Introduction to Quadratic Equations

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 direct answer is A: \(k=6\). For \(2x^2+kx+18=0\), the sum of roots is \(-k/2\), and the product is \(18/2=9\). The statement says product equals -3 times the sum, so \(9=-3(-k/2)=3k/2\). Multiplying by 2 gives \(18=3k\), hence \(k=6\). Option A is correct. Option B, \(-6\), makes the sum 3, so -3 times it is -9, not 9. Option C, 12, gives product relation 18, not 9. Option D, -12, gives the wrong sign and magnitude. The key is to use Vieta’s formulas with signs carefully. Memory cue: sum is \(-b/a\), product is \(c/a\).

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.

For a quadratic equation \(ax^2+bx+c=0\), the sum of the roots is \(-\frac{b}{a}\) and their product is \(\frac{c}{a}\). Here, \(a=1, b=-4, c=1\), so \(\alpha+\beta=4\) and \(\alpha\beta=1\). Therefore, \(\alpha+\beta+\alpha\beta=4+1=5\). Exam tip: Use the sum and product of roots directly instead of solving the quadratic equation.

For the quadratic equation \\(x^2+px+q=0\\), the product of the roots equals the constant term \\(q\\). Since the roots are 1 and \\(p\\), \\(q=1\cdot p=p\\). The sum condition also gives \\(1+p=-p\\), so \\(p=-\\frac{1}{2}\\); nevertheless, the required relation remains \\(q=p\\). Exam tip: In \\(x^2+bx+c=0\\), the product of the roots is always \\(c\\).

Bringing \(x(x-3)=4\) to one side gives \(x^2-3x-4=0\), where the coefficient of \(x^2\) is 1. Hence it is quadratic. Option D is an identity because both sides are identical. Exam tip: check that the highest power is 2.

Direct answer: Option A, \\(2x^2+3x-5=0\\). For \\(ax^2+bx+c=0\\), the product of roots is \\(c/a\\). We only need the sign. Option A has a=2 and c=-5, so the product is \\(-5/2\\), which is negative. Option B has c=5 and a=2, giving \\(5/2>0\\), so it is not negative. Option C also has a=2 and c=5, so its product is again positive; the changed sign of the x term does not affect the product. Option D has c=0, so its product is 0, not negative. Thus A is the only correct choice. Do not confuse the sign of b with the sign of the product: the product depends on c/a.

The area of a right triangle is \(\frac{1}{2}\times\text{base}\times\text{height}\). Hence, \(\frac{1}{2}(x+1)(x+3)=30\), giving \((x+1)(x+3)=60\). Expanding, \(x^2+4x+3=60\), so the quadratic equation is \(x^2+4x-57=0\). Option D results from mishandling the factor \(\frac{1}{2}\). Exam tip: For triangle-area problems, include the factor \(\frac{1}{2}\) before expanding the algebraic expression.

By Vieta’s formula, the sum of the roots of \(ax^2+bx+c=0\) is \(\alpha+\beta=-\frac{b}{a}\). Here, \(a=1\) and \(b=2\), so \(\alpha+\beta=-2\). Therefore, \((\alpha+\beta)^2=(-2)^2=4\). Exam tip: square the complete sum, including its sign; the negative sign disappears only after squaring.

A quadratic equation has no real roots when its discriminant \(D=b^2-4ac\) is negative. Here, \(a=1, b=4, c=k\), so \(D=4^2-4(1)(k)=16-4k\). Thus, \(16-4k<0\), which gives \(k>4\). At \(k=4\), the discriminant is zero and the equation has one repeated real root, so that option is incorrect. Exam tip: for ‘no real roots,’ always use the condition \(D<0\).

The equation can be rewritten as \((x-a)^2-9=0\). Thus, \((x-a)^2=9\), giving the roots \(x=a+3\) and \(x=a-3\). The difference between the larger and smaller roots is \((a+3)-(a-3)=6\). Option D represents only one root, while option C is not generally the difference between the roots. Exam tip: For this form, complete the square and identify the two roots directly.

Here ((6x+1)(x-4)=6x^2-23x-4), and subtracting (5x) gives (6x^2-28x-4=0). First expand and then bring all terms to one side.

Expand the squares: \\( (2x-3)^2=4x^2-12x+9 \\) and \\( (x+5)^2=x^2+10x+25 \\). This gives \\(5x^2-2x+34=34\\); subtracting 34 from both sides yields \\(5x^2-2x=0\\). This matches the standard quadratic form \\(ax^2+bx+c=0\\), with \\(c=0\\). Exam tip: after expansion, bring all terms to one side and equate the result to zero.

Multiplying the whole equation by (6) gives (2x^2+2-3x+6=30). Therefore the standard form is (2x^2-3x-22=0).

A given root must make the equation equal to zero. Substituting (x=-4) gives (2(-4)^2+p(-4)-12=0), or (32-4p-12=0). Thus (20-4p=0), so (p=5). Option B results from a sign error and gives (p=-5). Exam tip: Substitute the given root directly into the quadratic equation to find the unknown parameter.