Roots of a Quadratic Equation

Correct answer: A. x² − 6x + 8 = 0

Explanation: Use Vieta’s relations and the transformation of roots. For x² − 3x + 2 = 0, the original roots satisfy α + β = 3 and αβ = 2. The new roots are 2α and 2β, so their sum is 2α + 2β = 2(α + β) = 6. Their product is (2α)(2β) = 4αβ = 8. A monic quadratic with roots r and s is x² − (r + s)x + rs = 0. Substituting the new sum and product gives x² − 6x + 8 = 0, so option A is correct. Option C changes the sum correctly but leaves the product as 4 instead of 8. Option B leaves the sum unchanged, and option D uses the wrong sign for the x-term.

Correct answer: A. 9

Explanation: By Vieta’s formula, \(\alpha+\beta=\frac{-(-5)}{1}=5\). Therefore, \((\alpha+2)+(eta+2)=\alpha+\beta+4=5+4=9\). Hence, the correct answer is 9. Since 2 is added to both roots, the sum increases by 4, not by 2; therefore, 7 is incorrect. Exam tip: when the same number \(k\) is added to both roots, their sum increases by \(2k\).

Correct answer: A. 0

Explanation: The given equation can be written as \(x^2-10x+25=(x-5)^2=0\). Thus, both roots are equal: \(\alpha=\beta=5\). Therefore, \(\alpha-\beta=5-5=0\), so option A is correct. Exam tip: When the discriminant of a quadratic equation is zero, its two roots are equal.

Correct answer: D. \(\Delta>0\) and \(\Delta\) is not a perfect square

Explanation: Using \(x=\frac{-b\pm\sqrt{\Delta}}{2a}\), \(\Delta>0\) gives two distinct real roots. If positive \(\Delta\) is not a perfect square, \(\sqrt{\Delta}\) is irrational, so both roots are irrational. \(\Delta=0\) gives equal roots. Exam tip: check the discriminant first.

Correct answer: A. (-2)

Explanation: (\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta). From (20=16-2\alpha\beta), we get (\alpha\beta=-2).

Correct answer: A. 8

Explanation: Let the roots be t and (2t). By the sum-of-roots relation, t+2t=6, so t=2 and the roots are 2 and 4. Their product is c because the coefficient of x^2 is 1; hence c=2\times4=8. Exam tip: For x^2+bx+c, the sum of roots is -b and their product is c.

Correct answer: A. 8

Explanation: Let the roots be k and 3k because their ratio is 1:3. By Vieta’s product relation for x² + px + 12 = 0, their product is 12. Thus k(3k) = 12, so 3k² = 12 and k² = 4. Hence k = 2 or k = −2. The condition that both roots are negative forces k = −2, giving roots −2 and −6. Their sum is −8. Vieta’s sum relation says that the sum of the roots of x² + px + 12 = 0 is −p. Therefore −p = −8, so p = 8. Option A is correct. Choosing k = 2 would give positive roots and violate the stated condition; it would also produce the opposite sign for p.

Correct answer: A. -6

Explanation: Let both equal roots be \(r\). Their product is \(r^2=9\). Since the roots are positive, \(r=3\). Therefore, their sum is \(3+3=6\). For \(x^2+bx+9=0\), the sum of the roots is \(-b\), so \(-b=6\), giving \(b=-6\). In an exam, equal roots can also be checked using discriminant \(0\), but the positivity condition must still be verified.

Correct answer: A. (2)

Explanation: The sum is (-\frac{b}{a}=\frac{2}{3}) and the product is (\frac{k}{3}). From (\frac{k}{3}=\frac{2}{3}), (k=2).

Correct answer: A. \(m>0\)

Explanation: A quadratic equation has no real roots when its discriminant \(D=b^2-4ac\) is negative. Here, \(a=1, b=-2, c=m+1\), so \(D=(-2)^2-4(1)(m+1)=-4m\). Therefore, \(-4m<0\), which gives \(m>0\). Option D is incorrect because \(m=0\) gives equal real roots, while \(m<0\) gives two distinct real roots. Exam tip: Always determine the sign of the discriminant to identify the nature of quadratic roots.

Correct answer: D. No real \(m\)

Explanation: For equal roots, the discriminant \(D=b^2-4ac\) must be zero. Here, \(a=1\), \(b=-2(m+1)\), and \(c=m^2+2m\). Thus, \(D=[-2(m+1)]^2-4(m^2+2m)=4\), which is never zero for any real \(m\). Therefore, no real value of \(m\) makes the roots equal. Exam tip: For equal roots, immediately apply the condition \(D=0\).

Correct answer: A. Every real (m)

Explanation: The constant term is ((m+1)^2), and the equation becomes ((x-(m+1))^2=0). Hence roots are equal for every real (m).

Correct answer: A. 4

Explanation: For a quadratic equation cax^2+bx+c=0c, the sum of the roots is alpha+beta=-b/ac and their product is alphabeta=c/ac. Hence, here alpha+beta=7c and alphabeta=10c. Therefore, (alpha-1)(beta-1)=alphabeta-alpha-beta+1=10-7+1=4c. Exam tip: For expressions involving roots, first find their sum and product instead of solving for the roots individually.

Correct answer: A. −6

Explanation: For a quadratic equation \(ax^2+bx+c=0\), the product of the roots is \(\frac{c}{a}\). Here, \(a=3\) and \(c=q\), so the product of the roots is \(\frac{q}{3}\). The given product is \(2\times(-3)=-6\); therefore, \(\frac{q}{3}=-6\). Remember that the sum of roots is \(-\frac{p}{3}\), whereas their product is \(\frac{q}{3}\).

Correct answer: A. 1

Explanation: The sum of the roots is \(4+(-1)=3\). For \(x^2+px+q=0\), the sum of the roots is \(-p\), so \(-p=3\) and hence \(p=-3\). Their product is \(4\times(-1)=-4\), so \(q=-4\). Therefore, \(p-q=-3-(-4)=1\). Exam tip: For \(x^2+px+q=0\), remember that the sum of roots is \(-p\) and their product is \(q\).

Correct answer: A. 14

Explanation: By Vieta’s formulas,
alpha+beta=4
and
alpha beta=1
. Therefore,
alpha^2+beta^2=(alpha+beta)^2-2alpha beta=4^2-2(1)=14
. Hence, the correct answer is 14. Choosing 16 results from using only
(alpha+beta)^2
and missing the term
-2alpha beta
. Exam tip: For the sum of the squares of roots, use
alpha^2+beta^2=(alpha+beta)^2-2alpha beta
.

Correct answer: A. (2)

Explanation: Here (\alpha+\beta=-1) and (\alpha\beta=-1). (\alpha^2+\alpha\beta+\beta^2=(\alpha+\beta)^2-\alpha\beta=1-(-1)=2).

Correct answer: A. \(m\) and \(m+1\)

Explanation: The polynomial factors as \(x^2-(2m+1)x+m(m+1)=(x-m)(x-(m+1))\). Hence, \(x=m\) or \(x=m+1\), so the roots are \(m\) and \(m+1\). As a quick check, the sum of the roots must be \(2m+1\) and their product must be \(m(m+1)\); option A satisfies both conditions. Exam tip: compare the sum and product of the proposed roots with the coefficients.

Correct answer: A. (\frac{17}{8})

Explanation: The roots are (3) and (5), so (\alpha-1) and (\beta-1) are (2) and (4). The value is (\frac{2}{4}+\frac{4}{2}=\frac{5}{2}).

Correct answer: A. \(x^2-2x-1=0\)

Explanation: For \(x^2-2x-1=0\), the discriminant is \(D=b^2-4ac=(-2)^2-4(1)(-1)=8\). Since \(D>0\), the roots are real and distinct; since 8 is not a perfect square, they are irrational. Exam tip: use the discriminant to classify roots.

Correct answer: A. (1)

Explanation: The direct answer is option A: 1. For a quadratic equation, the roots are the values of x that make the expression zero. Here, x²−11x+30 can be factorised by finding two numbers whose product is 30 and whose sum is 11. Those numbers are 5 and 6, so x²−11x+30=(x−5)(x−6). Therefore, x=5 or x=6. The smaller root is α=5 and the larger root is β=6. Hence β−α=6−5=1. Option A is correct because it gives 1. Option B, 11, is the coefficient related to the sum of the roots, since 5+6=11; it is not their difference. Option C, 30, is the product of the roots, since 5×6=30; it is not the requested value. Option D, 5, is only the smaller root α, not β−α. A useful exam cue is to identify whether the question asks for a sum, product, or difference before choosing a number.

Correct answer: A. \(64\)

Explanation: For a quadratic equation \(ax^2+bx+c=0\), the product of the roots is \(\alpha\beta=\frac{c}{a}\). Here, \(a=1\) and \(c=8\), so \(\alpha\beta=8\). Therefore, \(\alpha^2\beta^2=(\alpha\beta)^2=8^2=64\). Option B is only the value of \(\alpha\beta\), not its square. Exam tip: Apply Vieta’s formula directly for the sum and product of the roots.

Correct answer: B. (\frac{7}{3}) or (-\frac{7}{3})

Explanation: The roots are (5) and (2), so the sum is (7) and the difference can be (\pm3). Hence the value is (\frac{7}{3}) or (-\frac{7}{3}).

Correct answer: A. 7

Explanation: By Vieta’s formulas, \(\alpha+\beta=2\) and \(\alpha\beta=-8\). Hence, \(\left(\alpha+3\right)\left(\beta+3\right)=\alpha\beta+3(\alpha+\beta)+9=-8+3(2)+9=7\). Therefore, the correct answer is 7. Exam tip: when the same number is added to both roots, expand the product and use the sum and product of roots; using only \(\alpha+\beta\) would be incomplete.

Correct answer: A. 12

Explanation: The product of the equal roots is \(36\), so each root must be \(-6\), since \((-6)(-6)=36\). Their sum is therefore \(-12\). By Vieta’s formula, the sum of the roots is \(-k\), so \(-k=-12\), giving \(k=12\). Exam tip: You can also use the equal-root condition \(k^2-144=0\); the requirement that the roots be negative selects \(k=12\), not \(-12\).