Roots of a Quadratic Equation

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.

For a quadratic equation ax² + bx + c = 0 with roots α and β, Vieta’s product relation is αβ = c/a. In this equation, a = 6 and c = −4. Therefore αβ = −4/6 = −2/3, so option A is correct. The sign must be retained when using c; dropping the negative sign would incorrectly produce option B. The values −5/6 and 5/6 involve the coefficient b = 5 and therefore confuse the product formula with the sum formula, since α + β = −b/a = −5/6. Thus the constant and leading coefficients, not the middle coefficient alone, determine the product.

A quadratic equation has real roots when its discriminant satisfies \(D=b^2-4ac\ge0\). Here, \(a=1\), \(b=6\), and \(c=k\), so \(D=36-4k\). Therefore, \(36-4k\ge0\) gives \(k\le9\). The value \(k=9\) is only the special case of equal roots, whereas \(k>9\) produces non-real roots. Exam tip: For real roots, begin by imposing the condition \(D\ge0\).

The roots are (-6) and (-7) because ((x+6)(x+7)=0). When the sum is negative and product is positive, both roots are negative.

For the quadratic equation \(x^2+ax+25=0\), the sum of the roots is \(-a\). The given sum is \(5+5=10\), so \(-a=10\), which gives \(a=-10\). Therefore, option A is correct. Exam tip: In \(x^2+bx+c=0\), the sum of the roots is \(-b\) and their product is \(c\).

Factor the quadratic: \(10x^2-17x+3=(2x-3)(5x-1)\). Thus, \((2x-3)(5x-1)=0\) gives \(x=\frac{3}{2}\) or \(x=\frac{1}{5}\), so option A is correct. Option B has the signs of both roots wrong. Exam tip: Set each linear factor equal to zero, then verify the roots using the sum and product of roots if needed.

For the quadratic equation \(ax^2+bx+d=0\), the product of the roots is \(\frac{d}{a}\). Here, the product is \(c\times(-c)=-c^2\), so \(\frac{d}{a}=-c^2\). Option B misses the negative sign. Exam tip: For \(ax^2+bx+d=0\), remember that the sum of roots is \(-\frac{b}{a}\) and the product is \(\frac{d}{a}\).

If \(\alpha\) and \(\beta\) are the roots of a monic quadratic equation, its form is \(x^2-(\alpha+\beta)x+\alpha\beta=0\). Substituting the given values gives \(x^2-(-7)x+(-18)=0\), which simplifies to \(x^2+7x-18=0\). Option B incorrectly uses the sign of the sum. Exam tip: the coefficient of \(x\) is the negative of the sum of the roots, while the constant term is their product.

Let the roots be \(\alpha\) and \(\beta\). By the relations between the roots and coefficients, \(\alpha+\beta=9\) and \(\alpha\beta=14\). Hence, \(\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta=9^2-2(14)=81-28=53\). Therefore, the correct answer is 53. Exam tip: use \(\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta\); \(81\) alone is only \((\alpha+\beta)^2\), not the required sum.

If a number is a root of a polynomial equation, substituting that number into the equation must make the expression equal to zero. Put x = 4 into ax² − 10x + 8 = 0: a(4)² − 10(4) + 8 = 0. This gives 16a − 40 + 8 = 0, or 16a − 32 = 0. Hence 16a = 32 and a = 2. Therefore option A is correct. The other options do not satisfy the equation: a = 1 gives −16, a = −2 gives −64, and a = −1 gives −48 after substitution. Direct substitution is the appropriate method because the unknown is a coefficient.

For the standard form \(ax^2+bx+c=0\), we have \(a=1\), \(b=-22\), and \(c=121\). The sum of the roots is \(-b/a=22\), and their product is \(c/a=121\). In fact, \(x^2-22x+121=(x-11)^2\), so both roots are 11. Option B has the wrong sign for the sum. Exam tip: use \(-b/a\) for the sum and \(c/a\) for the product.

Here (\alpha+\beta=\frac{5}{4}) and (\alpha\beta=-\frac{3}{2}). Therefore (\alpha+\beta-2\alpha\beta=\frac{5}{4}+3=\frac{17}{4}).

Here (\alpha+\beta=7) and (\alpha\beta=12). (\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{(\alpha+\beta)^2-2\alpha\beta}{\alpha\beta}=\frac{25}{12}).

((\alpha-\beta)^2=(\alpha+\beta)^2-4\alpha\beta=25-24=1). Remember this identity for difference of roots.

Here, \(a=1\), \(b=-2k\), and \(c=k^2\). The discriminant is \(D=b^2-4ac=(-2k)^2-4(1)(k^2)=0\). Hence the roots are real and equal; equivalently, the equation is \((x-k)^2=0\), so both roots are \(k\). Option D is true only for the special case \(k=0\), not for every value of \(k\). Exam tip: when \(D=0\), a quadratic equation has equal real roots.

For a quadratic equation \(ax^2+bx+c=0\), the product of its roots is \(\frac{c}{a}\). Here, the product of the roots is \(2\times3=6\) and \(c=6\). Thus, \(\frac{6}{a}=6\), giving \(a=1\). Therefore, option A is correct. Exam tip: remember that the sum of roots is \(-\frac{b}{a}\) and their product is \(\frac{c}{a}\).

By Vieta’s theorem, the sum of the roots of x² − (m + 2)x + 12 = 0 is m + 2. Since the given roots are 3 and 4, their sum is 7, so m + 2 = 7 and m = 5. Their product, 3 × 4 = 12, also confirms the constant term. Option B, 7, is the sum of the roots, not the value of m. Exam tip: For x² + bx + c = 0, the sum of the roots is −b.