Answer

(x^2=4) Dividing both sides by (5) gives (x^2=4). In exams, remove the coefficient first and then take square root.

Question 20

What are the solutions of the equation $5x^2=20$?

Accepted Answer

$x=\pm2$ Dividing both sides of $5x^2=20$ by 5 gives $x^2=4$. Taking square roots, $x=\pm\sqrt{4}=\pm2$, so the solutions are $x=2$ and $x=-2$. Options B and C give only one of the two roots, while option D results from an incorrect square root. Exam tip: remember to include both signs when taking the square root of a positive number.

Question 21

In which perfect square form will (x^2-12x+36=0) be written?

Accepted Answer

((x-6)^2=0) The direct answer is option A: (x − 6)² = 0. To form a square, use (x − a)² = x² − 2ax + a². Here the constant term is 36, so a = 6 because 6² = 36. The middle term then becomes −2·6·x = −12x, exactly as required. Therefore x² − 12x + 36 = (x − 6)², and the equation is (x − 6)² = 0. Option A has both the correct sign and the correct number. Option B, (x + 6)², expands to x² + 12x + 36, so its middle sign is wrong. Option C, (x − 12)², has constant term 144 and middle term −24x. Option D, (x + 12)², has constant term 144 and a positive middle term. The reliable check is to square the proposed binomial before choosing. Memory cue: a negative middle term means the binomial has a minus sign.

Question 22

What will be the root of (x^2-12x+36=0)?

Accepted Answer

(x=6) ((x-6)^2=0), so both equal roots are (x=6). In exams, a perfect square equation gives a repeated root.

Question 23

What is the correct first step to solve (x^2-16x=0)?

Accepted Answer

Write (x(x-16)=0) The common factor in (x^2-16x) is (x), so we write (x(x-16)=0). In exams, do not divide by the variable and lose (x=0).

Question 24

Using the factorisation method, what are the roots of x² − 9x + 20 = 0?

Accepted Answer

x = 4, 5 The governing concept is factorisation together with the zero-product property. We seek two numbers whose product is 20 and whose sum is 9; these numbers are 4 and 5. Therefore x² − 9x + 20 can be written as x² − 4x − 5x + 20 = (x − 4)(x − 5). The equation becomes (x − 4)(x − 5) = 0. By the zero-product property, either x − 4 = 0 or x − 5 = 0, giving x = 4 or x = 5. Option B would produce negative roots and a positive sum in the factorised expression, not the required middle coefficient. Options C and D have products or sums that do not match 20 and 9. Hence option A is correct.

Question 25

How will (x^2+11x+30=0) be written in factorised form?

Accepted Answer

((x+5)(x+6)=0) (5+6=11) and (5	imes6=30), so the correct factors are ((x+5)(x+6)). In exams, positive (c) and positive (b) give both positive signs.

Methods of Solving Quadratic Equations

Practice questions

01 In the given equation \(x^2+8x=0\), what does \(x^2+8x+16\) become when a perfect square is formed by completing the square?

02 What are the values of \(x\) when the equation \(x^2=49\) is solved by the square root method?

03 Which of the following is the factorised form of the equation \(4x^2-1=0\)?

04 What are the roots of (4x^2-1=0)?

05 Which rule directly applies to x² − 1 = 0?

06 Which are the correct roots for (x^2+6x+8=0)?

07 Why is (x=0) a root of (x^2-3x=0)?

08 If x² − 3x = 0 is divided by x to write x − 3 = 0, which root is missed?

09 What will be the factors of (x^2-8x+15=0)?

10 What is the correct splitting of the middle term in (2x^2-5x+3=0)?

11 Which is the factorised form of the quadratic equation \(2x^2-5x+3=0\)?

12 Which equation can be solved directly by square root method?

13 Which values of \\(x\\) are obtained by solving the equation \\(x-2\\)^2=9?

14 When solving the equation \(x^2-10x+7=0\) by completing the square method, which number should be added to make \(x^2-10x\) a perfect square?

15 The expression \(x^2-10x+25\) is equal to which of the following?

16 What are the correct factors for (x^2+x-6=0)?

17 What are the roots of (x^2+x-6=0)?

18 What is the correct statement about the real roots of (x^2+9=0)?

19 What is the correct simplified form to solve (5x^2=20)?

20 What are the solutions of the equation \(5x^2=20\)?

21 In which perfect square form will (x^2-12x+36=0) be written?

22 What will be the root of (x^2-12x+36=0)?

23 What is the correct first step to solve (x^2-16x=0)?

24 Using the factorisation method, what are the roots of x² − 9x + 20 = 0?

25 How will (x^2+11x+30=0) be written in factorised form?

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