Introduction to Quadratic Equations

A linear term is the term where the variable x has exponent 1. The terms in the equation are \(2x^2\) (degree 2 — quadratic), \(0x\) (degree 1 — linear), and \(-7\) (constant). So the linear term is \(0x\); its coefficient is zero so it contributes nothing numerically, which is why no visible linear part appears. Why the closest distractor is wrong: \(2x\) would be linear, but it is not present in this equation. Exam tip: identify the power of x to find linear terms — look for exponent 1, even if the coefficient is 0.

Expanding \((x+1)^2\) gives \((x+1)^2 = x^2+2x+1\), so \((x+1)^2=0\) is equivalent to the given equation. The closest distractor \((x-1)^2=0\) expands to \(x^2-2x+1\), which differs in the sign of the linear term. Exam tip: check if the quadratic is a perfect square by seeing if the middle coefficient equals twice some number (here 2 = 2·1), then write it as \((x±1)^2\) to solve quickly.

\(x^2-16\) is a difference of squares since \(16=4^2\). Use the identity \(a^2-b^2=(a-b)(a+b)\) with \(a=x\) and \(b=4\) to get \((x-4)(x+4)=0\). The other choices give different middle terms when expanded; for example \((x-4)^2\) yields a \(-8x\) term, so it does not match the original expression. Exam tip: spot the difference-of-squares pattern or expand candidate factors to check coefficients quickly.

Substitute \(x=1\): left-hand side is \(1^2+2\cdot1-3=1+2-3=0\). Hence the left-hand side equals \(0\), so option A is correct. The closest distractor (option B = \(1\)) is wrong because the arithmetic gives \(0\), not \(1\). Exam tip: when substituting, evaluate powers first, then multiplications, then additions/subtractions step by step to avoid simple mistakes.

Put the equation in standard form \(ax^2+bx+c=0\): \(7x^2+0x-14=0\). Thus \(a=7,\ b=0,\ c=-14\). Options B and D are wrong because they swap the positions of b and c (or a and c), which changes the coefficients. Exam tip: explicitly write any missing term as \(0x\) to identify a, b and c correctly.

In option (A) the highest power of x is 2 (the term x^2), so the equation is quadratic in x. y does not appear, so it is not quadratic in y (degree in y = 0). Option (C) might misleadingly look correct because it contains x^2, but it also contains y^2, so it is quadratic in both x and y and hence not the required answer. Option (D) is linear in x. Exam tip: Determine the degree with respect to a specific variable by checking the highest exponent of that variable; an absent variable has degree 0 and is not quadratic.

Distribute: \(3x(x-2)=3x\cdot x-3x\cdot 2=3x^2-6x\). Thus the standard form is \(3x^2-6x=0\). The closest distractor \(3x^2+6x=0\) simply has the sign wrong. Exam tip: expand using the distributive property and collect terms into \(ax^2+bx+c=0\); include \(+0\) if the constant term is zero to make the form explicit.

Compute: \(p(1)=1^2-1=0\). So the value is 0. When \(p(a)=0\), the number \(a\) is called a zero (root) of the polynomial. Choice C (−1) is a common slip — students often mishandle the subtraction; choices B (1) and D (2) result from simple arithmetic errors. Exam tip: substitute values carefully, evaluate powers before subtraction, and recheck your arithmetic.

The constant term is the term that does not contain \(x\) (equivalently the coefficient of \(x^0\)). In \(2x^2+3x+1=0\), only 1 has no \(x\), so 1 is the constant term. \(2x^2\) is the quadratic term and \(3x\) is the linear term — both contain \(x\). Option 0 is incorrect because 0 is not a term present in the polynomial. Exam tip: look for the term with \(x\) raised to the power 0 to find the constant term.

Write a quadratic in standard form \(ax^2+bx+c=0\). The given equation is \(x^2+10x+0=0\), so \(a=1, b=10, c=0\). The constant term \(c\) is zero and therefore absent. The quadratic term \(x^2\) and linear term \(10x\) are present, so options A and B are incorrect. Exam tip: always rewrite the polynomial as \(ax^2+bx+c=0\) and read off \(a,b,c\) to spot any missing term quickly.

A pure quadratic has only the \(x^2\) term and a constant term, with the coefficient of the linear \(x\) term equal to zero. Option A, \(x^2-9=0\), contains \(x^2\) and a constant and has no \(x\) term, so it is a pure quadratic. Option B contains a \(3x\) term (linear), so it is not pure; option C is linear and option D is cubic. Exam tip: check the coefficient of \(x\); for a pure quadratic it must be zero.

Substituting gives: \((-2)^2+4(-2)+4 = 4-8+4 = 0\). The left-hand side equals 0, so the equation is satisfied and \(x=-2\) is a solution. Option B is incorrect because the equation is not false; options C and D are wrong since the left side is not 4 or -4 but 0. Exam tip: When squaring negatives, remember \((-a)^2= a^2\); evaluate each term carefully when substituting.

Substitute the given values into the standard form: with a=2, b=-5, c=3 we get \(2x^2-5x+3=0\), so option A is correct. Option B is incorrect because it has b = +5 (wrong sign). Option C swaps the values of b and c (b would be -3 there), and option D changes the positions of the coefficients (a would be 5, not 2). Exam tip: always match each coefficient with its position (a with \(x^2\), b with \(x\), c constant) and check the signs carefully.

The standard form is \(ax^2+bx+c=0\). Substituting gives \(1\cdot x^2+0\cdot x+(-25)=0\), which simplifies to \(x^2-25=0\). Option B has the wrong sign for the constant term (\(+25\) instead of \(-25\)), option C wrongly assigns \(b=25\) and changes \(c\), and option D changes/scales the coefficients. Exam tip: Plug values into \(ax^2+bx+c=0\) and simplify term-by-term; always check signs.

The governing algebraic rule is that a term moved from one side of an equation to the other changes its sign. Start with the target standard form x² − 7x + 10 = 0. Move +10 from the left side to the right side; it becomes −10, giving x² − 7x = −10. Reversing this operation confirms the result: moving −10 back to the left changes it to +10 and produces x² − 7x + 10 = 0. Option B would give x² − 7x − 10 = 0 after rearrangement, so its constant sign is wrong. Options C and D have +7x rather than −7x, so their linear terms are also incorrect. Hence option A is correct.

A quadratic equation can have at most (2) real solutions. In easy questions, degree indicates the maximum possible solutions.

Match the equation to the standard form \(ax^2+bx+c=0\): here \(a=4\), \(b=4\), \(c=1\). So \(a+b+c=4+4+1=9\). Option B (8) arises from the common mistake of omitting the constant term \(c\) and adding only \(4+4=8\); therefore 8 is incorrect. Exam tip: always write the quadratic in the form \(ax^2+bx+c=0\) first, then read off and add the coefficients carefully.

From the standard form \(ax^2+bx+c=0\) we have \(a=2\), \(b=-3\), \(c=-5\). Substituting gives \(a-b+c=2-(-3)+(-5)=2+3-5=0\). Common mistake: treating \(b\) as +3 yields \(2-3-5=-6\), which is incorrect. Exam tip: always write the signs of coefficients explicitly before substituting to avoid sign errors.

The left-hand side is the polynomial \(5x^2+2x-1\). The highest power of x present is 2 (from the term \(x^2\)), so the degree of the equation is 2 — it is a quadratic equation. Option B (degree 1) would indicate a linear equation and is therefore incorrect; option A (0) refers to a constant polynomial. Exam tip: simplify the expression first if needed and then take the largest exponent of the variable (ignore terms with zero coefficients).

Expand the factors in A: \((x-5)(x+3)=x^2+3x-5x-15=x^2-2x-15\), so A matches the given quadratic. A common distractor B expands to \((x+5)(x-3)=x^2+2x-15\) which has +2x (not -2x), so it is incorrect. Exam tip: find two numbers whose product is -15 and sum is -2 (here -5 and 3) to factor quickly.

Area = length × width. With length \(x+3\) and width \(x\), the equation is \(x(x+3)=10\), which rearranges to the quadratic \(x^2+3x-10=0\) for solving. Option C (\(2x+3=10\)) might come from a mistaken linear relation (not the area formula); options B and D do not represent the product of length and width. Exam tip: Always form the product first, then bring all terms to one side to convert to standard quadratic form before solving.

If the smaller integer is \(x\), the next consecutive integer is \(x+1\). Since the relationship given is product, the correct equation is \(x(x+1)=30\). Option B uses a sum \(x+x+1=30\) instead of a product, so it is incorrect. Option C represents \(x\) being the larger integer (product \(x(x-1)\)), which contradicts the statement that \(x\) is the smaller one. Option D, \(2x=30\), models two equal integers (\(x+x\)), not consecutive integers. Exam tip: carefully note whether the problem states "product" or "sum" and which integer (smaller/larger) is denoted by the variable.

The correct standard form is \(x^2 - 5x + 6 = 0\). To convert, bring all terms to one side and set the equation equal to zero; moving \(-6\) from the right to the left changes its sign to \(+6\). Option (B) shows the common mistake where the sign of \(-6\) was not changed, so it is the wrong conversion and therefore the required 'common mistake'. The closest distractor (D) simply leaves the equation unchanged and is not in standard form; (C) unnecessarily writes \(0x\), which is not the standard conversion. Exam tip: always move all terms to one side, flip signs when moving terms, combine like terms and set equal to zero.

Putting (x=0) in option (A) gives (0=0). Direct substitution is the simplest way to check a solution.

The governing concept is comparison with the standard quadratic form ax² + bx + c = 0. In this form, a is the coefficient of x², b is the coefficient of x, and c is the constant term. In the given equation, the middle term is kx. Since the coefficient is the factor multiplying the variable, k is the coefficient of x in kx. The coefficient of x² is 1, while 12 is the constant term because it contains no variable. The zero on the right is the value to which the expression is equated, not the term represented by k. Therefore, k is the coefficient of the x term, so option B is correct. The other choices confuse the roles of a, b, and c.