Roots of a Quadratic Equation

By the factor theorem, if \(x-9\) is a factor, then setting the factor equal to zero gives \(x-9=0\), so \(x=9\). Hence, 9 is certainly a root. Exam tip: a factor of the form \(x-a\) gives the root \(a\), whereas \(x+a\) gives the root \(-a\).

Since \\(5\\) is a root, substitute \\(x=5\\) into the equation: \\(5^2+5t-30=0\\). Thus, \\(25+5t-30=0\\), so \\(5t=5\\) and \\(t=1\\). In such questions, substitute the given root directly and check the signs carefully.

Since \(11\neq 0\), dividing \(11x^2=0\) by 11 gives \(x^2=0\). Thus \(x=0\) occurs twice, so 0 is a repeated root. Exam tip: In an equation of the form \(ax^2=0\), where \(a\neq0\), the root is always \(x=0\) with multiplicity 2.

Rewriting \(x^2-81=0\) as \(x^2=81\) gives \(x=\pm 9\). Therefore, the two roots are \(9\) and \(-9\). Option B incorrectly treats 81 as a root, although 81 is the value of \(x^2\). In exams, remember to consider both the positive and negative values when taking the square root.

For \(ax^2+bx+c=0\), the discriminant is \(D=b^2-4ac\). Equal roots occur precisely when \(D=0\). If \(D>0\), there are two distinct real roots. Exam tip: identify the nature of roots by checking the discriminant first.

If the roots are 6 and −2, the equation is \((x-6)(x+2)=0\). Expanding it gives \(x^2-4x-12=0\), so option A is correct. Exam tip: the sum of the roots is 4, giving the coefficient −4 for \(x\), and their product is −12.

For a quadratic equation \(ax^2+bx+c=0\), the sum of the roots is \(-\frac{b}{a}\). Here, \(a=1\) and \(b=-13\), so the sum is \(-\frac{-13}{1}=13\). The value 42 is related to the product of the roots, \(\frac{c}{a}\), not their sum. In an exam, use \(-b/a\) directly to find the sum.

For a quadratic equation \\(ax^2+bx+c=0\\), the product of its roots is \\(\\frac{c}{a}\\). Here, \\(a=7\\) and \\(c=-14\\), so the product is \\(\\frac{-14}{7}=-2\\). Option B has the sign wrong. Exam tip: remember that the sum and product of roots are \\( -\\frac{b}{a}\\) and \\(\\frac{c}{a}\\), respectively.

For 0 to be a root, substituting \(x=0\) must satisfy the equation: \(5(0)^2+k(0)+c=0\). Hence, \(c=0\) is necessary. The coefficient \(k\) need not be zero. Exam tip: for a zero root, evaluate the polynomial at \(x=0\).

Substitute the given root \\(x=-3\\) into the equation: \\((-3)^2+u(-3)-18=0\\), which gives \\(9-3u-18=0\\). Thus, \\(-3u=9\\) and \\(u=-3\\). Therefore, option A is correct. Exam tip: When a root is given, substitute it directly into the polynomial and solve for the unknown parameter.

Since \(x^2+16x+64=(x+8)^2\), the equation becomes \((x+8)^2=0\). Hence \(x=-8\), so both roots are \((-8,-8)\). Option A results from an incorrect sign. Exam tip: when a quadratic is a perfect square, its repeated root can be found directly by setting the squared factor equal to zero.

Factoring the equation gives \(4x^2-20x=4x(x-5)=0\). Thus, \(4x=0\) gives \(x=0\), and \(x-5=0\) gives \(x=5\). Therefore, the roots are 0 and 5. Exam tip: If the product of two factors is zero, set each factor equal to zero separately.

Here, \(x^2-18x+81=(x-9)^2\). Thus, \((x-9)^2=0\) gives \(x=9\), and this root occurs twice. Therefore, the correct answer is 9. Exam tip: If a quadratic is a perfect square, set the linear expression inside the square equal to zero to find the repeated root.

The equation \(x^2-100=0\) gives \(x^2=100=10^2\). Thus, \((x-10)(x+10)=0\), so its roots are \(10\) and \(-10\). Substituting \(x=0\) gives \(-100\neq0\), so statement C is incorrect. Exam tip: check a proposed root by substituting it directly into the equation.

By the zero-product property, either \((x-4)=0\) or \((x+9)=0\). Thus, \(x=4\) or \(x=-9\), so the roots are \(4\) and \(-9\). Option B has both signs wrong; in particular, \(x+9=0\) gives \(x=-9\). In an exam, set each factor equal to zero and solve separately.

The sum of the roots is found by adding them directly: 4 + (−7) = −3. Therefore, option B is correct. The value 3 represents only the difference in magnitudes, not the signed sum. Exam tip: do not confuse the sum of the roots with their product.

For a monic quadratic equation with roots α and β, the standard form is x² − (α + β)x + αβ = 0. Here the sum α + β is 10 and the product αβ is 21. Substituting these values gives x² − 10x + 21 = 0, which is option B. The minus sign before the coefficient of x is essential: it represents the negative of the sum of the roots. Option A has the wrong sign for the sum, while options C and D interchange the sum and product or use incorrect signs. Thus option B is the unique correct equation.

Substitute the given root \(x=4\) into the equation: \(2(4)^2-9(4)+n=0\). Thus, \(32-36+n=0\), which gives \(n=4\). Exam tip: When a root is given, substitute it directly into the quadratic equation to find the unknown constant.

Since \(x^2-6x+9=(x-3)^2\), we get \((x-3)^2=0\), so \(x=3\). Therefore, both roots are 3 and 3; the equation has equal roots. In an exam, identifying the perfect square form is the quickest method.

By the factor theorem, if \(r\) is a root of a polynomial, then \((x-r)\) is its factor. Here, \(r=7\), so the required factor is \((x-7)\). The factor \((x+7)\) would correspond to the root \(-7\). Exam tip: substitute the given root directly into the form \(x-r\).