Roots of a Quadratic Equation

Using the given coefficients, the equation is \(x^2-16x+64=0\). It factors as \((x-8)^2=0\), so both roots are \(8\). Thus, the equation has equal roots. In an exam, you can also verify this using the discriminant \(D=b^2-4ac\); here, \(D=0\).

A negative product occurs only when the roots have opposite signs. Therefore one root will be positive and the other negative.

The equation \(x^2+x-20=0\) factors as \((x+5)(x-4)=0\). Hence, \(x=-5\) or \(x=4\), so the roots are 4 and -5. Option B reverses the signs of both roots. Exam tip: Set each linear factor equal to zero to obtain the roots.

For a quadratic equation \(ax^2+bx+c=0\), the sum of the roots is \(\alpha+\beta=-\frac{b}{a}\) and their product is \(\alpha\beta=\frac{c}{a}\). Here, \(a=1, b=-8, c=15\), so \(\alpha+\beta=8\) and \(\alpha\beta=15\). Therefore, \(\alpha+\beta+\alpha\beta=8+15=23\). The values 8 and 15 represent only the sum and product respectively, not their total. Exam tip: identify the sum and product of the roots first, then substitute them into the required expression.

For \(ax^2+bx+c=0\), the discriminant \(D=b^2-4ac\) determines root nature. If \(D>0\), roots are distinct real; if \(D=0\), they are equal. In exams, first write the equation in standard form.

The equation \(16x^2-8x+1=0\) can be written as \((4x-1)^2=0\). Hence, \(4x-1=0\), giving the repeated root \(x=\frac{1}{4}\). Exam tip: when a quadratic is a perfect square, its single root is repeated twice; do not confuse it with the coefficient 4 in the factor.

A root is a value that makes the left-hand side of the equation equal to zero. Substituting (x=0) gives a(0)^2+b(0)+c=c, so c=0 is necessary. The condition (a\ne0) only ensures that the equation is genuinely quadratic; it does not imply a=0 or b=0. Exam tip: if zero is a root of a quadratic equation, its constant term must be zero.

A monic quadratic equation has leading coefficient 1. If its roots are \(\alpha\) and \(\beta\), its form is \(x^2-(\alpha+\beta)x+\alpha\beta=0\). Here, the sum of the roots is \(2+7=9\), so the coefficient of \(x\) is \(-9\). Option B results from missing the negative sign. Exam tip: In a monic quadratic, the coefficient of \(x\) is the negative of the sum of the roots.

Factor the quadratic as \(3x^2+2x-8=(3x-4)(x+2)\). Thus, \((3x-4)(x+2)=0\) gives \(x=\frac{4}{3}\) or \(x=-2\). Therefore, option A is correct. In option B, the signs of both roots are incorrect. Exam tip: After factorising, set each factor equal to zero to obtain the roots.

For the quadratic equation \\(x^2+mx+20=0\\), the sum of the roots is \\(-m\\). The given roots have sum \\((-4)+(-5)=-9\\), so \\(-m=-9\\), which gives \\(m=9\\). Therefore, 9 is correct. Exam tip: for \\(x^2+bx+c=0\\), the sum of the roots is \\(-b\\).

The reciprocal of a non-zero number n is 1/n, and the product of a number and its reciprocal is 1. For the given roots, 5 × 1/5 = 1, so 1/5 is the reciprocal of 5. Therefore option A is correct. The roots are not equal, since 5 and 1/5 have different values. Their sum is 5 + 1/5 = 26/5, not 1, so option C is incorrect. Their product is 1, not 0, which rules out option D. The conclusion follows directly from the multiplicative-inverse definition and does not depend on solving or reconstructing the quadratic equation.

Factor the quadratic: \(x^2-5x-24=(x-8)(x+3)\). Hence, \(x=8\) or \(x=-3\). Only 8 is positive, so option A is correct. Exam tip: Always check the signs of both roots before selecting the positive root.

When the discriminant \(D=0\), the two roots of the quadratic equation are equal. If each root is \(r\), then \(r+r=12\), so \(2r=12\) and \(r=6\). Exam tip: When \(D=0\), assume equal roots first and use their sum or product accordingly.

The equation \(x^2-6x-7=0\) factors as \((x-7)(x+1)=0\). Hence, its roots are \(7\) and \(-1\), and their absolute difference is \(|7-(-1)|=8\). Exam tip: while finding the difference between roots, retain the signs of both roots before subtracting.

For a quadratic equation \(ax^2+bx+c=0\), the sum of the roots is \(\alpha+\beta=-\frac{b}{a}\). Here, \(a=3\) and \(b=-12\), so \(\alpha+\beta=-\frac{-12}{3}=4\). Option B has the wrong sign, while \(\frac{5}{3}\) and \(-\frac{5}{3}\) are related to the product of the roots, not their sum. Exam tip: Always check the sign of \(b\) before applying the formula.

The discriminant is \(D=b^2-4ac\). If \(D=0\), the formula gives the same value \(x=\frac{-b}{2a}\) twice, so the roots are equal. In contrast, \(D>0\) gives distinct real roots. Exam tip: check the discriminant sign first.

A quadratic equation has real roots when its discriminant satisfies \(D=b^2-4ac\ge0\). Here, \(a=1\), \(b=4\), and \(c=k\), so \(D=16-4k\). Therefore, \(16-4k\ge0\) gives \(k\le4\). Option D describes only a sufficient subset, not the complete condition, since roots are also real for \(0\le k\le4\). Exam tip: For questions about real roots, first apply \(D\ge0\).

The roots are (-3) and (-6) because ((x+3)(x+6)=0). When the sum is negative and product is positive, both roots are negative.

For the quadratic equation \(x^2+ax+16=0\), the sum of the roots is \(-\frac{a}{1}=-a\). Since both roots are 4, their sum is \(4+4=8\). Thus, \(-a=8\), giving \(a=-8\). Exam tip: In \(x^2+bx+c=0\), the sum of the roots is always \(-b\), not \(b\).

Factor the quadratic as \\(8x^2-10x+3=(2x-1)(4x-3)\\)。 Thus, \\(2x-1=0\\) gives \\(x=\\frac{1}{2}\\), and \\(4x-3=0\\) gives \\(x=\\frac{3}{4}\\)。 Hence option A is correct. Exam tip: After factorising, set each factor equal to zero and substitute the roots back into the original equation to verify them.