Roots of a Quadratic Equation

For the quadratic equation \(x^2+mx+12=0\), the sum of the roots is \(-m\). The given roots have sum \((-3)+(-4)=-7\), so \(-m=-7\), giving \(m=7\). Option B results from the sign error of taking the root sum as \(m\) instead of \(-m\). Exam tip: In \(x^2+bx+c=0\), the sum of the roots is \(-b\) and their product is \(c\).

Two non-zero numbers are reciprocals when their product is 1, because each is the multiplicative inverse of the other. For the given roots, 4 × 1/4 = 1. Hence 1/4 is the reciprocal of 4, and option A is correct. The roots are not equal because 4 is different from 1/4. Their sum is 4 + 1/4 = 17/4, not 1, so option C is false. Their product is 1 rather than 0, which rules out option D. The governing idea is the reciprocal relationship, not merely the fact that the numbers are both positive.

Factoring gives \(x^2-3x-18=(x-6)(x+3)\). Hence, the roots are \(x=6\) and \(x=-3\). Of these, only 6 is positive, so option A is correct. Exam tip: Check the sign of each root carefully; \(-3\) is a root but it is not positive.

When the discriminant \(D=0\), the two roots are equal. Let each root be \(x\). Then \(x+x=10\), so \(2x=10\) and \(x=5\). Therefore, each root is 5. Exam tip: When \(D=0\), divide the sum of the roots by 2 to find the repeated root.

Factoring gives \(x^2-4x-5=(x-5)(x+1)\), so the roots are \(5\) and \(-1\). Therefore, their absolute difference is \(|5-(-1)|=6\). Exam tip: when subtracting a negative root, carefully account for the two minus signs; the coefficient \(4\) is not the difference between the roots.

For a quadratic equation \(ax^2+bx+c=0\), the sum of the roots is \(\alpha+\beta=-\frac{b}{a}\). Here, \(a=2\) and \(b=-6\), so \(\alpha+\beta=-\frac{-6}{2}=3\). Option B results from mishandling the negative sign of \(b\). Exam tip: identify the coefficients carefully before applying \(-b/a\).

Here \(a=1, b=-6, c=9\). The discriminant is \(D=b^2-4ac=(-6)^2-4(1)(9)=0\), so the roots are real and equal. Distinct real roots occur only when \(D>0\). Exam tip: check the discriminant first to identify the nature of roots.

A quadratic equation has real roots when its discriminant satisfies \(D=b^2-4ac\ge0\). Here, \(a=1\), \(b=2\), and \(c=k\), so \(D=2^2-4(1)(k)=4-4k\). Therefore, \(4-4k\ge0\), which gives \(k\le1\). Hence, option A is correct. If \(k>1\), as stated in option B, the discriminant becomes negative, so the roots are not real. Exam tip: For real roots of a quadratic, always begin with the condition \(D\ge0\).

The roots are (-5) and (-2) because ((x+5)(x+2)=0). When the sum is negative and product is positive, both roots are negative.

For the quadratic equation \(x^2+ax+9=0\), the sum of the roots is \(-\frac{a}{1}=-a\). Since the roots are \(3\) and \(3\), their sum is \(6\). Thus, \(-a=6\), giving \(a=-6\). The value \(6\) results from missing the negative sign. In exams, remember that the sum of roots of \(x^2+bx+c=0\) is \(-b\).

Factorise the quadratic: \(6x^2-7x+2=(3x-2)(2x-1)\). Thus, \((3x-2)(2x-1)=0\) gives \(x=\frac{2}{3}\) or \(x=\frac{1}{2}\). Therefore, option A is correct. The values in option D do not satisfy the two linear factors. In an exam, set each linear factor equal to zero after factorisation to obtain the roots.

For a quadratic equation (a_1x^2+bx+c=0), the product of its roots equals (\frac{c}{a_1}). Here, the product is (a\times(-a)=-a^2), so (\frac{c}{a_1}=-a^2). Therefore, option A is correct. Exam tip: When the roots have equal magnitudes and opposite signs, their product is the negative of the square of either root.

For roots (alpha) and (beta), the monic quadratic equation is (x^2-(alpha+beta)x+alpha beta=0). Substituting the given values gives (x^2-(-3)x-10=0), which simplifies to (x^2+3x-10=0). Option B has the wrong sign for the sum of the roots. Exam tip: use the pattern (x^2-(sum of roots)x+product of roots=0).

Let the roots be \(\alpha\) and \(\beta\). By Vieta’s formulas, \(\alpha+\beta=6\) and \(\alpha\beta=5\). Therefore, \(\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta=6^2-2(5)=36-10=26\). Hence, option A is correct. Remember that 36 is the square of the sum of the roots, not the sum of their squares.

Putting (x=2) gives (4a-10+2=0), so (4a=8) and (a=2). Substitute in questions with an unknown coefficient.

Here, \(a=1\), \(b=-14\), and \(c=49\). For a quadratic equation, the sum of the roots is \(-b/a=14\), while their product is \(c/a=49\). Indeed, \(x^2-14x+49=(x-7)^2\), so both roots are 7; their sum is 14 and their product is 49. Exam tip: directly apply the relations sum \(=-b/a\) and product \(=c/a\). Option C gives the individual repeated root instead of its sum and product.

For a quadratic ax^2+bx+c=0 the discriminant is \(D=b^2-4ac\). If \(D<0\), there are no real roots; instead the equation has two complex conjugate roots. The closest distractor C (2) would be correct only when \(D>0\), which gives two distinct real roots; when \(D=0\) there is one repeated real root. Exam tip: compute the discriminant first — its sign immediately tells you the number of real roots (D>0 → 2, D=0 → 1, D<0 → 0).

Here (\alpha+\beta=\frac{4}{3}) and (\alpha\beta=-\frac{4}{3}). Therefore (\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}=-1).

Since the equation becomes true after substituting (x=a), (a) is a root. In exams check a root by direct substitution.

To check whether a number is a root, substitute \(x=4\) into the equation: \(4^2-9(4)+20=16-36+20=0\). Since the value of the polynomial is zero, \(x=4\) is a root. In fact, \(x^2-9x+20=(x-4)(x-5)\), so the two roots are \(4\) and \(5\); therefore, the statement that only \(x=5\) is a root is incorrect. Exam tip: a number is a root of a polynomial equation exactly when substitution gives zero.