Roots of a Quadratic Equation

Two distinct real roots require \(b^2-4ac>0\). For option C, \(25-24=1>0\), so it has distinct real roots. Option B has discriminant 0 and hence equal roots. Exam tip: check the sign of the discriminant first.

If α and β are the roots of a monic quadratic equation, the standard relationship is x² − (α + β)x + αβ = 0. The word monic means that the coefficient of x² is 1. Here α + β = 11 and 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 makes option B incorrect. The constant term must equal the product, so options C and D incorrectly interchange the sum and product; option D also has incorrect signs. As a check, the equation factors as (x − 4)(x − 7) = 0, whose roots have sum 11 and product 28.

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.

The direct answer is option A: 34. Apply the identity α² + β² = (α + β)² - 2αβ. The reason is that (α + β)² expands to α² + 2αβ + β², so removing 2αβ leaves α² + β². Here α + β = 8, hence (α + β)² = 8² = 64. Also αβ = 15, hence 2αβ = 2 × 15 = 30. Therefore α² + β² = 64 - 30 = 34. Thus option A is correct. Option B, 49, is not the required value; it may come from using an unrelated square or an incorrect calculation. Option C, 64, is only the square of the sum and fails to remove the middle term 2αβ. Option D, 30, is twice the product, not the sum of the squares. The important point is that the sum alone is insufficient; the product must also be used. Because both are supplied, the identity gives the answer quickly and safely. Exam cue: square the given sum first, calculate twice the product second, then subtract.

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 governing concept is the zero-product principle: if a product is zero, at least one factor must be zero. Since the factors are x + 6 and x − 2, set each one equal to zero separately. From x + 6 = 0, subtracting 6 from both sides gives x = −6. From x − 2 = 0, adding 2 gives x = 2. Therefore the two roots are −6 and 2, so option A is correct. The signs must change when a factor is equated to zero; simply copying the constants would give the incorrect pair 6 and 2. The order in which roots are written is immaterial, but their signs are essential. Both values make one of the given factors zero.

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 \(-b\).

The governing property is the definition of reciprocal numbers. Two nonzero numbers are reciprocals when their product is 1; equivalently, the reciprocal of a nonzero number t is 1/t. For the given roots, 6 × (1/6) = 1, so 6 and 1/6 are reciprocals of one another. Therefore option A is correct. They are not equal, because 6 is much larger than 1/6, so option B is false. Their sum is 6 + 1/6 = 36/6 + 1/6 = 37/6, not 1, ruling out option C. Their product is 1 rather than 0, ruling out option D. No complete quadratic equation is needed because the question asks only about a direct relationship between the stated roots.

Factorise the quadratic: \(x^2-7x-30=(x-10)(x+3)\). Hence, \(x=10\) or \(x=-3\). Only \(10\) is positive, so option A is correct. Exam tip: When asked for a positive root, find both roots and check their signs.

When the discriminant is \(D=0\), the two roots of the quadratic equation are equal. If each root is \(r\), then \(r+r=14\), so \(2r=14\) and \(r=7\). Therefore, each root is 7. Exam tip: When \(D=0\), divide the sum of the roots by 2 to find the repeated root.

Factoring gives \(x^2-8x-9=(x-9)(x+1)\). Therefore, the roots are \(9\) and \(-1\), and their positive difference is \(9-(-1)=10\). Exam tip: for roots \(\alpha\) and \(\beta\), use \(|\alpha-\beta|\) when the question asks for the difference between the roots.

A quadratic has equal real roots when its discriminant \(b^2-4ac=0\). For option A, \(36-4\times1\times9=0\). Option B has a positive discriminant, so its roots are distinct. Exam tip: check the discriminant before solving the equation.