Q: Which middle step is correct while solving $7x^2-22x+7=0$ by completing the square?

$\left(x-\frac{11}{7}\right)^2=\frac{72}{49}$ First $x^2-\frac{22}{7}x+1=0$ is obtained, then $\left(x-\frac{11}{7}\right)^2=\frac{72}{49}$. In exams, divide by (a) first when $a\neq1$.

Q: What will be the roots of $7x^2-22x+7=0$?

$x=\frac{11\pm6\sqrt{2}}{7}$ Since $\left(x-\frac{11}{7}\right)^2=\frac{72}{49}$, $x=\frac{11\pm6\sqrt{2}}{7}$. In exams, simplify $\sqrt{72}=6\sqrt{2}$.

Q: If one root of (x^2-15x+q=0) is (6), what is the value of (q)?

(54) The other root is (9), so (q=6\times9=54). In exams, when (a=1), (c) equals the product of roots.

Question 1

If the roots of (4x^2-(p-1)x+9=0) are (\frac{3}{2}) and (\frac{3}{2}), what is (p)?

Accepted Answer

(p=13) The sum of roots is (3), and (\frac{p-1}{4}=3), so (p=13). In exams, use (-\frac{b}{a}) for the sum of roots.

Question 2

Which middle step is correct while solving $7x^2-22x+7=0$ by completing the square?

Accepted Answer

$\left(x-\frac{11}{7}\right)^2=\frac{72}{49}$ First $x^2-\frac{22}{7}x+1=0$ is obtained, then $\left(x-\frac{11}{7}\right)^2=\frac{72}{49}$. In exams, divide by (a) first when $a\neq1$.

Question 3

What will be the roots of $7x^2-22x+7=0$?

Accepted Answer

$x=\frac{11\pm6\sqrt{2}}{7}$ Since $\left(x-\frac{11}{7}\right)^2=\frac{72}{49}$, $x=\frac{11\pm6\sqrt{2}}{7}$. In exams, simplify $\sqrt{72}=6\sqrt{2}$.

Question 4

If (x^2-18x+m=0) has real and equal roots, what is the value of (m)?

Accepted Answer

(m=81) The direct answer is option A, m=81. For a quadratic equation ax^2+bx+c=0, the discriminant is D=b^2-4ac. Real and equal roots occur exactly when D=0, because the square-root part of the quadratic formula must become zero. Here a=1, b=-18, and c=m. Therefore D=(-18)^2-4(1)(m)=324-4m. Set it equal to zero: 324-4m=0, so 4m=324 and m=81. Thus option A is correct. Option B, m=18, gives D=324-72=252, so the roots are real but unequal. Option C, m=36, gives D=324-144=180, also unequal roots. Option D, m=324, gives D=324-1296=-972, so the roots are not real. A quick alternative is to use the equal-root form: x^2-18x+81=(x-9)^2. Memory cue: equal roots always mean discriminant zero.

Question 5

Which root is common to (5x^2-16x+12=0) and (6x^2-17x+12=0)?

Accepted Answer

(x=\frac{3}{2}) The first equation has roots (2,\frac{6}{5}), and the second has roots (\frac{3}{2},\frac{4}{3}), so there is no common root among the given values. In exams, solve both equations before comparing.

Question 6

Which root is common to (4x^2-12x+5=0) and (6x^2-17x+12=0)?

Accepted Answer

(x=\frac{3}{2}) The first equation has roots (\frac{1}{2},\frac{5}{2}), and the second has roots (\frac{3}{2},\frac{4}{3}), so none of the listed values is common. In exams, solve both equations correctly before comparing.

Question 7

If one root of (x^2-15x+q=0) is (6), what will be the other root?

Accepted Answer

(9) The sum of roots is (15), so the other root is (15-6=9). In exams, use the sum when one root is given.

Question 8

If one root of (x^2-15x+q=0) is (6), what is the value of (q)?

Accepted Answer

(54) The other root is (9), so (q=6	imes9=54). In exams, when (a=1), (c) equals the product of roots.

Question 9

A student wrote (x=\frac{1}{2},-\frac{3}{2}) for (4x^2+4x-3=0). How is this answer?

Accepted Answer

Correct (4x^2+4x-3=(2x-1)(2x+3)), so (x=\frac{1}{2},-\frac{3}{2}) is correct. In exams, change signs carefully from factors.

Question 10

In which form can (x^2+2\sqrt{7}x+7=0) be written?

Accepted Answer

((x+\sqrt{7})^2=0) Since (7=(\sqrt{7})^2) and the middle term is (2\sqrt{7}x), it is ((x+\sqrt{7})^2). In exams, identify perfect squares even with irrational coefficients.

Question 11

What is the root of (x^2+2\sqrt{7}x+7=0)?

Accepted Answer

(x=-\sqrt{7}) The direct answer is option A: the repeated root is $x=-\sqrt{7}$. Since $(\sqrt{7})^2=7$ and the middle term is $2x\sqrt{7}$, the quadratic becomes $x^2+2\sqrt{7}x+7=(x+\sqrt{7})^2$. Hence $(x+\sqrt{7})^2=0$, so $x+ \sqrt{7}=0$ and $x=-\sqrt{7}$. Option A is correct. Option B has the wrong sign; substituting $\sqrt{7}$ does not make the expression zero. Option C, $-7$, treats the square root as the number itself and is not a solution. Option D, 7, has both the wrong sign and wrong magnitude. Because the square is zero, both algebraic roots are the same, so this is a repeated root. Memory cue: match $x^2+2ax+a^2$ with $(x+a)^2$, then change the sign of a.

Question 12

What are the roots of (x^2-(u+v)x+uv=0)?

Accepted Answer

(x=u,v) The equation is equivalent to ((x-u)(x-v)=0), so the roots are (u) and (v). In exams, apply the same rule to symbolic factors.

Question 13

If (x^2-2mx+m^2-n^2=0), what are its roots?

Accepted Answer

(x=m+n,m-n) It is ((x-m)^2-n^2=0), so (x-m=\pm n) and (x=m\pm n). In exams, recognize the difference of squares.

Question 14

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

Accepted Answer

(x=a,b) Dividing the whole equation by (16) gives (x^2-(a+b)x+ab=0). In exams, removing the common factor first makes solving easier.

Question 15

If (y=x^2) is put in (x^4-13x^2+36=0), which equation is obtained?

Accepted Answer

(y^2-13y+36=0) Since (x^4=(x^2)^2=y^2), the new equation is (y^2-13y+36=0). In exams, substitution simplifies a difficult form.

Question 16

What are the real solutions of (x^4-13x^2+36=0)?

Accepted Answer

(x=\pm2,\pm3) From (y^2-13y+36=0), (y=4,9), so (x^2=4,9) and (x=\pm2,\pm3). In exams, do not forget to return to (x).

Question 17

For (\frac{1}{x}+x=\frac{17}{4}), (x
eq0), what quadratic form is obtained?

Accepted Answer

(4x^2-17x+4=0) Multiplying both sides by (4x) gives (4+4x^2=17x), that is (4x^2-17x+4=0). In exams, remember the condition (x
eq0).

Question 18

What are the solutions of (\frac{1}{x}+x=\frac{17}{4}), (x
eq0)?

Accepted Answer

(x=4,\frac{1}{4}) (4x^2-17x+4=(4x-1)(x-4)), so (x=\frac{1}{4}) and (4). In exams, check whether obtained roots are valid in the original equation.

Question 19

What is the correct quadratic form of (\frac{x+3}{x}=\frac{16}{x+3}), (x
eq0,-3)?

Accepted Answer

(x^2-10x+9=0) Cross multiplication gives ((x+3)^2=16x), so (x^2+6x+9-16x=0), and (x^2-10x+9=0). In exams, cross multiply carefully.

Question 20

What are the solutions of (\frac{x+3}{x}=\frac{16}{x+3}), (x
eq0,-3)?

Accepted Answer

(x=1,9) (x^2-10x+9=(x-1)(x-9)), so (x=1) and (x=9). In exams, check solutions against excluded denominator values.

Question 21

What are the roots of (x^2-8x+3=0) by quadratic formula?

Accepted Answer

(x=4\pm\sqrt{13}) (D=(-8)^2-4(1)(3)=52), so (x=\frac{8\pm2\sqrt{13}}{2}=4\pm\sqrt{13}). In exams, simplify the square root.

Question 22

If the sum of roots of (5x^2+px+20=0) is (-7), what is (p)?

Accepted Answer

(35) The sum of roots is (-\frac{p}{5}), so (-\frac{p}{5}=-7) gives (p=35). In exams, remember the sum formula (-\frac{b}{a}).

Question 23

If the product of roots of (6x^2-13x+p=0) is (\frac{1}{2}), what is (p)?

Accepted Answer

(3) The product of roots is (\frac{p}{6}), so (\frac{p}{6}=\frac{1}{2}) gives (p=3). In exams, use the product formula (\frac{c}{a}).

Question 24

If the roots of (x^2-17x+70=0) are (\alpha) and (\beta), what is (\alpha^2+\beta^2)?

Accepted Answer

(149) (\alpha+\beta=17) and (\alpha\beta=70), so (\alpha^2+\beta^2=17^2-2(70)=149). In exams, remember ((\alpha+\beta)^2-2\alpha\beta).

Question 25

If the roots of (x^2-12x+35=0) are (\alpha) and (\beta), what is (\frac{1}{\alpha}+\frac{1}{\beta})?

Accepted Answer

( \frac{12}{35}) / (\frac{12}{35}) (\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}=\frac{12}{35}). In exams, first write sum and product in reciprocal questions.

Methods of Solving Quadratic Equations

Practice questions

01 If the roots of (4x^2-(p-1)x+9=0) are (\frac{3}{2}) and (\frac{3}{2}), what is (p)?

02 Which middle step is correct while solving \(7x^2-22x+7=0\) by completing the square?

03 What will be the roots of \(7x^2-22x+7=0\)?

04 If (x^2-18x+m=0) has real and equal roots, what is the value of (m)?

05 Which root is common to (5x^2-16x+12=0) and (6x^2-17x+12=0)?

06 Which root is common to (4x^2-12x+5=0) and (6x^2-17x+12=0)?

07 If one root of (x^2-15x+q=0) is (6), what will be the other root?

08 If one root of (x^2-15x+q=0) is (6), what is the value of (q)?

09 A student wrote (x=\frac{1}{2},-\frac{3}{2}) for (4x^2+4x-3=0). How is this answer?

10 In which form can (x^2+2\sqrt{7}x+7=0) be written?

11 What is the root of (x^2+2\sqrt{7}x+7=0)?

12 What are the roots of (x^2-(u+v)x+uv=0)?

13 If (x^2-2mx+m^2-n^2=0), what are its roots?

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

15 If (y=x^2) is put in (x^4-13x^2+36=0), which equation is obtained?

16 What are the real solutions of (x^4-13x^2+36=0)?

17 For (\frac{1}{x}+x=\frac{17}{4}), (x\neq0), what quadratic form is obtained?

18 What are the solutions of (\frac{1}{x}+x=\frac{17}{4}), (x\neq0)?

19 What is the correct quadratic form of (\frac{x+3}{x}=\frac{16}{x+3}), (x\neq0,-3)?

20 What are the solutions of (\frac{x+3}{x}=\frac{16}{x+3}), (x\neq0,-3)?

21 What are the roots of (x^2-8x+3=0) by quadratic formula?

22 If the sum of roots of (5x^2+px+20=0) is (-7), what is (p)?

23 If the product of roots of (6x^2-13x+p=0) is (\frac{1}{2}), what is (p)?

24 If the roots of (x^2-17x+70=0) are (\alpha) and (\beta), what is (\alpha^2+\beta^2)?

25 If the roots of (x^2-12x+35=0) are (\alpha) and (\beta), what is (\frac{1}{\alpha}+\frac{1}{\beta})?

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