Introduction to Quadratic Equations

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 \(-3+8=5\), so \(-b=5\), giving \(b=-5\). The product is \((-3)(8)=-24\), so \(c=-24\). Therefore, \(b+c=-5-24=-29\). Exam tip: For \(x^2+bx+c=0\), use \(b=-(\text{sum of roots})\) and \(c=\text{product of roots}\).

In option A, \(x^2+7x=x(x+7)\). Hence, the roots are \(x=0\) and \(x=-7\), so the second root is negative. Option B has roots \(0\) and \(7\), and therefore does not satisfy the condition. Exam tip: For an equation of the form \(x^2+bx=0\), the roots are \(0\) and \(-b\).

Substituting \(a=2\) gives \((2-2)x^2+(2^2-4)x+5=0\), or \(0x^2+0x+5=0\), which reduces to \(5=0\). This statement is false and is satisfied by no value of \(x\), so it is a contradictory statement. It is not linear because the coefficient of \(x\) is also zero, and it is not always true. Exam tip: After substituting a parameter value, check the highest non-zero power and whether the resulting constant statement is true or false.

Expanding gives left side (2x^2-6x+29) and right side (x^2+2x+1). Subtracting gives (x^2-8x+28=0).

Since \(x=-2\) is a root, substitute it into the equation: \((-2)^2+(3k+1)(-2)+2k=0\). This gives \(4-6k-2+2k=0\), so \(2-4k=0\) and hence \(k=\frac{1}{2}\). The nearby distractor \(k=-\frac{1}{2}\) does not make the equation equal to zero. Exam tip: when a root is given, substitute it directly into the polynomial equation.

For a quadratic equation \\(ax^2+bx+c=0\\), the sum of the roots is \\( -b/a \\) and their product is \\(c/a\\). Here, \\(a=1,b=-5,c=2\\), so \\(\\alpha+\\beta=5\\) and \\(\\alpha\\beta=2\\). Therefore, \\(\\alpha+\\beta+2\\alpha\\beta=5+2(2)=9\\). Exam tip: use the sum and product of roots directly instead of solving for the individual roots.

For the monic quadratic equation (x^2+px+q=0), the product of the roots equals the constant term (q). Since the roots are (2) and (p), (q=2\times p=2p). Also, their sum gives (2+p=-p), so (p=-1) and (q=-2), confirming the relation. Exam tip: In (x^2+ax+b=0), the sum of roots is (-a) and their product is (b).

By Vieta’s formulas, \alpha+\beta=9 and \alpha\beta=20. Therefore, (\alpha-4)(\beta-4)=\alpha\beta-4(\alpha+\beta)+16=20-4(9)+16=0. Hence, option A is correct. Choosing 4 or -16 usually results from omitting the middle term -4(\alpha+\beta) during expansion. Exam tip: for (ax^2+bx+c=0), the sum of roots is -b/a and their product is c/a.

The area of a triangle is \(\frac{1}{2}\times\text{base}\times\text{height}\). Therefore, \(\frac{1}{2}(x+2)(x+6)=40\), giving \((x+2)(x+6)=80\). Expanding, \(x^2+8x+12=80\), so the required equation is \(x^2+8x-68=0\). Option B incorrectly uses 40 instead of 80 after removing the \(\frac{1}{2}\) factor. Exam tip: write the area formula first, then expand and collect all terms on one side.

For a quadratic equation \(ax^2+bx+c=0\), the sum of the roots is \(\alpha+\beta=-\frac{b}{a}\). Here, \(a=1\) and \(b=-9\), so \(\alpha+\beta=9\). Therefore, \((\alpha+eta)^2=9^2=81\). The value 14 is the product \(\alpha\beta\), not the sum of the roots. Exam tip: remember that the sum of roots is \(-b/a\), while their product is \(c/a\).

Here (\alpha+\beta=4) and (\alpha\beta=-21). Thus (\alpha\beta+2\alpha+2\beta=-21+2(4)=-13).

A quadratic equation has no real roots only when its discriminant \(D=b^2-4ac\) is negative. Here, \(a=1, b=8, c=k\), so \(D=8^2-4(1)(k)=64-4k\). Thus, \(64-4k<0\), which gives \(k>16\). Note that \(k=16\) makes the discriminant zero and produces one repeated real root, so it is not correct.

The equation can be rewritten as \((x-a)^2-16=0\). Thus, \((x-a)^2=16\), giving the roots \(x=a+4\) and \(x=a-4\). Their difference is \((a+4)-(a-4)=8\). The value 4 represents the distance of each root from \(a\), not the distance between the two roots. In the exam, completing the square is a quick method for such questions.

The sum of roots is (3+p), and in the equation the sum is (-p). Thus (3+p=-p), giving (p=-\frac{3}{2}).

Expanding the two products gives \((x-2)(x+3)=x^2+x-6\) and \((2x-1)(x-4)=2x^2-9x+4\). Combining like terms, \(x^2+2x^2=3x^2\), \(x-9x=-8x\), and \(-6+4=-2\). Hence the standard form is \(3x^2-8x-2=0\). Option B has the wrong sign for the linear term. Exam tip: expand each bracket carefully, then combine the quadratic, linear, and constant terms separately.

Here ((2x+5)(3x-4)=6x^2+7x-20) and (4(x-1)=4x-4). Bringing all terms to one side gives (6x^2+3x-16=0).

For the equation to be quadratic, (k^2-25\neq0) is required. So both (k\neq5) and (k\neq-5) are necessary.

Expand the squares: \((x+2)^2=x^2+4x+4\) and \(2(x-3)^2=2x^2-12x+18\). Thus, \(3x^2-8x+22=35\), and moving 35 to the left gives \(3x^2-8x-13=0\). Therefore, option A is correct. Exam tip: when converting to standard form, keep all terms on one side and make the other side zero; the constant becomes \(22-35=-13\).

Multiplying the whole equation by (10) gives (4x^2-2+5x+15=70). Therefore the standard form is (4x^2+5x-57=0).

A root must satisfy the equation. Substituting \(x=-3\) gives \(4(-3)^2+p(-3)-9=0\), so \(36-3p-9=0\). Hence, \(27-3p=0\) and \(p=9\). Therefore, option A is correct. Exam tip: Substitute the given root directly into the equation and solve for the unknown parameter.