Q: Which middle step is correct while solving $3x^2-10x+3=0$ by completing the square?

$\left(x-\frac{5}{3}\right)^2=\frac{16}{9}$ First we get $x^2-\frac{10}{3}x+1=0$, then $\left(x-\frac{5}{3}\right)^2=\frac{16}{9}$. In exams, divide by (a) first when $a\neq1$.

Question 1

What will be the roots of (6x^2-7x-3=0) by factorisation method?

Accepted Answer

(x=\frac{3}{2},-\frac{1}{3}) (6x^2-7x-3=(3x+1)(2x-3)), so the roots are (-\frac{1}{3}) and (\frac{3}{2}). In exams, solve both linear factors separately.

Question 2

What is the correct splitting of the middle term in (12x^2-17x+6=0)?

Accepted Answer

(12x^2-9x-8x+6=0) Here (ac=72) and (-9+(-8)=-17), so the correct split is (-9x-8x). In exams, check both sum (b) and product (ac).

Question 3

What are the roots of (12x^2-17x+6=0)?

Accepted Answer

(x=\frac{2}{3},\frac{3}{4}) (12x^2-17x+6=(3x-2)(4x-3)), so the roots are (\frac{2}{3}) and (\frac{3}{4}). In exams, write fractional roots in simplest form.

Question 4

If (x^2-2(k+1)x+k^2=0) has equal roots, what is the value of (k)?

Accepted Answer

(k=-\frac{1}{2}) For equal roots, (D=0), so (4(k+1)^2-4k^2=0) gives (2k+1=0). In exams, write (D) first in parameter questions.

Question 5

If the roots of (2x^2-(p+3)x+8=0) are (2) and (4), what is (p)?

Accepted Answer

(p=9) The sum of roots is (2+4=6), and (\frac{p+3}{2}=6), so (p=9). In exams, use (-\frac{b}{a}) for sum of roots.

Question 6

Which middle step is correct while solving $3x^2-10x+3=0$ by completing the square?

Accepted Answer

$\left(x-\frac{5}{3}\right)^2=\frac{16}{9}$ First we get $x^2-\frac{10}{3}x+1=0$, then $\left(x-\frac{5}{3}\right)^2=\frac{16}{9}$. In exams, divide by (a) first when $a\neq1$.

Question 7

What will be the roots of (3x^2-10x+3=0)?

Accepted Answer

(x=3,\frac{1}{3}) (3x^2-10x+3=(3x-1)(x-3)), so the roots are (\frac{1}{3}) and (3). In exams, you may verify by completing square or factoring.

Question 8

If (x^2-6x+m=0) has real and equal roots, what is the value of (m)?

Accepted Answer

(m=9) For real and equal roots, (D=0), so (36-4m=0) gives (m=9). In exams, remember equal roots mean (D=0).

Question 9

Which root is common to (2x^2-5x+2=0) and (3x^2-8x+4=0)?

Accepted Answer

(x=2) The roots of the first equation are (2,\frac{1}{2}), and the roots of the second are (2,\frac{2}{3}). In exams, solve both equations separately for common root.

Question 10

If one root of (x^2-5x+q=0) is (2), what will be the other root?

Accepted Answer

(3) The sum of roots is (5), so the other root is (5-2=3). In exams, use sum or product when one root is given.

Question 11

If one root of the quadratic equation $x^2-5x+q=0$ is 2, what is the value of $q$?

Accepted Answer

6 Since 2 is a root, substitute $x=2$ in the equation: $2^2-5(2)+q=0$, giving $4-10+q=0$ and hence $q=6$. Alternatively, the sum of the roots is 5, so the other root is 3 and their product is $q=2\times3=6$. Exam tip: for $x^2+bx+c=0$, the product of the roots is $c$; 10 is related to neither the required constant term nor the correct substitution result.

Question 12

A student wrote (x=\frac{1}{2},-2) for (2x^2+3x-2=0). How is this answer?

Accepted Answer

Correct (2x^2+3x-2=(2x-1)(x+2)), so (x=\frac{1}{2},-2) is correct. In exams, change signs carefully from factors.

Question 13

In which form can (x^2+2\sqrt{3}x+3=0) be written?

Accepted Answer

((x+\sqrt{3})^2=0) Since (3=(\sqrt{3})^2) and the middle term is (2\sqrt{3}x), it is ((x+\sqrt{3})^2). In exams, identify perfect squares even with irrational coefficients.

Question 14

What is the root of (x^2+2\sqrt{3}x+3=0)?

Accepted Answer

(x=-\sqrt{3}) The direct answer is option A: the root is $x=-\sqrt{3}$, repeated twice. Start with $x^2+2\sqrt{3}x+3$. Since $(\sqrt{3})^2=3$ and the middle term is twice $x$ times $\sqrt{3}$, the expression becomes $(x+\sqrt{3})^2=0$. Therefore $x+\sqrt{3}=0$, so $x=-\sqrt{3}$. Option A is correct because it gives this repeated root. Option B, $x=\sqrt{3}$, has the wrong sign; substituting it gives a positive expression, not zero. Option C, $x=-3$, confuses $\sqrt{3}$ with 3 and is not a solution. Option D, $x=3$, has both the wrong sign and wrong value. Memory cue: $(x+a)^2=0$ always gives the repeated root $x=-a$.

Question 15

If (x^2-2ax+a^2-b^2=0), what are its roots?

Accepted Answer

(x=a+b,a-b) The direct answer is option A: the roots are $x=a+b$ and $x=a-b$. Start with $x^2-2ax+a^2-b^2=0$. The first three terms form $(x-a)^2$, so the equation becomes $(x-a)^2-b^2=0$. This is a difference of squares: $((x-a)-b)((x-a)+b)=0$. Therefore either $x-a=b$, giving $x=a+b$, or $x-a=-b$, giving $x=a-b$. Option A is correct because it states both roots. Option B gives expressions beginning with $-a$, which do not follow from the factorisation. Option C lists $b+a$, which is the same as $a+b$, but $b-a$ is generally not the same as $a-b$; it misses the required sign. Option D gives $a^2$ and $b^2$, but squaring the constants is not the method for finding these roots. Remember: when $u^2-v^2=0$, write $u=\pm v$.

Question 16

After simplifying (4x^2-4(a+b)x+4ab=0), what will be the roots?

Accepted Answer

(x=a,b) Dividing the whole equation by (4) gives (x^2-(a+b)x+ab=0). In exams, removing the common factor first is easier.

Question 17

If (y=x^2) is put in (x^4-5x^2+4=0), which equation is obtained?

Accepted Answer

(y^2-5y+4=0) Since (x^4=(x^2)^2=y^2), the new equation is (y^2-5y+4=0). In exams, substitution can simplify difficult forms.

Question 18

What are the real solutions of (x^4-5x^2+4=0)?

Accepted Answer

(x=\pm1,\pm2) From (y^2-5y+4=0), (y=1,4), so (x^2=1,4) and (x=\pm1,\pm2). In exams, do not forget to return to (x).

Question 19

For (\frac{1}{x}+x=\frac{5}{2}), (x
eq0), what quadratic form is obtained?

Accepted Answer

(2x^2-5x+2=0) Multiplying both sides by (2x) gives (2+2x^2=5x), that is (2x^2-5x+2=0). In exams, remember the condition (x
eq0).

Question 20

What are the solutions of (\frac{1}{x}+x=\frac{5}{2}), (x
eq0)?

Accepted Answer

(x=2,\frac{1}{2}) (2x^2-5x+2=(2x-1)(x-2)), so (x=\frac{1}{2}) and (2). In exams, check whether obtained roots are valid in the original equation.

Question 21

For (\frac{x+1}{x}=\frac{6}{x+1}), (x
eq0,-1), what quadratic form is obtained?

Accepted Answer

(x^2+2x-5=0) Cross multiplication gives ((x+1)^2=6x), so (x^2+2x+1=6x), and the correct form is (x^2-4x+1=0). In exams, cross multiply very carefully.

Question 22

What is the correct quadratic form of (\frac{x+1}{x}=\frac{6}{x+1}), (x
eq0,-1)?

Accepted Answer

(x^2-4x+1=0) The direct answer is option A: $x^2-4x+1=0$. The original fractions are defined only when $x\neq0,-1$, so multiplying by $x(x+1)$ is allowed under the stated restriction. Starting with $\frac{x+1}{x}=\frac{6}{x+1}$, cross-multiply: $(x+1)(x+1)=6x$. Expand the square: $x^2+2x+1=6x$. Move everything to the left: $x^2+2x+1-6x=0$, hence $x^2-4x+1=0$. Option A matches this exact result. Option B has the wrong middle coefficient: it would require $+2x$ after simplification, which does not happen. Option C incorrectly keeps the number 6 as the middle coefficient and forgets the $2x$ produced by expanding $(x+1)^2$. Option D has the wrong sign for the middle term. The excluded values remain excluded in the original equation, even if an algebraic transformation ever produces them. Memory cue: cross-multiply first, expand the square, then combine the middle terms.

Question 23

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

Accepted Answer

(x=2\pm\sqrt{3}) (D=(-4)^2-4(1)(1)=12), so (x=\frac{4\pm2\sqrt{3}}{2}=2\pm\sqrt{3}). In exams, simplify the square root.

Question 24

If the sum of roots of (3x^2+px+12=0) is (-5), what is (p)?

Accepted Answer

(15) The sum of roots is (-\frac{p}{3}), so (-\frac{p}{3}=-5) gives (p=15). In exams, remember the sum formula (-\frac{b}{a}).

Question 25

If the product of roots of (4x^2-7x+p=0) is (\frac{3}{2}), what is (p)?

Accepted Answer

(6) The product of roots is (\frac{p}{4}), so (\frac{p}{4}=\frac{3}{2}) gives (p=6). In exams, use the product formula (\frac{c}{a}).

Methods of Solving Quadratic Equations

Practice questions

01 What will be the roots of (6x^2-7x-3=0) by factorisation method?

02 What is the correct splitting of the middle term in (12x^2-17x+6=0)?

03 What are the roots of (12x^2-17x+6=0)?

04 If (x^2-2(k+1)x+k^2=0) has equal roots, what is the value of (k)?

05 If the roots of (2x^2-(p+3)x+8=0) are (2) and (4), what is (p)?

06 Which middle step is correct while solving \(3x^2-10x+3=0\) by completing the square?

07 What will be the roots of (3x^2-10x+3=0)?

08 If (x^2-6x+m=0) has real and equal roots, what is the value of (m)?

09 Which root is common to (2x^2-5x+2=0) and (3x^2-8x+4=0)?

10 If one root of (x^2-5x+q=0) is (2), what will be the other root?

11 If one root of the quadratic equation \(x^2-5x+q=0\) is 2, what is the value of \(q\)?

12 A student wrote (x=\frac{1}{2},-2) for (2x^2+3x-2=0). How is this answer?

13 In which form can (x^2+2\sqrt{3}x+3=0) be written?

14 What is the root of (x^2+2\sqrt{3}x+3=0)?

15 If (x^2-2ax+a^2-b^2=0), what are its roots?

16 After simplifying (4x^2-4(a+b)x+4ab=0), what will be the roots?

17 If (y=x^2) is put in (x^4-5x^2+4=0), which equation is obtained?

18 What are the real solutions of (x^4-5x^2+4=0)?

19 For (\frac{1}{x}+x=\frac{5}{2}), (x\neq0), what quadratic form is obtained?

20 What are the solutions of (\frac{1}{x}+x=\frac{5}{2}), (x\neq0)?

21 For (\frac{x+1}{x}=\frac{6}{x+1}), (x\neq0,-1), what quadratic form is obtained?

22 What is the correct quadratic form of (\frac{x+1}{x}=\frac{6}{x+1}), (x\neq0,-1)?

23 What are the roots of (x^2-4x+1=0) by quadratic formula?

24 If the sum of roots of (3x^2+px+12=0) is (-5), what is (p)?

25 If the product of roots of (4x^2-7x+p=0) is (\frac{3}{2}), what is (p)?

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