Q: If $\alpha,\beta$ are roots of $7x^2-25x+12=0$, what is $(\alpha-\beta)^2$?

$ \frac{289}{49}$ / $\frac{289}{49}$ $(\alpha-\beta)^2=(\alpha+\beta)^2-4\alpha\beta=\left(\frac{25}{7}\right)^2-\frac{48}{7}=\frac{289}{49}$. In exams, convert fractions to a common denominator.

Q: If (x^2-14x+n=0) has two distinct real roots, which condition on (n) is correct?

(n 0), so (196-4n>0) and (n 0) with distinct real roots.

Q: Which factorised form of the equation $7x^2-19x-6=0$ is correct for solving it?

$(7x+2)(x-3)=0$ Expanding option A gives $(7x+2)(x-3)=7x^2-21x+2x-6=7x^2-19x-6$, so it is the correct factorised form. Option B produces a middle term of $+19x$, while options C and D produce $-11x$ and $+11x$, respectively. Exam tip: always expand the factors to verify the middle term and constant term.

Q: What are the roots of the equation $7x^2-19x-6=0$?

$x=3,\,-\frac{2}{7}$ The quadratic factors as $7x^2-19x-6=(7x+2)(x-3)$. Therefore, $(7x+2)(x-3)=0$ gives $x=-\frac{2}{7}$ or $x=3$. In option B, the signs of both roots are incorrect. Exam tip: After factorisation, set each factor equal to zero and carefully reverse the sign when isolating the root.

Question 1

If the product of roots of (9x^2-20x+p=0) is (\frac{1}{3}), what is (p)?

Accepted Answer

(3) The product of roots is (\frac{p}{9}), so (\frac{p}{9}=\frac{1}{3}) gives (p=3). In exams, use the product formula (\frac{c}{a}).

Question 2

If the roots of (x^2-23x+126=0) are (\alpha) and (\beta), what is (\alpha^2+\beta^2)?

Accepted Answer

(277) (\alpha+\beta=23) and (\alpha\beta=126), so (\alpha^2+\beta^2=23^2-2(126)=277). In exams, remember ((\alpha+\beta)^2-2\alpha\beta).

Question 3

If the roots of (x^2-20x+91=0) are (\alpha) and (\beta), what is (\frac{1}{\alpha}+\frac{1}{\beta})?

Accepted Answer

( \frac{20}{91}) / (\frac{20}{91}) (\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}=\frac{20}{91}). In exams, first write sum and product in reciprocal questions.

Question 4

If (\alpha,\beta) are roots of (7x^2-25x+12=0), what is (\alpha+\beta)?

Accepted Answer

( \frac{25}{7}) / (\frac{25}{7}) The sum of roots is (-\frac{b}{a}=-\frac{-25}{7}=\frac{25}{7}). In exams, keep the sign of (b) carefully.

Question 5

If (\alpha,\beta) are roots of (7x^2-25x+12=0), what is (\alpha\beta)?

Accepted Answer

( \frac{12}{7}) / (\frac{12}{7}) The product of roots is (\frac{c}{a}=\frac{12}{7}). In exams, use (\frac{c}{a}) for the product.

Question 6

If $\alpha,\beta$ are roots of $7x^2-25x+12=0$, what is $(\alpha-\beta)^2$?

Accepted Answer

$ \frac{289}{49}$ / $\frac{289}{49}$ $(\alpha-\beta)^2=(\alpha+\beta)^2-4\alpha\beta=\left(\frac{25}{7}\right)^2-\frac{48}{7}=\frac{289}{49}$. In exams, convert fractions to a common denominator.

Question 7

If (x^2-14x+n=0) has no real roots, which condition on (n) is correct?

Accepted Answer

(n>49) For no real roots, (D<0), so (196-4n<0) and (n>49). In exams, connect (D<0) with no real roots.

Question 8

If (x^2-14x+n=0) has two distinct real roots, which condition on (n) is correct?

Accepted Answer

(n<49) For two distinct real roots, (D>0), so (196-4n>0) and (n<49). In exams, connect (D>0) with distinct real roots.

Question 9

Which factorised form of the equation $7x^2-19x-6=0$ is correct for solving it?

Accepted Answer

$(7x+2)(x-3)=0$ Expanding option A gives $(7x+2)(x-3)=7x^2-21x+2x-6=7x^2-19x-6$, so it is the correct factorised form. Option B produces a middle term of $+19x$, while options C and D produce $-11x$ and $+11x$, respectively. Exam tip: always expand the factors to verify the middle term and constant term.

Question 10

What are the roots of the equation $7x^2-19x-6=0$?

Accepted Answer

$x=3,\,-\frac{2}{7}$ The quadratic factors as $7x^2-19x-6=(7x+2)(x-3)$. Therefore, $(7x+2)(x-3)=0$ gives $x=-\frac{2}{7}$ or $x=3$. In option B, the signs of both roots are incorrect. Exam tip: After factorisation, set each factor equal to zero and carefully reverse the sign when isolating the root.

Question 11

If (x^2+12x+8=0), which form is correct by completing square?

Accepted Answer

((x+6)^2=28) Direct answer: option A, (x+6)²=28. Start with x²+12x+8=0 and move 8 to the other side: x²+12x=−8. Half of 12 is 6, and 6²=36. Add 36 to both sides: x²+12x+36=−8+36=28. The left side is (x+6)², so the required form is (x+6)²=28. B has the wrong sign: (x−6)² gives −12x, not +12x. C uses x+12 instead of x+6 and does not complete the square correctly. D has the correct square on the left but fails to add 36 on the right, so 8 is not correct. The same quantity must be added to both sides. Memory cue: use half of the x coefficient, not the full coefficient.

Question 12

What are the roots of the equation $x^2+12x+8=0$?

Accepted Answer

$x=-6\pm2\sqrt{7}$ Completing the square gives $x^2+12x+8=0\Rightarrow (x+6)^2=28$. Therefore, $x+6=\pm\sqrt{28}=\pm2\sqrt{7}$, so the roots are $x=-6\pm2\sqrt{7}$. Option B has the wrong sign for the constant term in the roots. Exam tip: simplify $\sqrt{28}$ to $2\sqrt{7}$.

Question 13

If the roots of (x^2-10x-24=0) are (\alpha,\beta), what is (\alpha^2\beta+\alpha\beta^2)?

Accepted Answer

(-240) (\alpha^2\beta+\alpha\beta^2=\alpha\beta(\alpha+\beta)), where (\alpha+\beta=10) and (\alpha\beta=-24), so the value is (-240). In exams, factor the expression first.

Question 14

If one root of (x^2+px+49=0) is double the other and both are negative, what is (p)?

Accepted Answer

( \frac{21\sqrt{2}}{2}) / (\frac{21\sqrt{2}}{2}) Let the roots be (-r) and (-2r), then (2r^2=49) and (p=3r=\frac{21\sqrt{2}}{2}). In exams, do not forget to rationalize the denominator.

Question 15

What is the difference between the roots of (x^2-11x+6=0)?

Accepted Answer

(\sqrt{97}) Here (D=(-11)^2-4(1)(6)=97), so the difference of roots is (\frac{\sqrt{D}}{|a|}=\sqrt{97}). In exams, the difference of roots can be found directly from (D).

Question 16

What is the correct statement about real roots of (6x^2-12x+17=0)?

Accepted Answer

There are no real roots Here (D=(-12)^2-4(6)(17)=-264<0), so there are no real roots. In exams, (D<0) means no real roots.

Question 17

Which completed square form is correct for (6x^2-12x+17=0)?

Accepted Answer

(6(x-1)^2+11=0) (6x^2-12x+17=6(x-1)^2+11), so it cannot be zero for real (x). In exams, completed square form also shows the nature of roots.

Question 18

If roots of (x^2-14x+33=0) are (\alpha,\beta), which equation has roots (\alpha+5,\beta+5)?

Accepted Answer

(x^2-24x+128=0) The roots are (3,11), so new roots are (8,16), and the equation is ((x-8)(x-16)=0). In exams, form the new roots and then the new equation.

Question 19

If the roots of (x^2-22x+117=0) are (\alpha,\beta), what is (\frac{\alpha}{\beta}+\frac{\beta}{\alpha})?

Accepted Answer

( \frac{250}{117}) / (\frac{250}{117}) (\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha\beta}), where (\alpha+\beta=22) and (\alpha\beta=117), so the value is (\frac{484-234}{117}=\frac{250}{117}). In exams, convert expressions into sum and product.

Question 20

If the quadratic equation $kx^2-22x+121=0$ has equal roots, what is the value of $k$?

Accepted Answer

1 For equal roots, the discriminant $D=b^2-4ac$ must be zero. Here, $a=k$, $b=-22$, and $c=121$, so $D=(-22)^2-4(k)(121)=484-484k$. Setting this equal to zero gives $k=1$. Therefore, option A is correct. Exam tip: identify $a$, $b$, and $c$ carefully before applying $D=0$.

Question 21

What is (k) for equal roots of (x^2-2(k+6)x+k^2=0)?

Accepted Answer

(k=-3) (D=4(k+6)^2-4k^2=0) gives ((k+6)^2=k^2), so (12k+36=0) and (k=-3). In exams, expand squares carefully.

Question 22

If ((x-6)(x-13)=0), what are the solutions by zero product rule?

Accepted Answer

(x=6,13) Direct answer: option A, x=6 and x=13. Apply the zero-product rule to (x−6)(x−13)=0. Either x−6=0 or x−13=0. From the first equation, adding 6 gives x=6. From the second, adding 13 gives x=13. Both values work: x=6 makes the first factor zero, and x=13 makes the second factor zero. B has incorrect negative signs; the equations contain x minus positive numbers, so their zeros are positive 6 and 13. C, 0 and 19, are not obtained by setting either factor to zero. D, 1 and 78, also do not make either factor zero. A product can be zero when one factor is zero; both factors need not be zero at the same time. Memory cue: set each bracket separately equal to zero, then solve.

Question 23

If ((x-6)(x-13)=22), what is the standard quadratic equation?

Accepted Answer

(x^2-19x+56=0) ((x-6)(x-13)=x^2-19x+78), so (x^2-19x+78=22) gives (x^2-19x+56=0). In exams, bring all terms to one side after expansion.

Question 24

What are the roots of (x^2-19x+56=0) by quadratic formula?

Accepted Answer

(x=\frac{19\pm\sqrt{137}}{2}) Here (D=(-19)^2-4(1)(56)=137), so (x=\frac{19\pm\sqrt{137}}{2}). In exams, finding (D) correctly is important.

Question 25

If (3x^2+18x+5=0) is solved by completing the square, which middle step is correct?

Accepted Answer

((x+3)^2=\frac{22}{3}) First (x^2+6x+\frac{5}{3}=0) is obtained, then adding (9) gives ((x+3)^2=\frac{22}{3}). In exams, divide by (a) first when (a
eq1).

Methods of Solving Quadratic Equations

Practice questions

01 If the product of roots of (9x^2-20x+p=0) is (\frac{1}{3}), what is (p)?

02 If the roots of (x^2-23x+126=0) are (\alpha) and (\beta), what is (\alpha^2+\beta^2)?

03 If the roots of (x^2-20x+91=0) are (\alpha) and (\beta), what is (\frac{1}{\alpha}+\frac{1}{\beta})?

04 If (\alpha,\beta) are roots of (7x^2-25x+12=0), what is (\alpha+\beta)?

05 If (\alpha,\beta) are roots of (7x^2-25x+12=0), what is (\alpha\beta)?

06 If \(\alpha,\beta\) are roots of \(7x^2-25x+12=0\), what is \((\alpha-\beta)^2\)?

07 If (x^2-14x+n=0) has no real roots, which condition on (n) is correct?

08 If (x^2-14x+n=0) has two distinct real roots, which condition on (n) is correct?

09 Which factorised form of the equation \(7x^2-19x-6=0\) is correct for solving it?

10 What are the roots of the equation \(7x^2-19x-6=0\)?

11 If (x^2+12x+8=0), which form is correct by completing square?

12 What are the roots of the equation \(x^2+12x+8=0\)?

13 If the roots of (x^2-10x-24=0) are (\alpha,\beta), what is (\alpha^2\beta+\alpha\beta^2)?

14 If one root of (x^2+px+49=0) is double the other and both are negative, what is (p)?

15 What is the difference between the roots of (x^2-11x+6=0)?

16 What is the correct statement about real roots of (6x^2-12x+17=0)?

17 Which completed square form is correct for (6x^2-12x+17=0)?

18 If roots of (x^2-14x+33=0) are (\alpha,\beta), which equation has roots (\alpha+5,\beta+5)?

19 If the roots of (x^2-22x+117=0) are (\alpha,\beta), what is (\frac{\alpha}{\beta}+\frac{\beta}{\alpha})?

20 If the quadratic equation \(kx^2-22x+121=0\) has equal roots, what is the value of \(k\)?

21 What is (k) for equal roots of (x^2-2(k+6)x+k^2=0)?

22 If ((x-6)(x-13)=0), what are the solutions by zero product rule?

23 If ((x-6)(x-13)=22), what is the standard quadratic equation?

24 What are the roots of (x^2-19x+56=0) by quadratic formula?

25 If (3x^2+18x+5=0) is solved by completing the square, which middle step is correct?

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