Roots of a Quadratic Equation

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.

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} \\).