Introduction to Quadratic Equations

Factoring the equation as \((3x-4)(x-2)=0\) gives the roots \(x=\frac{4}{3}\) and \(x=2\). Therefore, their positive difference is \(2-\frac{4}{3}=\frac{2}{3}\). In an exam, factoring first and subtracting the smaller root from the larger one is the quickest method.

For the quadratic expression to be a perfect square, \(x^2+6x+c\) must equal \((x+3)^2\). Since \((x+3)^2=x^2+6x+9\), we get \(c=9\), and the equation becomes \((x+3)^2=0\). In the exam, identify the pattern \(x^2+2ax+a^2=(x+a)^2\).

The sum of the two roots is \((-5)+(-5)=-10\). For \(x^2+bx+25=0\), the sum of the roots is \(-b\). Hence, \(-b=-10\), giving \(b=10\). Exam tip: for \(x^2+bx+c=0\), remember that the sum of the roots is \(-b\) and their product is \(c\).

For \(ax^2+bx+c=0\), the sum of roots is \(-\frac ba\), and their product is \(\frac ca\). In option A, the equation is \(x^2-5x+6=0\), so \(a=1\), \(b=-5\), and \(c=6\). The sum is \(-(-5)/1=5\), and the product is \(6/1=6\). Both values are positive.

For comparison, option B has a negative root sum, while options C and D have negative root products because their constant terms are negative. Therefore only option A satisfies both requested conditions. The actual roots are 2 and 3, which also confirms a positive sum and positive product. Thus the supplied answer and explanation are correct.

The area of a rectangle is length × breadth, so ((x+5)(x-2)=84). Expanding the product gives (x^2+3x-10=84). Bringing 84 to the left side results in (x^2+3x-94=0), so option A is correct. Option B incorrectly changes the constant term. Exam tip: expand the binomials carefully and then rearrange the equation into the standard form equal to zero.

If the sum of roots is (11) and product is (30), the equation is (x^2-11x+30=0). Remember the monic form formula.

(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}). Here the value is (\frac{5}{6}).

In option A, x² − 12x + 36 = (x − 6)², so the repeated root is x = 6, which is positive. Option B has equal roots of −6, option C has two distinct roots, 6 and −6, and option D has no real roots. Exam tip: equal roots require a discriminant of zero, and the repeated root must be positive.

Expanding gives \\(x+a)^2=x^2+2ax+a^2\\). Comparing the coefficients of \\(x\\) on both sides, \\(2a=10\\), so \\(a=5\\). Option B is incorrect because \\(a=-5\\) would make the coefficient of \\(x\\) equal to \\(-10\\). Exam tip: In an identity, compare the coefficients of like powers of the variable.

The equation can be written as \(4x^2-4kx+k^2=(2x-k)^2=0\). Hence \(2x-k=0\), giving the repeated root \(x=\frac{k}{2}\). Therefore, the roots are equal and real. Exam tip: use the discriminant \(D=b^2-4ac\); here \(D=0\), so the roots are equal. Thus option B is incorrect because the roots are not distinct.

If one root is the reciprocal of the other, the product of roots must be (1). Here the product is (16), so it is not possible.

For equal roots, (p^2-36=0) gives (p=\pm6). For equal negative roots, (-\frac{p}{2}<0), so (p=6).

The monic equation is (x^2-(\text{sum})x+\text{product}=0). Using sum (0) and product (-49) gives (x^2-49=0).

The product is (9) and the sum is \(-\frac{k}{2}\). From \(9=-3\left(-\frac{k}{2}\right)\), we get \(k=6\).

The sum of roots is (2) and the product is (-3). (\alpha^3+\beta^3=(\alpha+\beta)^3-3\alpha\beta(\alpha+\beta)=8+18=26).

For the monic quadratic equation \\(x^2+bx+c=0\\), the sum of the roots is \\(-b\\) and their product is \\(c\\). Here, the sum is \\(-2+7=5\\), so \\(-b=5\\), giving \\(b=-5\\). The product is \\((-2)\\times7=-14\\), so \\(c=-14\\). Therefore, \\(b+c=-5-14=-19\\). Exam tip: For \\(ax^2+bx+c=0\\), use the root sum \\(-b/a\\) and root product \\(c/a\\).

In option A, \(x^2-6x=x(x-6)\), so the roots are \(x=0\) and \(x=6\). Since \(6\) is positive, this option is correct. Option B has roots \(0\) and \(-6\), so its other root is negative; option C has no real roots, and option D does not have zero as a root. A quick exam check is that an equation with \(x=0\) as a root must have a zero constant term.

Substituting \\(a=-1\\) gives \\(a+1=0\\) and \\(a^2-1=1-1=0\\). Hence the equation becomes \\(0x^2+0x+2=0\\), or \\(2=0\\). This is a false, contradictory statement and has no solution. It is not a linear equation because the coefficient of \\(x\\) is also zero. In parameter-based equations, first check the coefficients of \\(x^2\\), \\(x\\), and the constant term separately.

Expanding gives left side (2x^2-6x+5) and right side (x^2-6x+9). Subtracting gives (x^2-4=0).

A root must make the left-hand side of the equation equal to zero. Substituting \(x=1\) gives \(1+(2k-1)+k=0\), so \(3k=0\) and hence \(k=0\). The values \(\pm\frac{1}{3}\) result from incorrect simplification. Exam tip: substitute the given root carefully and collect the parameter terms before solving.