Introduction to Quadratic Equations

The highest power of (y) is (2), so it is quadratic in (y). The variable may be different, but the degree must be (2).

In the standard form \(ax^2+bx+c=0\), the coefficient of \(x^2\) is \(a\). The coefficient is the number multiplying that power of \(x\); if \(a=0\) the equation would not be quadratic. Option A (\(b\)) is incorrect because \(b\) is the coefficient of \(x\); option C (\(c\)) is the constant term; option D (\(x\)) is the variable itself. Exam tip: always check that the leading coefficient \(a\) is nonzero when identifying a quadratic equation.

In (x^2+4=0), the coefficient of (x^2) is (1), so it is quadratic. Having real roots is not a condition for being quadratic.

A quadratic equation has highest power 2, so its standard form is written as \\(ax^2+bx+c=0\\), where \\(a\\ne0\\). This form can contain three kinds of terms: the quadratic term \\(ax^2\\), the linear term \\(bx\\), and the constant term \\(c\\). If a coefficient is zero, one of these terms may disappear in a particular equation, but the maximum possible number of different terms remains three.

For example, \\(2x^2+5x+7=0\\) contains all three terms. The symbols \\(a\\), \\(b\\), and \\(c\\) are coefficients or constants attached to these terms, not extra terms themselves. Hence the maximum is 3, which is option C. Option 2 would apply only when one term is absent, while option 4 incorrectly counts something that is not a separate algebraic term.

In (2x^2=0), the coefficient of (x^2) is (2\neq 0). It can be quadratic even without linear and constant terms.

The area is (x(x+3)=10), which gives (x^2+3x-10=0). In geometry questions, form the equation using the formula.

Multiplying gives (2x^2+8x-3x-12=0), so (2x^2+5x-12=0). Adding middle terms correctly is important in exams.

For a quadratic equation, the coefficient of (x^2) must not be (0). Thus (k+1\neq 0), so (k\neq -1).

In standard form \(ax^2+bx+c=0\), we have \(a=3,\; b=-2,\; c=7\). Thus \(a+b+c=3+(-2)+7=8\). A common mistake is to drop the negative sign and take \(b=+2\), giving the incorrect sum \(3+2+7=12\) (option B). Exam tip: Always write down coefficients with their signs from the standard form before calculating sums.

Bringing all terms to one side gives (-2x^2+5x-9=0), and multiplying by (-1) gives (2x^2-5x+9=0). Change signs carefully in standard form.

Check a candidate root by substituting it into the expression. Putting \(x=3\) gives \(3^2-4\cdot3+3=9-12+3=0\), so \(x=3\) satisfies the equation and is a root. Factoring gives \(x^2-4x+3=(x-1)(x-3)\), showing the roots are \(x=1\) and \(x=3\); thus option C (only \(x=1\)) is incorrect. Exam tip: verify roots by direct substitution or by factoring the quadratic to list all roots quickly.

Substitute the given root into the equation: \((-2)^2 + p(-2) - 6 = 0\) ⇒ \(4 - 2p - 6 = 0\) ⇒ \(-2 - 2p = 0\) ⇒ \(-2p = 2\) ⇒ \(p = -1\). A common mistake is to drop the negative sign on the \(px\) term, which would incorrectly give \(p=1\) (option A). Options C and D do not satisfy the equation when substituted. Exam tip: always substitute the root and simplify carefully, paying attention to signs.

(x^2=5x-4) can be written as (x^2-5x+4=0), so it is quadratic. In standard form, the right side must be (0).

((x-5)^2=x^2-10x+25), so (x^2-10x+25-16=0). Do not forget the middle term while using square identities.

Using the standard form \(ax^2+bx+c=0\) and substituting \(a=2, b=-7, c=3\) yields \(2x^2-7x+3=0\). Option A is wrong because it uses +7 instead of −7 (wrong sign for b). Option C swaps coefficients (a and c are interchanged). Option D alters the coefficients arbitrarily. Exam tip: write \(ax^2+bx+c=0\) first and then substitute the given a, b, c to avoid sign errors.

The discriminant tells us the nature of the roots of a quadratic equation \\(ax^2+bx+c=0\\). It is calculated as \\(D=b^2-4ac\\). If \\(D>0\\), the equation has two real and unequal roots. If \\(D=0\\), the roots are real and equal, and if \\(D<0\\), there are no real roots. Here \\(D=25\\), and 25 is positive.

Therefore, the equation has two distinct real roots, so option A is correct. The actual values of the roots are not needed because the question asks only about their nature. A positive discriminant does not reduce the degree of the equation; the equation remains quadratic as long as the coefficient of \\(x^2\\) is nonzero. Thus options B, C, and D do not follow from \\(D=25\\).

The discriminant is defined by \(D=b^2-4ac\). Here \(a=2,\; b=3,\; c=5\). So \(D=3^2-4\cdot2\cdot5=9-40=-31\). Since \(D<0\), the discriminant is negative and the equation has no real roots (it has complex conjugate roots). Option B (zero) is wrong because \(D=0\) would give equal real roots; option A (positive) is wrong because a positive \(D\) would give two distinct real roots. Exam tip: compute the \(4ac\) term first and track signs carefully to avoid arithmetic errors.

The factors are ((x+2)) and ((x-5)), giving (x^2-3x-10=0). Make factors with signs opposite to the roots.

Let the unknown number be \(x\). Its square is \(x^2\). The statement says that the number plus its square equals 42, so the direct equation is \(x+x^2=42\). Rearranging all terms to one side gives \(x^2+x-42=0\), which is option A.

The order of the first two terms does not matter because addition is commutative, but none of the other options represents the stated sum correctly. Option B changes the sign of the number, option C doubles the square instead of adding the number, and option D uses 42 as a coefficient of \(x\). The correct method is to translate each phrase carefully into an algebraic expression before rearranging it. Therefore the supplied answer is correct.

Let the smaller integer be \(x\); the next consecutive integer is \(x+1\). Their product gives \(x(x+1)=56\). Expanding yields \(x^2+x-56=0\), so option A is correct. Distractor B (\(x^2-x-56=0\)) corresponds to integers \(x\) and \(x-1\), not the given consecutive pair. Option C (\(x^2+2x-56=0\)) would arise from \(x\) and \(x+2\), and D is a linear equation unrelated to the product of two integers. Exam tip: after forming the equation, check factor pairs of 56 (e.g., 7 and 8) to verify your result quickly.