Roots of a Quadratic Equation

To check whether a number is a root, substitute it into the equation and verify whether the left-hand side becomes zero. For x = -2, 3(-2)² + 2(-2) - 8 = 12 - 4 - 8 = 0. Therefore, x = -2 is a root of the equation. Option C is incorrect because substituting x = 2 gives 12 + 4 - 8 = 8, not zero. Exam tip: the square of a negative number is positive, but its linear term remains negative.

Factorise the quadratic: \(x^2-12x+35=(x-5)(x-7)\). Thus, \((x-5)(x-7)=0\) gives \(x=5\) or \(x=7\), so the roots are \((5,7)\). Option B has numbers whose product is 35, but their sum is not 12. Exam tip: For \(x^2+bx+c\), look for two numbers whose sum is \(-b\) and product is \(c\).

The quadratic factors as \(3x^2-13x+4=(3x-1)(x-4)\). Thus, \((3x-1)(x-4)=0\) gives \(x=\frac{1}{3}\) or \(x=4\), so option A is correct. The values in option D have sum \(\frac{7}{3}\), whereas the sum of the roots must be \(\frac{13}{3}\). Exam tip: verify the roots using the sum \(-\frac{b}{a}\) and product \(\frac{c}{a}\).

Factoring the quadratic gives \(2x^2+9x+4=(2x+1)(x+4)\). Thus, \((2x+1)(x+4)=0\) gives \(x=-\frac{1}{2}\) or \(x=-4\), so option A is correct. In option B, the signs of both roots are incorrect. Exam tip: Set each linear factor equal to zero to obtain the roots directly.

If α and β are the roots of a quadratic equation, its equation is (x − α)(x − β) = 0. Here α = −3 and β = 6, so (x + 3)(x − 6) = 0. Expanding gives x² − 3x − 18 = 0, making option A correct. Exam tip: the sum of the roots is 3, so the coefficient of x is −3, while their product is −18, the constant term.

Since \(x=3\) is a root, substitute it into the equation: \(3^2+3k-18=0\). Thus, \(9+3k-18=0\), so \(3k=9\) and \(k=3\). Hence, option A is correct. Option B results from an incorrect sign. Exam tip: substitute the given root directly into the quadratic equation and solve for the parameter.

For a quadratic equation \(ax^2+bx+c=0\), the sum of the roots is \(-\frac{b}{a}\). Here, \(a=4\) and \(b=-11\), so the sum is \(-\frac{-11}{4}=\frac{11}{4}\). Option B has the wrong sign, while the options containing \(\frac{7}{4}\) reflect confusion with the constant term. Exam tip: use \(-b/a\) directly to find the sum of the roots.

For a quadratic equation \\(ax^2+bx+c=0\\), the product of its roots is \\(\\frac{c}{a}\\). Here, \\(a=5\\) and \\(c=-8\\), so the product is \\(\\frac{-8}{5}=-\\frac{8}{5}\\). Option B has the wrong sign. Exam tip: remember that the sum of the roots is \\( -\\frac{b}{a} \\), while their product is \\( \\frac{c}{a} \\).

For a quadratic equation \(ax^2+bx+c=0\), here \(a=3\), \(b=-6\), and \(c=3\). Thus, \(D=b^2-4ac=(-6)^2-4(3)(3)=36-36=0\). Therefore, the two roots are real and equal. Exam tip: when \(D=0\), a quadratic equation has two equal real roots.

For the given quadratic equation, \(a=1\), \(b=-4\), and \(c=13\). Its discriminant is \(D=b^2-4ac=(-4)^2-4(1)(13)=16-52=-36\). Since \(D<0\), the equation has no real roots. Equal real roots would require \(D=0\), so option B is incorrect. Exam tip: use the sign of the discriminant to determine the nature of the roots quickly.

For a monic quadratic with root sum \(S\) and product \(P\), the equation is \(x^2-Sx+P=0\). Substituting \(S=5\) and \(P=6\) gives option A. Option B has root sum \(-5\). Exam tip: check the sign of the middle term carefully.

If the sum of the roots is \(S\) and their product is \(P\), the monic quadratic equation is \(x^2-Sx+P=0\). Substituting \(S=9\) and \(P=18\) gives \(x^2-9x+18=0\). Option B has the wrong sign for the sum, while options C and D interchange the sum and product. Exam tip: use a negative coefficient for the sum of roots and a positive constant equal to their product.

For a quadratic equation \(ax^2+bx+c=0\), the product of the roots is \(\frac{c}{a}\). Here, \(a=1\) and \(c=54\), so the product of the two roots is \(54\). If one root is \(6\), the other root is \(\frac{54}{6}=9\). Option 6 merely repeats the given root, while 15 is the sum of the roots, not their product. Exam tip: use the product-of-roots relation and divide by the known root.

The sum of new roots is (3\alpha+3\beta=3(\alpha+\beta)=12). When roots are multiplied by a factor, the sum is multiplied by the same factor.

(\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta=49-24=25). Use identities in such questions.

Here (\alpha+\beta=-5) and (\alpha\beta=-14). Therefore (\frac{1}{\alpha}+\frac{1}{\beta}=\frac{-5}{-14}=\frac{5}{14}).

The sum is (-\frac{b}{a}=\frac{14}{3}) and the product is (\frac{c}{a}=\frac{8}{3}). Identify (a), (b), and (c) first.

The roots are obtained by setting each linear factor equal to zero. For the factor \\(x-4\\), solve \\(x-4=0\\); adding 4 to both sides gives \\(x=4\\). For the factor \\(x+7\\), solve \\(x+7=0\\); subtracting 7 gives \\(x=-7\\). Therefore the roots are 4 and -7.

Equivalently, the quadratic can be represented, apart from a nonzero constant multiplier, by \\((x-4)(x+7)=0\\). The product is zero when either factor is zero, so these are exactly the two roots. A factor written as \\(x-r\\) has root \\(r\\), while a factor written as \\(x+r\\) has root \\(-r\\). Thus option A is correct. The alternatives with both signs positive or both negative do not satisfy the two given factors.

For a quadratic equation \(x^2+px+25=0\), the sum of the roots is \(-p\). The given roots have sum \(-5+(-5)=-10\). Hence, \(-p=-10\), so \(p=10\). Exam tip: In \(x^2+bx+c=0\), the sum of the roots is \(-b\), not \(b\).

Putting (x=\frac{4}{3}) gives (\frac{16}{3}-\frac{28}{3}+c=0), so (c=4). For fractional roots calculate each term separately.