Roots of a Quadratic Equation

If α and β are the roots of a monic quadratic, then the equation formed from their sum and product is x² − (α + β)x + αβ = 0. The given sum is α + β = 11, and the product is αβ = 28. Substitution gives x² − 11x + 28 = 0, so option A is correct. The coefficient of x must be the negative of the sum, which rules out option B. The constant term must be the product, so options C and D interchange the roles of the sum and product or use an incorrect sign. The equation is therefore uniquely determined.

For a quadratic equation \(ax^2+bx+c=0\), the product of its roots is \(\frac{c}{a}\). Here, \(a=1\) and \(c=70\), so the product of the two roots is 70. Since one root is 7, the other root is \(\frac{70}{7}=10\). Therefore, option A is correct. In exams, remember that the sum of roots is \(-\frac{b}{a}\) and their product is \(\frac{c}{a}\).

The sum of new roots is (4\alpha+4\beta=4(\alpha+\beta)=12). When roots are multiplied by a factor, the sum is also multiplied by that factor.

(\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta=64-30=34). Use identities in such questions.

Here (\alpha+\beta=-5) and (\alpha\beta=-36). Therefore (\frac{1}{\alpha}+\frac{1}{\beta}=\frac{-5}{-36}=\frac{5}{36}).

The sum is (-\frac{b}{a}=\frac{18}{5}) and the product is (\frac{c}{a}=\frac{9}{5}). Identify (a), (b), and (c) first.

The factor theorem and the zero-product principle say that a root is obtained by setting each linear factor equal to zero. From x + 6 = 0, we get x = −6. From x − 2 = 0, we get x = 2. Thus the roots are −6 and 2, making option A correct. A common mistake is to copy the constants without changing their signs, which leads to option C, or to reverse both signs, which leads to option B or D. The order of listing the two roots does not matter, but their signs do. Therefore only option A represents the zeros of the given factors.

For the quadratic equation \(x^2+px+36=0\), the sum of the roots is \(-p\). Since both roots are \(-6\), their sum is \(-6+(-6)=-12\). Therefore, \(-p=-12\), giving \(p=12\). The value \(-12\) is the sum of the roots, not the value of \(p\). Exam tip: In \(x^2+px+c=0\), the sum of the roots is always \(-p\).

Putting (x=\frac{5}{2}) gives (\frac{25}{2}-\frac{45}{2}+c=0), so (c=10). Calculate each term carefully with a fractional root.

Using the given coefficients, the equation is \(x^2-20x+100=0\), which factors as \((x-10)^2=0\). Hence the equation has one repeated root, \(x=10\), so both roots are 10 and 10. Option B has the wrong sign for the roots. Exam tip: when the discriminant \(D=b^2-4ac\) is zero, the two roots are equal.

For \(ax^2+bx+c=0\), the discriminant is \(D=b^2-4ac\). If \(D=0\), both roots equal \(-b/(2a)\). With \(D>0\), the real roots are distinct. Exam tip: check the sign of the discriminant first.

Factoring the equation gives \\(x^2+2x-35=(x+7)(x-5)\\). Thus, \\(x+7=0\\) gives \\(x=-7\\), and \\(x-5=0\\) gives \\(x=5\\). Therefore, the correct pair of roots is \\(5,-7\\). Exam tip: the two factors should have a product of \\(-35\\) and a sum of \\(2\\).

For the quadratic equation (x^2-9x+20=0), the sum of the roots is (\alpha+\beta=9) and their product is (\alpha\beta=20). Hence, (\alpha+\beta+\alpha\beta=9+20=29). Therefore, 29 is correct. Exam tip: for (x^2+bx+c=0), the sum of the roots is (-b) and their product is (c).

If the sum of the roots is S and their product is P, the monic quadratic equation is x² − Sx + P = 0. Substituting S = 0 and P = −36 gives x² − 36 = 0, so option A is correct. Option B represents a product of +36, while options C and D have a non-zero coefficient of x despite the zero sum of roots. Exam tip: change the sign of the root sum for the x-term and use the product as the constant term.

The equation can be factorised as \(25x^2-10x+1=(5x-1)^2=0\). Thus, \(5x-1=0\), giving \(x=\frac{1}{5}\). This is the repeated root. In an exam, identifying a perfect square or checking that the discriminant is zero is the quickest approach.

The equation must be satisfied when a root is substituted. Putting \(x=0\) gives \(2(0)^2+m(0)+n=0\), so \(n=0\). Exam tip: if zero is a root of a quadratic equation, its constant term must be zero.

For a monic quadratic with roots \(r_1\) and \(r_2\), the equation is \(x^2-(r_1+r_2)x+r_1r_2=0\). Here the roots are 3 and 8, so their sum is \(3+8=11\). Consequently, the coefficient of \(x\) is the negative of this sum, namely \(-11\). Therefore option A is correct.

The word “monic” means that the coefficient of \(x^2\) is 1, so no division is needed. The corresponding equation is \(x^2-11x+24=0\), since the product is \(3\cdot8=24\). This confirms that the coefficient of \(x\) is indeed \(-11\), not 11 or either value involving 24. The supplied explanation correctly applies the root-sum relation.

Factor the quadratic as \\(4x^2+3x-10=(4x-5)(x+2)\\). Thus, \\((4x-5)(x+2)=0\\) gives \\(x=\frac{5}{4}\\) or \\(x=-2\\). Hence, option A is correct. In option B, the signs of both roots are reversed. Exam tip: after factorisation, set each factor equal to zero to obtain the roots.

The sum of the roots is \((-5)+(-6)=-11\). For a quadratic equation of the form \(x^2+mx+30=0\), the sum of the roots is \(-m\). Thus, \(-m=-11\), giving \(m=11\). Option B results from a sign error. Exam tip: For \(x^2+bx+c=0\), remember that the sum of the roots is always \(-b\).

Two nonzero numbers are reciprocals when their product is 1, or equivalently when one is the reciprocal of the other. Here 6 × 1/6 = 1, so the roots 6 and 1/6 are reciprocal numbers. Hence option A is correct. They are not equal because 6 is different from 1/6. Their sum is 6 + 1/6 = 37/6, not 1, so option C is false. Their product is 1 rather than 0, which rules out option D. The reciprocal relationship is a direct property of the two stated roots and does not require finding the complete quadratic equation.