Q: What are the roots of the equation $x^2-12x+20=0$?

$x=2,10$ Write the equation by completing the square: $x^2-12x+20=(x-6)^2-16=0$. Thus, $(x-6)^2=16$, so $x-6=\pm4$, giving $x=2$ or $x=10$. The values in option C have sum 12 but product 32, whereas the product of the roots must be 20. Exam tip: always include both values obtained from $\pm$ in a quadratic equation.

Q: If the roots of the quadratic equation $x^2+px+12=0$ are $-3$ and $-4$, what is the value of $p$?

7 The factors corresponding to the roots are $(x+3)(x+4)$. Expanding gives $x^2+7x+12=0$, so $p=7$. Option B incorrectly uses the sum of the roots, $-7$, whereas the coefficient is its negative. Exam tip: for $x^2+px+q=0$, the sum of the roots is $-p$.

Q: What are the roots of the equation \$x^2+4x-12=0\$?

\$x=2,-6\$ Factor the equation as \$x^2+6x-2x-12=0\$, giving \$ (x+6)(x-2)=0 \$. Therefore, \$x=-6\$ or \$x=2\$, so option A is correct. In option B, the signs of both roots are incorrect. As an exam check, the roots should have sum \$-4\$ and product \$-12\$.

Question 1

What are the roots of the equation $x^2-12x+20=0$?

Accepted Answer

$x=2,10$ Write the equation by completing the square: $x^2-12x+20=(x-6)^2-16=0$. Thus, $(x-6)^2=16$, so $x-6=\pm4$, giving $x=2$ or $x=10$. The values in option C have sum 12 but product 32, whereas the product of the roots must be 20. Exam tip: always include both values obtained from $\pm$ in a quadratic equation.

Question 2

If the roots of the quadratic equation $x^2+px+12=0$ are $-3$ and $-4$, what is the value of $p$?

Accepted Answer

7 The factors corresponding to the roots are $(x+3)(x+4)$. Expanding gives $x^2+7x+12=0$, so $p=7$. Option B incorrectly uses the sum of the roots, $-7$, whereas the coefficient is its negative. Exam tip: for $x^2+px+q=0$, the sum of the roots is $-p$.

Question 3

If (2) and (5) are roots of a quadratic equation, which equation can it be?

Accepted Answer

(x^2-7x+10=0) If the roots are (2) and (5), the equation is ((x-2)(x-5)=0), that is (x^2-7x+10=0). In exams, form factors with opposite signs of roots.

Question 4

What will be the roots after solving (2x^2+7x+6=0)?

Accepted Answer

(x=-\frac{3}{2},-2) (2x^2+7x+6=(2x+3)(x+2)), so (x=-\frac{3}{2}) and (-2). In exams, positive factors give negative roots.

Question 5

Which statement is correct for (x^2-3x-10=0)?

Accepted Answer

The roots are (5) and (-2) (x^2-3x-10=(x-5)(x+2)), so the roots are (5) and (-2). In exams, in mixed signs the larger value decides the middle-term sign.

Question 6

What roots are obtained by putting (a=1,b=-6,c=8) in the quadratic formula?

Accepted Answer

(x=2,4) (D=(-6)^2-4(1)(8)=4), so (x=\frac{6\pm2}{2}) gives (2) and (4). In exams, keep the sign of (-b) correct.

Question 7

Which number pair helps in making the middle term in (x^2-14x+45=0)?

Accepted Answer

(-5) and (-9) (-5+(-9)=-14) and ((-5)(-9)=45), so this pair is correct. In exams, if (c) is positive and (b) is negative, both numbers may be negative.

Question 8

What is the correct way to solve (7x^2=28x)?

Accepted Answer

Write (7x(x-4)=0) From (7x^2-28x=0), (7x(x-4)=0), so (x=0) and (x=4). In exams, dividing by the variable can miss one root.

Question 9

Which step is correct in solving x² + 4x − 12 = 0 by completing the square?

Accepted Answer

(x + 2)² = 16 Completing the square requires preserving equality while converting the quadratic expression into a perfect square. First move the constant term: x² + 4x = 12. The coefficient of x is 4, so take half of it, 2, and add its square, 2² = 4, to both sides. This gives x² + 4x + 4 = 12 + 4 = 16. The left side factors as (x + 2)², so the correct middle step is (x + 2)² = 16. Therefore option A is correct. Option B has the wrong sign inside the square. Option C uses 4 instead of half the coefficient and also changes the equality improperly. Option D adds the square to the left but fails to add it to the right, violating the balance of the equation.

Question 10

What are the roots of the equation $x^2+4x-12=0$?

Accepted Answer

$x=2,-6$ Factor the equation as $x^2+6x-2x-12=0$, giving $ (x+6)(x-2)=0 $. Therefore, $x=-6$ or $x=2$, so option A is correct. In option B, the signs of both roots are incorrect. As an exam check, the roots should have sum $-4$ and product $-12$.

Question 11

What is the correct splitting of the middle term in (2x^2+x-6=0)?

Accepted Answer

(2x^2+4x-3x-6=0) (ac=-12) and (4+(-3)=1), so (x) is split as (4x-3x). In exams, check the sign of (ac) carefully.

Question 12

If (D=0) and (a=2,b=-8), what will be the equal root?

Accepted Answer

(x=2) The equal root is (x=\frac{-b}{2a}), so (x=\frac{8}{4}=2). In exams, this short formula is useful when (D=0).

Question 13

What is the correct statement about real roots of (x^2-2x+5=0)?

Accepted Answer

No real roots Here (D=(-2)^2-4(1)(5)=-16<0), so there are no real roots. In exams, (D<0) means no real solution.

Question 14

What form is obtained by completing the square in (x^2-2x+5=0)?

Accepted Answer

((x-1)^2+4=0) (x^2-2x+5=(x-1)^2+4), so no real roots are obtained. In exams, the completed square form also shows the nature of roots.

Question 15

What are the roots of the equation $3x^2=12$ when it is solved by the square-root method?

Accepted Answer

$x=\pm2$ Dividing the equation by 3 gives $x^2=4$. Taking square roots, $x=\pm\sqrt{4}=\pm2$, so the two roots are 2 and −2. Options B and C show only one of the two roots, while option D results from an incorrect square-root calculation. Exam tip: whenever $x^2=a$, write $x=\pm\sqrt{a}$ to include both roots.

Question 16

When solving (x − 5)² = 3, what is the value of x?

Accepted Answer

x = 5 ± √3 The governing principle is the square-root property: if u² = c, then u = ±√c. The plus-or-minus sign is necessary because both a positive and a negative number can have the same nonzero square. Here u = x − 5 and c = 3, so x − 5 = ±√3. Adding 5 to both sides gives x = 5 ± √3. Consequently, the two real solutions are x = 5 + √3 and x = 5 − √3, so option A is correct. Option B changes the sign of 5 incorrectly. Option C replaces √3 with 3, which is not valid, and option D changes both the algebraic structure and the values. Substitution of either value into the original equation confirms the result.

Question 17

Which is the correct solution for (x^2+10x+21=0)?

Accepted Answer

(x=-3,-7) (x^2+10x+21=(x+3)(x+7)), so (x=-3,-7). In exams, a positive middle term and positive constant can give both negative roots.

Question 18

Which is the correct factorised form of the equation $5x^2+6x+1=0$?

Accepted Answer

$(5x+1)(x+1)=0$ $(5x+1)(x+1)=5x^2+5x+x+1=5x^2+6x+1$, so option A is correct. Option D expands to $5x^2+11x+6$, so it does not match the given equation. In an exam, multiply the factors back to verify them.

Question 19

For which equation is the quadratic formula more convenient than factorisation?

Accepted Answer

(x^2+x-1=0) (x^2+x-1=0) has no simple integer factors, so the formula method is easier. In exams, the quadratic formula is safe in such cases.

Question 20

What are the roots of (x^2+x-1=0) by the quadratic formula?

Accepted Answer

(x=\frac{-1\pm\sqrt{5}}{2}) Here (D=1-4(1)(-1)=5), so (x=\frac{-1\pm\sqrt{5}}{2}). In exams, keep the sign of (c=-1) correct.

Question 21

Which factorised form is correct for (2x^2-3x-2=0)?

Accepted Answer

((2x+1)(x-2)=0) Direct answer: Option A, $ (2x+1)(x-2)=0 $. To check a factorisation, multiply every term: $2x\cdot x=2x^2$, $2x\cdot(-2)=-4x$, $1\cdot x=x$, and $1\cdot(-2)=-2$. Combining the middle terms gives $2x^2-4x+x-2=2x^2-3x-2$, exactly the original equation. Option A is therefore correct. Option B gives $2x^2+4x-x-2=2x^2+3x-2$, with the wrong sign of the middle term. Option C gives $2x^2+2x-2x-2=2x^2-2$, missing the required linear term. Option D gives $x^2-4$, which has the wrong leading coefficient and constant. Exam cue: multiply the two brackets and check all three terms.

Question 22

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

Accepted Answer

$x=2,-\frac{1}{2}$ Factorise the expression as $2x^2-3x-2=(2x+1)(x-2)$. Thus, $2x+1=0$ gives $x=-\frac{1}{2}$, and $x-2=0$ gives $x=2$. Therefore, option A is correct. Exam tip: Set each factor equal to zero and verify both roots in the original equation; option B has the signs reversed and is therefore incorrect.

Question 23

If a student wrote only (x=5) from (x^2=25), what is the correction?

Accepted Answer

One should write (x=\pm5) From (x^2=25), (x=\pm\sqrt{25}=\pm5). In exams, both signs are necessary in the square root method.

Question 24

If x² − 8x + 15 = 0 is solved by completing the square, which middle step is correct?

Accepted Answer

(x − 4)² = 1 To complete the square, first isolate the variable terms by moving 15 to the right: x² − 8x = −15. The coefficient of x is −8, and half of it is −4. Squaring this half gives (−4)² = 16. Add 16 to both sides: x² − 8x + 16 = −15 + 16 = 1. The left side is the perfect square (x − 4)², because (x − 4)² expands to x² − 8x + 16. Therefore the correct middle step is (x − 4)² = 1, so option A is correct. Option B has the wrong sign, option C does not complete the square, and option D fails to add 16 to the right side.

Question 25

What are the roots obtained by solving $x^2-8x+15=0$ using the completing-the-square method?

Accepted Answer

$x=3$ और $x=5$ Rearranging the equation gives $x^2-8x+16=1$, so $(x-4)^2=1$. Therefore, $x-4=\pm1$, which gives $x=5$ or $x=3$. In option C, the sum of the roots is correct, but their product is $7$, not the required $15$. In exams, remember to consider both signs of $\pm$ when completing the square.

Methods of Solving Quadratic Equations

Practice questions

01 What are the roots of the equation \(x^2-12x+20=0\)?

02 If the roots of the quadratic equation \(x^2+px+12=0\) are \(-3\) and \(-4\), what is the value of \(p\)?

03 If (2) and (5) are roots of a quadratic equation, which equation can it be?

04 What will be the roots after solving (2x^2+7x+6=0)?

05 Which statement is correct for (x^2-3x-10=0)?

06 What roots are obtained by putting (a=1,b=-6,c=8) in the quadratic formula?

07 Which number pair helps in making the middle term in (x^2-14x+45=0)?

08 What is the correct way to solve (7x^2=28x)?

09 Which step is correct in solving x² + 4x − 12 = 0 by completing the square?

10 What are the roots of the equation \\(x^2+4x-12=0\\)?

11 What is the correct splitting of the middle term in (2x^2+x-6=0)?

12 If (D=0) and (a=2,b=-8), what will be the equal root?

13 What is the correct statement about real roots of (x^2-2x+5=0)?

14 What form is obtained by completing the square in (x^2-2x+5=0)?

15 What are the roots of the equation \(3x^2=12\) when it is solved by the square-root method?

16 When solving (x − 5)² = 3, what is the value of x?

17 Which is the correct solution for (x^2+10x+21=0)?

18 Which is the correct factorised form of the equation \(5x^2+6x+1=0\)?

19 For which equation is the quadratic formula more convenient than factorisation?

20 What are the roots of (x^2+x-1=0) by the quadratic formula?

21 Which factorised form is correct for (2x^2-3x-2=0)?

22 What are the roots of the quadratic equation \\(2x^2-3x-2=0\\)?

23 If a student wrote only (x=5) from (x^2=25), what is the correction?

24 If x² − 8x + 15 = 0 is solved by completing the square, which middle step is correct?

25 What are the roots obtained by solving \(x^2-8x+15=0\) using the completing-the-square method?

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