Q: If in $6x^2+kx+66=0$, the product of roots is (-2) times the sum of roots, what is (k)?

(33) The product is (11) and the sum is $-\frac{k}{6}$. From $11=-2\left(-\frac{k}{6}\right)$, we get $k=33$.

Q: If the roots of $x^2+bx+c=0$ are $-6$ and $13$, what is the value of $b+c$?

-85 For the monic quadratic $x^2+bx+c=0$, Vieta's formulas give sum of roots $\alpha+\beta=-b$ and product $\alpha\beta=c$. Here $\alpha+\beta=-6+13=7$, so $-b=7$ and hence $b=-7$. The product is $-6\times13=-78$, so $c=-78$. Therefore $b+c=-7+(-78)=-85$. A common error (leading to option C = 71) is to take the product as +78 instead of -78. Exam tip: apply Vieta’s relations directly and watch sign conventions for sum and product of roots in monic quadratics.

Q: If $a=-4$, what type of statement does $(a+4)x^2+(a^2-16)x+13=0$ become?

Contradictory statement Substituting $a=-4$ gives $a+4=0$ and $a^2-16=0$, so the equation reduces to $0\cdot x^2+0\cdot x+13=0$, i.e. $13=0$. This is impossible, so the statement is contradictory. Option B is wrong because the quadratic term vanishes, C is wrong because the linear term also vanishes, and D is wrong because the equality is not true for any x. Exam tip: when a parameter is given, first check whether coefficients of highest-degree terms become zero — that indicates a degenerate (constant) equation leading to either contradiction or identity.

Q: If $x=-1$ is a root of $x^2+(6k-3)x+5k=0$, what is the value of $k$?

4 Substitute $x=-1$: $1+(6k-3)(-1)+5k=0$ which simplifies to $1-6k+3+5k=0$, i.e. $4-k=0$. Therefore $k=4$. Option B ($-4$) often results from a sign error during substitution; the correct simplification yields a positive 4. Exam tip: Always substitute the root directly and simplify terms carefully, watching for sign changes when $x$ is negative.

Q: The roots of $x^2-9x+4=0$ are $\alpha,\beta$. What is $\alpha+\beta+5\alpha\beta$?

29 For $x^2-9x+4=0$ we have $a=1,\;b=-9,\;c=4$. By Vieta, $\alpha+\beta=-b/a=9$ and $\alpha\beta=c/a=4$. Therefore $\alpha+\beta+5\alpha\beta=9+5\times4=29$. Option B (13) is wrong because it uses $\alpha+\beta+\alpha\beta=9+4$ instead of adding $5\alpha\beta$; option C (20) equals only $5\alpha\beta$; option D (36) equals $(\alpha+\beta)(\alpha\beta)=9\times4$. Exam tip: apply Vieta’s formulas quickly to get sum and product of roots for any quadratic $ax^2+bx+c=0$.

Q: If $\alpha,\beta$ are the roots of $x^2-20x+96=0$, what is the value of $(\alpha-8)(\beta-8)$?

0 By Vieta's formulas for $x^2-20x+96=0$, we have $\alpha+\beta=20$ and $\alpha\beta=96$. Thus $(\alpha-8)(\beta-8)=\alpha\beta-8(\alpha+\beta)+64=96-8\times20+64=96-160+64=0.$ Notes on distractors: 96 is a common mistake from taking $\alpha\beta$ directly; -64 can arise from a sign error when handling the constant term. Exam tip: Always compute sum and product of roots first (Vieta) and then substitute into the expanded expression to avoid algebraic slips.

Q: A right triangle has base $(x+5)$, height $(x+9)$, and area 95 square units. Which quadratic equation represents this situation?

$x^2+14x-145=0$ Area of the triangle is given by $\frac{1}{2}\times\text{base}\times\text{height}$. So $\frac{1}{2}(x+5)(x+9)=95$. Multiplying both sides by 2 gives $(x+5)(x+9)=190$. Expanding yields $x^2+14x+45=190$, which simplifies to $x^2+14x-145=0$; hence option A is correct. Option B is the common mistake of setting $(x+5)(x+9)=95$ (forgetting the factor 1/2). Options C and D have incorrect constant terms. Exam tip: write the area formula first and multiply by 2 immediately to avoid sign/constant errors when forming the quadratic.

Question 1

If in $6x^2+kx+66=0$, the product of roots is (-2) times the sum of roots, what is (k)?

Accepted Answer

(33) The product is (11) and the sum is $-\frac{k}{6}$. From $11=-2\left(-\frac{k}{6}\right)$, we get $k=33$.

Question 2

What is the sum of cubes of the roots of (x^2-7x-18=0)?

Accepted Answer

(721) The sum of roots is (7) and the product is (-18). (\alpha^3+\beta^3=(\alpha+\beta)^3-3\alpha\beta(\alpha+\beta)=343+378=721).

Question 3

If the roots of $x^2+bx+c=0$ are $-6$ and $13$, what is the value of $b+c$?

Accepted Answer

-85 For the monic quadratic $x^2+bx+c=0$, Vieta's formulas give sum of roots $\alpha+\beta=-b$ and product $\alpha\beta=c$. Here $\alpha+\beta=-6+13=7$, so $-b=7$ and hence $b=-7$. The product is $-6\times13=-78$, so $c=-78$. Therefore $b+c=-7+(-78)=-85$. A common error (leading to option C = 71) is to take the product as +78 instead of -78. Exam tip: apply Vieta’s relations directly and watch sign conventions for sum and product of roots in monic quadratics.

Question 4

If $a=-4$, what type of statement does $(a+4)x^2+(a^2-16)x+13=0$ become?

Accepted Answer

Contradictory statement Substituting $a=-4$ gives $a+4=0$ and $a^2-16=0$, so the equation reduces to $0\cdot x^2+0\cdot x+13=0$, i.e. $13=0$. This is impossible, so the statement is contradictory. Option B is wrong because the quadratic term vanishes, C is wrong because the linear term also vanishes, and D is wrong because the equality is not true for any x. Exam tip: when a parameter is given, first check whether coefficients of highest-degree terms become zero — that indicates a degenerate (constant) equation leading to either contradiction or identity.

Question 5

What is the standard form of ((x+6)^2+(x-7)^2=(x+3)^2)?

Accepted Answer

(x^2-14x+76=0) Expanding gives left side (2x^2-2x+85) and right side (x^2+6x+9). Subtracting gives (x^2-8x+76=0).

Question 6

If $x=-1$ is a root of $x^2+(6k-3)x+5k=0$, what is the value of $k$?

Accepted Answer

4 Substitute $x=-1$: $1+(6k-3)(-1)+5k=0$ which simplifies to $1-6k+3+5k=0$, i.e. $4-k=0$. Therefore $k=4$. Option B ($-4$) often results from a sign error during substitution; the correct simplification yields a positive 4. Exam tip: Always substitute the root directly and simplify terms carefully, watching for sign changes when $x$ is negative.

Question 7

The roots of $x^2-9x+4=0$ are $\alpha,\beta$. What is $\alpha+\beta+5\alpha\beta$?

Accepted Answer

29 For $x^2-9x+4=0$ we have $a=1,\;b=-9,\;c=4$. By Vieta, $\alpha+\beta=-b/a=9$ and $\alpha\beta=c/a=4$. Therefore $\alpha+\beta+5\alpha\beta=9+5\times4=29$. Option B (13) is wrong because it uses $\alpha+\beta+\alpha\beta=9+4$ instead of adding $5\alpha\beta$; option C (20) equals only $5\alpha\beta$; option D (36) equals $(\alpha+\beta)(\alpha\beta)=9\times4$. Exam tip: apply Vieta’s formulas quickly to get sum and product of roots for any quadratic $ax^2+bx+c=0$.

Question 8

If the roots of (x^2+px+q=0) are (6) and (p), what is the value of (p)?

Accepted Answer

(-3) The direct answer is option A: p = -3. In a quadratic x² + px + q = 0, the sum of the roots is -p because the coefficient of x is p. The stated roots are 6 and p, so their sum is 6 + p. Equating these gives 6 + p = -p. Add p to both sides: 6 + 2p = 0. Therefore 2p = -6 and p = -3. Hence option A is correct. Option B, p = 3, would make the root sum 9, while the coefficient rule gives -3, so it is inconsistent. Option C, p = 6, would make the two roots both 6 and their sum 12, but the rule gives -6. Option D, p = -6, would make the sum 0, while the rule gives 6. The value of q is not needed because the question asks only for p and the sum-of-roots relation already determines it. A common mistake is to write 6 + p = p; the correct coefficient rule is 6 + p = -p, including the minus sign. Exam cue: whenever the equation is x² + px + q, immediately write root sum = -p.

Question 9

If $\alpha,\beta$ are the roots of $x^2-20x+96=0$, what is the value of $(\alpha-8)(\beta-8)$?

Accepted Answer

0 By Vieta's formulas for $x^2-20x+96=0$, we have $\alpha+\beta=20$ and $\alpha\beta=96$. Thus
$(\alpha-8)(\beta-8)=\alpha\beta-8(\alpha+\beta)+64=96-8\times20+64=96-160+64=0.$
Notes on distractors: 96 is a common mistake from taking $\alpha\beta$ directly; -64 can arise from a sign error when handling the constant term. Exam tip: Always compute sum and product of roots first (Vieta) and then substitute into the expanded expression to avoid algebraic slips.

Question 10

A right triangle has base $(x+5)$, height $(x+9)$, and area 95 square units. Which quadratic equation represents this situation?

Accepted Answer

$x^2+14x-145=0$ Area of the triangle is given by $\frac{1}{2}\times\text{base}\times\text{height}$. So $\frac{1}{2}(x+5)(x+9)=95$. Multiplying both sides by 2 gives $(x+5)(x+9)=190$. Expanding yields $x^2+14x+45=190$, which simplifies to $x^2+14x-145=0$; hence option A is correct. Option B is the common mistake of setting $(x+5)(x+9)=95$ (forgetting the factor 1/2). Options C and D have incorrect constant terms. Exam tip: write the area formula first and multiply by 2 immediately to avoid sign/constant errors when forming the quadratic.

Question 11

If $(\alpha,\beta)$ are the roots of $x^2-19x+84=0$, what is the value of $(\alpha+\beta)^2$?

Accepted Answer

361 For a quadratic $ax^2+bx+c=0$, the sum of roots is $\alpha+\beta=-\dfrac{b}{a}$ and the product is $\alpha\beta=\dfrac{c}{a}$. In this equation $a=1,\; b=-19,\; c=84$, so $\alpha+\beta=-\dfrac{-19}{1}=19$. Therefore $(\alpha+\beta)^2=19^2=361$. Option B (84) is the product $\alpha\beta$, not the square of the sum. Options C (256) and D (324) are $16^2$ and $18^2$ respectively and are incorrect because the sum is 19, not 16 or 18. Exam tip: identify $a,b,c$ quickly and use $\alpha+\beta=-b/a$; watch the sign of $b$.

Question 12

The equation $x^2+14x+k=0$ has no real roots. What is the correct condition on $k$?

Accepted Answer

$k>49$ For no real roots the discriminant $D=b^2-4ac$ must be negative. Here $a=1,\; b=14,\; c=k$, so $D=14^2-4\cdot1\cdot k=196-4k$. Requiring $D<0$ gives $196-4k<0\Rightarrow k>49$. Thus $k>49$ is correct. The closest distractor $k=49$ yields $D=0$, giving one repeated real root, not “no real roots.” Exam tip: compute $D$ first and use its sign to decide the nature of roots ($D<0,=0,>0$).

Question 13

If $x^2-2ax+a^2-49=0$, what is the difference between the two roots of this quadratic equation?

Accepted Answer

14 Complete the square: $x^2-2ax+a^2-49=(x-a)^2-49=0$. So $(x-a)^2=49$ and the roots are $a+7$ and $a-7$. Their difference is $(a+7)-(a-7)=14$. Option B (7) is incorrect — it is half the actual difference. Option C (2a) is unrelated to the difference here (it would be a mistaken use of coefficients), and option D (a+7) is just one root, not the difference. Exam tip: try completing the square or factor as $(x-a-7)(x-a+7)=0$ to read off roots quickly.

Question 14

What is the standard quadratic form of (\frac{x+3}{x-2}+\frac{x-2}{x+3}=\frac{13}{6})?

Accepted Answer

(x^2+x-156=0) The direct answer is A: $x^2+x-156=0$, with the restrictions $x\ne2,-3$. Multiply by the common denominator $6(x-2)(x+3)$: $6(x+3)^2+6(x-2)^2=13(x-2)(x+3)$. Expand the left side: $6(x^2+6x+9+x^2-4x+4)=12x^2+12x+78$. The right side is $13(x^2+x-6)=13x^2+13x-78$. Subtracting gives $x^2+x-156=0$. Option A is correct. Option B has the wrong sign of the linear term. Option C has the wrong sign of the constant term. Option D incorrectly keeps a coefficient 2 on $x^2$. The denominator restrictions must be remembered because multiplying can otherwise introduce invalid values.

Question 15

If (\alpha,\beta) are roots of (3x^2-11x+6=0), what is the value of (\frac{\alpha}{\beta}+\frac{\beta}{\alpha})?

Accepted Answer

( \frac{85}{18} ) We use (\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha\beta}). Here (\alpha+\beta=\frac{11}{3}) and (\alpha\beta=2), so the value is (\frac{85}{18}).

Question 16

If the roots of (x^2-(2k+5)x+(k^2+5k+6)=0) are consecutive integers, what will those roots be?

Accepted Answer

(k+2) and (k+3) The sum of roots is (2k+5) and the product is (k^2+5k+6). ((k+2)+(k+3)=2k+5) and ((k+2)(k+3)=k^2+5k+6) are correct.

Question 17

If the roots of $x^2+px+q=0$ are $p+2$ and $q-2$, and $p+q=8$, what is the value of $p$?

Accepted Answer

-8 By Vieta's relations for $x^2+px+q=0$, the sum of roots equals $-p$. The given roots sum to $(p+2)+(q-2)=p+q=8$, so $-p=8$ and hence $p=-8$. Note: checking the product condition $(p+2)(q-2)=q$ with $p=-8$ (so $q=16$) gives $-84\neq16$, so the full set of data is inconsistent; nevertheless the value of $p$ is determined by the sum relation asked in the question. Exam tip: check both sum and product; a common mistake is to miss the negative sign and pick $8$.

Introduction to Quadratic Equations

Practice questions

01 If in \(6x^2+kx+66=0\), the product of roots is (-2) times the sum of roots, what is (k)?

02 What is the sum of cubes of the roots of (x^2-7x-18=0)?

03 If the roots of \(x^2+bx+c=0\) are \(-6\) and \(13\), what is the value of \(b+c\)?

04 If \(a=-4\), what type of statement does \((a+4)x^2+(a^2-16)x+13=0\) become?

05 What is the standard form of ((x+6)^2+(x-7)^2=(x+3)^2)?

06 If \(x=-1\) is a root of \(x^2+(6k-3)x+5k=0\), what is the value of \(k\)?

07 The roots of \(x^2-9x+4=0\) are \(\alpha,\beta\). What is \(\alpha+\beta+5\alpha\beta\)?

08 If the roots of (x^2+px+q=0) are (6) and (p), what is the value of (p)?

09 If \(\alpha,\beta\) are the roots of \(x^2-20x+96=0\), what is the value of \((\alpha-8)(\beta-8)\)?

10 A right triangle has base \((x+5)\), height \((x+9)\), and area 95 square units. Which quadratic equation represents this situation?

11 If \((\alpha,\beta)\) are the roots of \(x^2-19x+84=0\), what is the value of \((\alpha+\beta)^2\)?

12 The equation \(x^2+14x+k=0\) has no real roots. What is the correct condition on \(k\)?

13 If \(x^2-2ax+a^2-49=0\), what is the difference between the two roots of this quadratic equation?

14 What is the standard quadratic form of (\frac{x+3}{x-2}+\frac{x-2}{x+3}=\frac{13}{6})?

15 If (\alpha,\beta) are roots of (3x^2-11x+6=0), what is the value of (\frac{\alpha}{\beta}+\frac{\beta}{\alpha})?

16 If the roots of (x^2-(2k+5)x+(k^2+5k+6)=0) are consecutive integers, what will those roots be?

17 If the roots of \(x^2+px+q=0\) are \(p+2\) and \(q-2\), and \(p+q=8\), what is the value of \(p\)?

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