Roots of a Quadratic Equation

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.

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

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.

Let the two roots in the ratio 1:3 be k and 3k. Their product is 3k². Since the constant term of x² + px + 12 = 0 is 12, Vieta’s relation gives 3k² = 12, so k² = 4 and k = ±2. Both roots are specified to be negative, so k = −2. The roots are therefore −2 and −6, whose sum is −8. For a monic quadratic x² + px + 12 = 0, the sum of the roots is −p. Thus −p = −8, giving p = 8. Option A is correct. Choosing positive roots would violate the stated condition and lead to the wrong sign.

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.

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).

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.

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 \(D=0\).

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

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.

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 \(\frac{q}{3}\).

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 \(q\).

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
.

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

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.

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}).

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.

The roots are (5) and (6). Thus the smaller root is (5) and the larger root is (6), so (\beta-\alpha=1).

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.

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}).