Question 1

What will be the roots of (16x^2-38x+15=0) by factorisation method?

Accepted Answer

(x=\frac{3}{8},\frac{5}{2}) (16x^2-38x+15=(8x-3)(2x-5)), so the roots are (\frac{3}{8}) and (\frac{5}{2}). In exams, do not invert fractional roots.

Question 2

What is the correct splitting of the middle term in (24x^2-50x+25=0)?

Accepted Answer

(24x^2-30x-20x+25=0) Here (ac=600) and (-30+(-20)=-50), so the correct split is (-30x-20x). In exams, even for large (ac), match both sum and product.

Question 3

What are the roots of (24x^2-50x+25=0)?

Accepted Answer

(x=\frac{5}{6},\frac{5}{4}) (24x^2-50x+25=(6x-5)(4x-5)), so the roots are (\frac{5}{6}) and (\frac{5}{4}). In exams, keep the denominator coefficients correctly.

Question 4

If (x^2-2(k-4)x+k^2-25=0) has equal roots, what is the value of (k)?

Accepted Answer

(k=\frac{41}{8}) For equal roots, (D=0), so ((k-4)^2=k^2-25) and (k=\frac{41}{8}). In exams, handle constant terms carefully while expanding squares.

Question 5

If the roots of (6x^2-(p+5)x+24=0) are (2) and (2), what is (p)?

Accepted Answer

(p=19) The sum of roots is (4), and (\frac{p+5}{6}=4), so (p=19). In exams, use (-\frac{b}{a}) for the sum.

Question 6

Which middle step is correct while solving (11x^2-44x+7=0) by completing the square?

Accepted Answer

((x-2)^2=\frac{37}{11}) First (x^2-4x+\frac{7}{11}=0) is obtained, then ((x-2)^2=\frac{37}{11}). In exams, divide by (a) first when (a
eq1).

Question 7

What roots are obtained for (11x^2-44x+7=0) by completing square method?

Accepted Answer

(x=2\pm\frac{\sqrt{407}}{11}) Since ((x-2)^2=\frac{37}{11}), (x=2\pm\sqrt{\frac{37}{11}}=2\pm\frac{\sqrt{407}}{11}). In exams, rationalize the denominator.

Question 8

If (x^2-22x+m=0) has real and equal roots, what is the value of (m)?

Accepted Answer

(m=121) For real and equal roots, (D=0), so (484-4m=0) gives (m=121). In exams, equal roots indicate (D=0).

Question 9

Which root is common to (8x^2-30x+27=0) and (12x^2-31x+20=0)?

Accepted Answer

(x=\frac{3}{2}) The first equation has roots (\frac{3}{2},\frac{9}{4}), and the second has roots (\frac{3}{2},\frac{10}{9}). In exams, solve both equations separately for the common root.

Question 10

If one root of (x^2-23x+q=0) is (9), what will be the other root?

Accepted Answer

(14) The sum of roots is (23), so the other root is (23-9=14). In exams, use the sum when one root is given.

Question 11

If one root of (x^2-23x+q=0) is (9), what is the value of (q)?

Accepted Answer

(126) The other root is (14), so (q=9	imes14=126). In exams, when (a=1), the constant term is the product of roots.

Question 12

A student wrote (x=\frac{1}{2},-\frac{3}{5}) for (10x^2+x-3=0). How is this answer?

Accepted Answer

Correct (10x^2+x-3=(5x+3)(2x-1)), so (x=\frac{1}{2},-\frac{3}{5}) is correct. In exams, change signs carefully from factors.

Question 13

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

Accepted Answer

((x+\sqrt{13})^2=0) Since (13=(\sqrt{13})^2) and the middle term is (2\sqrt{13}x), it is ((x+\sqrt{13})^2). In exams, identify perfect squares even with irrational coefficients.

Question 14

What is the root of (x^2+2\sqrt{13}x+13=0)?

Accepted Answer

(x=-\sqrt{13}) The direct answer is option A, x=-√13. The equation is x^2+2√13x+13=0. Compare it with the identity (x+a)^2=x^2+2ax+a^2. Taking a=√13 gives (x+√13)^2=x^2+2√13x+(√13)^2=x^2+2√13x+13. Therefore the equation becomes (x+√13)^2=0. A square is zero only when its inside is zero, so x+√13=0 and x=-√13. This is a repeated root, meaning both roots have the same value. Option B, x=√13, has the wrong sign; substituting it gives a positive nonzero expression. Option C, x=-13, is not the value obtained from the square factor and has the wrong magnitude. Option D, x=13, has both wrong sign and magnitude. Memory cue: (x+a)^2=0 always gives x=-a.

Question 15

What are the roots of (x^2-(s+t)x+st=0)?

Accepted Answer

(x=s,t) The equation is equivalent to ((x-s)(x-t)=0), so the roots are (s) and (t). In exams, apply zero product rule to symbolic factors too.

Question 16

If (x^2-2rx+r^2-s^2=0), what are its roots?

Accepted Answer

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

Question 17

After simplifying (36x^2-36(m+n)x+36mn=0), what will be the roots?

Accepted Answer

(x=m,n) Dividing the whole equation by (36) gives (x^2-(m+n)x+mn=0). In exams, removing the common factor first shortens the solution.

Question 18

If (y=x^2) is put in (x^4-20x^2+64=0), which equation is obtained?

Accepted Answer

(y^2-20y+64=0) Since (x^4=(x^2)^2=y^2), the new equation is (y^2-20y+64=0). In exams, use substitution to form a quadratic.

Question 19

What are the real solutions of (x^4-20x^2+64=0)?

Accepted Answer

(x=\pm4,\pm2) From (y^2-20y+64=0), (y=4,16), so (x^2=4,16) and (x=\pm2,\pm4). In exams, do not forget to return to (x).

Question 20

For (\frac{1}{x}+x=\frac{37}{6}), (x
eq0), what quadratic form is obtained?

Accepted Answer

(6x^2-37x+6=0) Multiplying both sides by (6x) gives (6+6x^2=37x), that is (6x^2-37x+6=0). In exams, remember the condition (x
eq0).

Question 21

What are the solutions of (\frac{1}{x}+x=\frac{37}{6}), (x
eq0)?

Accepted Answer

(x=6,\frac{1}{6}) (6x^2-37x+6=(6x-1)(x-6)), so (x=\frac{1}{6}) and (6). In exams, check whether obtained roots are valid in the original equation.

Question 22

What is the correct quadratic form of (\frac{x+5}{x}=\frac{36}{x+5}), (x
eq0,-5)?

Accepted Answer

(x^2-26x+25=0) Cross multiplication gives ((x+5)^2=36x), so (x^2+10x+25-36x=0), and (x^2-26x+25=0). In exams, cross multiply carefully.

Question 23

What are the solutions of (\frac{x+5}{x}=\frac{36}{x+5}), (x
eq0,-5)?

Accepted Answer

(x=1,25) (x^2-26x+25=(x-1)(x-25)), so (x=1) and (x=25). In exams, check solutions against excluded denominator values.

Question 24

What are the roots of (x^2-12x+11=0) by quadratic formula?

Accepted Answer

(x=1,11) (D=(-12)^2-4(1)(11)=100), so (x=\frac{12\pm10}{2}) gives (1) and (11). In exams, if (D) is a perfect square, simplify quickly.

Question 25

If the sum of roots of (8x^2+px+32=0) is (-9), what is (p)?

Accepted Answer

(72) The sum of roots is (-\frac{p}{8}), so (-\frac{p}{8}=-9) gives (p=72). In exams, remember the sum formula (-\frac{b}{a}).

Methods of Solving Quadratic Equations

Practice questions

01 What will be the roots of (16x^2-38x+15=0) by factorisation method?

02 What is the correct splitting of the middle term in (24x^2-50x+25=0)?

03 What are the roots of (24x^2-50x+25=0)?

04 If (x^2-2(k-4)x+k^2-25=0) has equal roots, what is the value of (k)?

05 If the roots of (6x^2-(p+5)x+24=0) are (2) and (2), what is (p)?

06 Which middle step is correct while solving (11x^2-44x+7=0) by completing the square?

07 What roots are obtained for (11x^2-44x+7=0) by completing square method?

08 If (x^2-22x+m=0) has real and equal roots, what is the value of (m)?

09 Which root is common to (8x^2-30x+27=0) and (12x^2-31x+20=0)?

10 If one root of (x^2-23x+q=0) is (9), what will be the other root?

11 If one root of (x^2-23x+q=0) is (9), what is the value of (q)?

12 A student wrote (x=\frac{1}{2},-\frac{3}{5}) for (10x^2+x-3=0). How is this answer?

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

14 What is the root of (x^2+2\sqrt{13}x+13=0)?

15 What are the roots of (x^2-(s+t)x+st=0)?

16 If (x^2-2rx+r^2-s^2=0), what are its roots?

17 After simplifying (36x^2-36(m+n)x+36mn=0), what will be the roots?

18 If (y=x^2) is put in (x^4-20x^2+64=0), which equation is obtained?

19 What are the real solutions of (x^4-20x^2+64=0)?

20 For (\frac{1}{x}+x=\frac{37}{6}), (x\neq0), what quadratic form is obtained?

21 What are the solutions of (\frac{1}{x}+x=\frac{37}{6}), (x\neq0)?

22 What is the correct quadratic form of (\frac{x+5}{x}=\frac{36}{x+5}), (x\neq0,-5)?

23 What are the solutions of (\frac{x+5}{x}=\frac{36}{x+5}), (x\neq0,-5)?

24 What are the roots of (x^2-12x+11=0) by quadratic formula?

25 If the sum of roots of (8x^2+px+32=0) is (-9), what is (p)?

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