Introduction to Quadratic Equations

In the standard form \(ax^2+bx+c=0\), \(a\) is the coefficient of \(x^2\). In \(-x^2+3x+10=0\) the \(x^2\) term is \(-x^2\), so its coefficient is \(-1\). Therefore \(a=-1\). Note: 3 is the coefficient \(b\) of \(x\), 10 is the constant \(c\), and 1 is the wrong sign. Exam tip: rewrite any quadratic as \(ax^2+bx+c=0\) and compare coefficients term by term.

\(x^2+6x+9\) is a perfect-square trinomial. Comparing with \((x+a)^2=x^2+2ax+a^2\) gives \(2a=6\), so \(a=3\) and the trinomial equals \((x+3)^2\). Hence the equation factors as \((x+3)^2=0\). Option B is incorrect because \((x-3)^2\) would produce a middle term of \(-6x\); options C and D have wrong constant terms. Exam tip: identify perfect squares by checking if the constant equals the square of half the middle coefficient (i.e., check if constant = (middle/2)^2).

Factor the quadratic: \(x^2+x-6=(x+3)(x-2)\). Thus the roots are \(x=-3\) and \(x=2\). The value 2 is among the options, so it is the correct answer. Checking the nearest distractor 1 gives \(1^2+1-6=-4\), not zero, so it is incorrect. Exam tip: either factor the quadratic or substitute the option values to verify quickly.

The governing definition is the degree of a polynomial equation. The degree is the greatest exponent of the variable that has a non-zero coefficient. When the highest power of x is 2, the equation is called a quadratic equation. Its usual standard form is ax² + bx + c = 0, where a is not zero; this condition ensures that the x² term is actually present. A linear equation has degree 1, a cubic equation has degree 3, and a constant equation has degree 0 because it contains no variable term of positive degree. Therefore, an equation of degree 2 is precisely a quadratic equation, making option B correct. The other choices refer to different degrees.

A quadratic (second-degree) expression must include a nonzero \(x^2\) term. Option C, \(x^2-2=0\), clearly contains the \(x^2\) term so it is correct. The other options, \(5x+2=0\), \(7x-9=0\), and \(3x+4=7\), are linear equations whose highest power of \(x\) is 1 and therefore do not contain \(x^2\). Exam tip: To identify a quadratic quickly, check the highest power of the variable — if it is 2, the equation is quadratic.

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.