Q: In factorisation method, how will (x^2+7x+12=0) be written?

((x+3)(x+4)=0) Because (3+4=7) and (3\times4=12), the correct factors are ((x+3)(x+4)). In exams, pay close attention to signs.

Q: In completing square method for $x^2+6x+5=0$, which number is added and subtracted?

(9) Half of the coefficient (6) is (3), and $3^2=9$. In exams, use $\left(\frac{b}{2}\right)^2$.

Q: What are the correct factors for (x^2-2x-8=0)?

((x-4)(x+2)) Since (-4+2=-2) and (-4\times2=-8), ((x-4)(x+2)) is correct. In exams, match both product and sum.

Question 1

Using factorisation method, what are the roots of (x^2-5x+6=0)?

Accepted Answer

(x=2,3) (x^2-5x+6=(x-2)(x-3)), so the roots are (2) and (3). In exams, first find two numbers whose product is (6) and sum is (-5).

Question 2

In factorisation method, how will (x^2+7x+12=0) be written?

Accepted Answer

((x+3)(x+4)=0) Because (3+4=7) and (3	imes4=12), the correct factors are ((x+3)(x+4)). In exams, pay close attention to signs.

Question 3

Which is the simplest method to solve (x^2-9=0)?

Accepted Answer

Difference of squares method (x^2-9=x^2-3^2=(x-3)(x+3)), so it is solved quickly by difference of squares. In exams, recognizing (a^2-b^2) is useful.

Question 4

If (x^2=16), what is the value of (x) by square root method?

Accepted Answer

(x=\pm4) The direct answer is option A: x=±4. The square root method starts with x²=16. Taking square roots gives x=±√16=±4, because both 4² and (-4)² equal 16. The plus-or-minus sign is essential whenever a positive number is the square of an unknown real number. Option A is correct because it includes both solutions. Option B, x=4, is only one solution and wrongly excludes -4. Option C, x=-4, is also only one solution and excludes 4. Option D, x=±8, is wrong because 8²=64, not 16. To check, substitute: 4²=16 and (-4)²=16. A common mistake is to write only the positive square root; remember that squaring removes the sign, so both signs must be restored.

Question 5

Using the zero-product rule, what are the solutions of x(x−4)=0?

Accepted Answer

x=0, 4 The zero-product rule states that if the product of two real factors is zero, then at least one factor must be zero. In x(x−4)=0, set the first factor equal to zero: x=0. Then set the second factor equal to zero: x−4=0, which gives x=4. Hence the solution set is {0,4}, so option A is correct. Substitution confirms both results: 0·(0−4)=0 and 4·(4−4)=0. Option B comes from changing the sign incorrectly and solving x+4=0, which is not the given factor. Option C and option D do not make either factor zero in the required way. Factoring has already been done, so applying the quadratic formula is unnecessary.

Question 6

In completing square method for $x^2+6x+5=0$, which number is added and subtracted?

Accepted Answer

(9) Half of the coefficient (6) is (3), and $3^2=9$. In exams, use $\left(\frac{b}{2}\right)^2$.

Question 7

Which is the correct formula to find roots of (ax^2+bx+c=0) by quadratic formula?

Accepted Answer

(x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}) The quadratic formula is (x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}). In exams, identifying (a), (b), and (c) correctly is most important.

Question 8

What do we get first after simplifying (2x^2-8=0)?

Accepted Answer

(x^2-4=0) Dividing both sides by (2) gives (x^2-4=0). In exams, simplifying the equation first saves time.

Question 9

What are the correct factors for (x^2-2x-8=0)?

Accepted Answer

((x-4)(x+2)) Since (-4+2=-2) and (-4	imes2=-8), ((x-4)(x+2)) is correct. In exams, match both product and sum.

Question 10

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

Accepted Answer

((x+2)^2=0) (x^2+4x+4) is a perfect square and equals ((x+2)^2). In exams, recognize the pattern (a^2+2ab+b^2).

Question 11

What is the repeated root of (x^2-6x+9=0)?

Accepted Answer

(x=3) The direct answer is option A: x = 3, and it occurs twice as a repeated root. Start with x² − 6x + 9 = 0. The first and last terms are x² and 9 = 3², and the middle term is −6x = −2·3·x. Therefore the expression is the perfect square (x − 3)² = 0. A square is zero only when its inside expression is zero, so x − 3 = 0 and x = 3. Both roots are equal, so 3 is repeated. Option A is correct. Option B, x = −3, would belong to (x + 3)², whose middle term is +6x. Option C, x = 6, does not produce the required factorization. Option D, x = 9, confuses the constant term with the root. You can also check using the quadratic formula: the discriminant is 36 − 36 = 0, so the roots coincide at 6/2 = 3. Memory cue: when the discriminant is zero, there is one repeated root.

Question 12

Before applying the quadratic formula to 3x²+5x−2=0, what are a, b, and c?

Accepted Answer

a=3, b=5, c=−2 The quadratic formula is applied to an equation in standard form ax²+bx+c=0. Compare 3x²+5x−2=0 term by term with that form. The coefficient of x² is a, so a=3; the coefficient of x is b, so b=5; and the constant term is c, so c=−2. Therefore option A is correct. The negative sign belongs to the constant term and must not be lost. Option B swaps the coefficients of x² and x. Option C changes both signs of the linear and constant terms, although the original equation has +5x and −2. Option D assigns the constant to a and the leading coefficient to c, reversing the standard roles. Correct identification of these coefficients is essential before substitution into the formula x=[−b±√(b²−4ac)]/(2a).

Question 13

What is the correct first step to solve the equation x² + 5x = 0?

Accepted Answer

x(x + 5) = 0 The terms x² and 5x have x as a common factor. Factoring it out gives x² + 5x = x(x + 5), so the equation becomes x(x + 5) = 0. Option B has the wrong sign; the correct rearrangement would be x² = -5x. Exam tip: factor out the common term first, then apply the zero-product property to set each factor equal to zero.

Question 14

If the equation is (x^2+10x+25=0), which method is fastest?

Accepted Answer

Recognizing perfect square It is ((x+5)^2=0), so recognizing the perfect square is fastest. In exams, look for patterns to save time.

Question 15

What are the roots of (x^2-7x+10=0)?

Accepted Answer

(x=2,5) The direct answer is option A: x = 2 and x = 5. We need two numbers whose product is 10 and whose sum is 7. The numbers 2 and 5 satisfy both conditions. Hence x² − 7x + 10 factors as (x − 2)(x − 5) = 0. By the zero-product rule, either x − 2 = 0 or x − 5 = 0, giving x = 2 or x = 5. Option A is correct. Option B, −2 and −5, has product 10 but sum −7, so it would produce a positive middle coefficient, not −7x. Option C, 1 and 10, has product 10 but sum 11. Option D, 3 and 4, has sum 7 but product 12, not 10. A quick check is useful: for x² − Sx + P, the roots have sum S and product P. Here their sum must be 7 and product 10. Exam cue: factor the constant term and check both sum and product, including signs.

Question 16

Which number pair helps in factoring (x^2+2x-15=0)?

Accepted Answer

(5) and (-3) (5+(-3)=2) and (5	imes(-3)=-15), so this pair is correct. In exams, split the middle term using such a pair.

Question 17

In the middle-term splitting method for 2x² + 7x + 3 = 0, what is the value of ac?

Accepted Answer

6 For a quadratic equation written in the standard form ax² + bx + c = 0, the letters a, b and c represent the coefficients of x², x and the constant term respectively. Comparing 2x² + 7x + 3 = 0 with this form gives a = 2, b = 7 and c = 3. Therefore, ac means the product of the first and last coefficients: ac = 2 × 3 = 6. This product is useful because the middle term 7x can then be split into two terms whose coefficients have product 6 and sum 7, namely 6x and x. Hence option A is correct. The values 7 and 3 are individual coefficients, while 10 is obtained by adding a and c, not multiplying them.

Question 18

For (2x^2+7x+3=0), which is the correct splitting of the middle term?

Accepted Answer

(2x^2+6x+x+3=0) Since (6+1=7) and (6	imes1=6), split (7x) as (6x+x). In exams, keep the sum (b) and product (ac).

Question 19

What are the solutions of the equation $x^2-25=0$?

Accepted Answer

$x=\pm 5$ The equation gives $x^2=25=5^2$. Therefore, $x=5$ or $x=-5$, that is, $x=\pm5$. Options B and C incorrectly treat 25 as a root, while option D uses $\pm25$ instead of $\pm\sqrt{25}$. In an exam, remember that $x^2=a^2$ has the two solutions $x=\pm a$.

Question 20

What is obtained by factoring 3x² − 12x = 0?

Accepted Answer

3x(x − 4) = 0 The governing idea is to take the greatest common factor from every term. In 3x² − 12x = 0, both terms contain 3x. Dividing each term by 3x leaves x and −4, so 3x² − 12x = 3x(x − 4). Retaining the equation gives 3x(x − 4) = 0, which is option A. This form is especially useful because the zero-product property can later be applied: either 3x = 0 or x − 4 = 0. Option B incorrectly removes x from the common factor, option C changes the sign of the second term, and option D gives the wrong sign inside the bracket. Factoring must preserve the original expression exactly.

Question 21

What is (b^2-4ac) called in the quadratic formula?

Accepted Answer

Discriminant (b^2-4ac) is called the discriminant and it tells the nature of roots. In exams, it is also written as (D).

Question 22

If the discriminant of a quadratic equation is $D=b^2-4ac=0$, what type of roots does it have?

Accepted Answer

Equal real roots For a quadratic equation $ax^2+bx+c=0$, the nature of the roots is determined by the discriminant $D=b^2-4ac$. When $D=0$, $\sqrt{D}=0$, so the quadratic formula gives both roots as $-\frac{b}{2a}$; hence they are equal and real. Two distinct real roots occur when $D>0$, so option B is incorrect. Exam tip: remember $D=0$ as ‘equal real roots’.

Question 23

What is the discriminant $D$ of the equation $x^2-4x+4=0$?

Accepted Answer

0 For a quadratic equation $ax^2+bx+c=0$, the discriminant is given by $D=b^2-4ac$. Here, $a=1$, $b=-4$, and $c=4$, so $D=(-4)^2-4(1)(4)=16-16=0$. Therefore, the correct answer is 0. Exam tip: Substitute the value of $b$ with its sign included.

Question 24

Which easy method is most suitable to solve (x^2+3x+2=0)?

Accepted Answer

Factorisation method It easily factors as ((x+1)(x+2)=0). In exams, factorisation is fast for questions with small coefficients.

Question 25

What are the roots of the equation $x^2+3x+2=0$?

Accepted Answer

$x=-1,\,-2$ Factorising the equation gives $x^2+3x+2=(x+1)(x+2)$. Therefore, $(x+1)(x+2)=0$ gives $x=-1$ or $x=-2$. Option B has the signs wrong, while options C and D do not satisfy the equation. Exam tip: if $(x+a)=0$, the corresponding root is $x=-a$.

Methods of Solving Quadratic Equations

Practice questions

01 Using factorisation method, what are the roots of (x^2-5x+6=0)?

02 In factorisation method, how will (x^2+7x+12=0) be written?

03 Which is the simplest method to solve (x^2-9=0)?

04 If (x^2=16), what is the value of (x) by square root method?

05 Using the zero-product rule, what are the solutions of x(x−4)=0?

06 In completing square method for \(x^2+6x+5=0\), which number is added and subtracted?

07 Which is the correct formula to find roots of (ax^2+bx+c=0) by quadratic formula?

08 What do we get first after simplifying (2x^2-8=0)?

09 What are the correct factors for (x^2-2x-8=0)?

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

11 What is the repeated root of (x^2-6x+9=0)?

12 Before applying the quadratic formula to 3x²+5x−2=0, what are a, b, and c?

13 What is the correct first step to solve the equation x² + 5x = 0?

14 If the equation is (x^2+10x+25=0), which method is fastest?

15 What are the roots of (x^2-7x+10=0)?

16 Which number pair helps in factoring (x^2+2x-15=0)?

17 In the middle-term splitting method for 2x² + 7x + 3 = 0, what is the value of ac?

18 For (2x^2+7x+3=0), which is the correct splitting of the middle term?

19 What are the solutions of the equation \(x^2-25=0\)?

20 What is obtained by factoring 3x² − 12x = 0?

21 What is (b^2-4ac) called in the quadratic formula?

22 If the discriminant of a quadratic equation is \(D=b^2-4ac=0\), what type of roots does it have?

23 What is the discriminant \(D\) of the equation \(x^2-4x+4=0\)?

24 Which easy method is most suitable to solve (x^2+3x+2=0)?

25 What are the roots of the equation \(x^2+3x+2=0\)?

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