Introduction to Quadratic Equations

The equation can be written as \((4x-3k)^2=0\). Hence \(4x-3k=0\), giving the repeated root \(x=\frac{3k}{4}\). Equivalently, the discriminant is \(D=b^2-4ac=0\), which confirms that the roots are equal and real. Exam tip: for a quadratic equation, \(D=0\) indicates two equal real roots.

If one root is the reciprocal of the other, the product of roots must be (1). Here the product is (36), so it is not possible.

For equal roots, (p^2-100=0) gives (p=\pm10). For equal positive roots, (-\frac{p}{2}>0), so (p=-10).

Putting (x=2), the left side becomes (4k-14+3=4k-11). For it not to be a root, (4k-11\neq0), so (k\neq\frac{11}{4}).

The monic equation is (x^2-(\text{sum})x+\text{product}=0). Using sum (0) and product (-121) gives (x^2-121=0).

The product is (7) and the sum is \(-\frac{k}{4}\). From \(7=-2\left(-\frac{k}{4}\right)\), we get \(k=14\).

The sum of roots is (5) and the product is (-14). (\alpha^3+\beta^3=(\alpha+\beta)^3-3\alpha\beta(\alpha+\beta)=125+210=335).

For area, write \((x+3)(x-2)=40\). Expanding gives \(x^2+x-6=40\), so \(x^2+x-46=0\). Do not forget to move \(-6\) to the other side.

Substituting \(a=-2\) gives \(a+2=0\) and \(a^2-4=4-4=0\). Therefore, the expression becomes \(0x^2+0x+7=0\), or \(7=0\). Since this statement is impossible, it is contradictory. Exam tip: when the coefficients of all variable terms become zero, classify the remaining constant statement rather than calling it linear or quadratic.

Expanding gives left side (2x^2+2x+25) and right side (x^2-2x+1). Subtracting gives (x^2+4x+24=0).

Since \(x=-1\) is a root, substitute it into the equation: \((-1)^2+(4k-1)(-1)+3k=0\). This gives \(1-4k+1+3k=0\), or \(2-k=0\), so \(k=2\). Exam tip: When a root is given, substitute it directly into the polynomial to form an equation in the parameter.

By Vieta’s formulas, for \(ax^2+bx+c=0\), the sum of the roots is \(-\frac{b}{a}\) and their product is \(\frac{c}{a}\). Here, \(a=1,b=-7,c=3\), so \(\alpha+\beta=7\) and \(\alpha\beta=3\). Therefore, \(\alpha+\beta+3\alpha\beta=7+3(3)=16\). Option 7 is only the sum of the roots, while option 3 is only their product. Exam tip: Use Vieta’s formulas directly instead of solving for the roots.

The sum of roots is (4+p), and in the equation the sum is (-p). Thus (4+p=-p), giving (p=-2).

By Vieta’s formulas, \(\alpha+\beta=12\) and \(\alpha\beta=35\). Therefore, \((\alpha-5)(\beta-5)=\alpha\beta-5(\alpha+\beta)+25=35-5(12)+25=0\). Hence, option A is correct. Exam tip: For expressions involving the roots, first use the sum and product of roots; finding the roots individually is not necessary.

The area of a triangle is \(\frac{1}{2}\times\text{base}\times\text{height}\). Hence, \(\frac{1}{2}(x+3)(x+7)=55\), so \((x+3)(x+7)=110\). Expanding gives \(x^2+10x+21=110\), and therefore \(x^2+10x-89=0\). Option B results from mishandling the factor \(\frac{1}{2}\) in the area formula. Exam tip: In triangle-area problems, first multiply both sides by 2 and then expand the product.

For a quadratic equation \(ax^2+bx+c=0\), the sum of the roots is \(\alpha+\beta=-\frac{b}{a}\). Here, \(a=1\) and \(b=-13\), so \(\alpha+\beta=13\). Therefore, \((\alpha+\beta)^2=13^2=169\), making option A correct. Option B is the constant term, not the sum of the roots. Exam tip: Use Vieta’s formulas directly: the sum of roots is \(-\frac{b}{a}\) and their product is \(\frac{c}{a}\).

A quadratic equation has no real roots when its discriminant satisfies \\(D<0\\). Here, \\(D=b^2-4ac=10^2-4(1)(k)=100-4k\\). Thus, \\(100-4k<0\\), which gives \\(k>25\\). When \\(k=25\\), the discriminant is zero and the equation has one repeated real root, so that option is not correct. Exam tip: For “no real roots,” apply the condition \\(D<0\\).

The equation can be rewritten as \((x-a)^2-25=0\). Thus, \((x-a)^2=25\), giving the roots \(a+5\) and \(a-5\). Therefore, the difference between the larger and smaller roots is \((a+5)-(a-5)=10\). In such questions, completing the square is the quickest method.

Multiplying the whole equation by (6) gives (3(x+1)^2+2(x-3)^2=60). Simplifying gives the correct form (5x^2-6x-39=0).

A quadratic equation has two real and distinct roots only when its discriminant satisfies \(D>0\). Here, \(a=1\), \(b=-2(k+1)\), and \(c=k^2\), so \(D=b^2-4ac=4(k+1)^2-4k^2=4(2k+1)\). Therefore, \(4(2k+1)>0\), giving \(k>-rac{1}{2}\). At \(k=-\frac{1}{2}\), the roots are equal, so option B is not sufficient. Exam tip: use \(D>0\) for distinct real roots, \(D=0\) for equal roots, and \(D<0\) for non-real roots.