Introduction to Quadratic Equations

Rewriting the polynomial in descending powers gives \(x^2-7x+12=0\). The coefficient of \(x\) is therefore \(b=-7\). Option C (7) is a common mistake — it ignores the negative sign of the \(x\)-term. Exam tip: arrange terms in order of descending powers (\(x^2, x,\) constant) to read off coefficients directly.

Substitute \(x=5\): \(5^2-6\cdot5+5=25-30+5=0\). Since substitution yields zero, \(x=5\) is a root. Option C is incorrect because the quadratic factors as \((x-1)(x-5)=0\), so both \(x=1\) and \(x=5\) are roots. Exam tip: Verify a proposed root by direct substitution or factor the quadratic to find all roots.

Expand the left side: \((x-6)^2 = x^2 - 12x + 36\). Move the right side \(4x+5\) to the left and combine like terms: \(x^2-12x+36-(4x+5)=x^2-16x+31\). Thus the standard form is \(x^2-16x+31=0\). The closest distractor C (\(x^2-12x+41=0\)) keeps the middle term but has an incorrect constant — a typical arithmetic slip when subtracting the constant. Exam tip: always bring all terms to one side and simplify carefully, checking signs and arithmetic.

A quadratic equation in standard form is ax² + bx + c = 0. The values supplied are a = −2, b = 9 and c = −7. Substituting them in their respective positions gives (−2)x² + 9x + (−7) = 0, which simplifies to −2x² + 9x − 7 = 0. Therefore option B is correct. The coefficient a belongs to x², b belongs to x, and c is the constant term; changing their positions produces a different equation. Option A loses the negative sign of a, option C changes the sign of b, and option D swaps the roles of a and c. Keeping the sign and position of every coefficient is essential.

Use the discriminant formula \(D=b^2-4ac\). Here \(a=25,\; b=-20,\; c=4\), so \(D=(-20)^2-4\cdot25\cdot4=400-400=0\). \(D=0\) means the quadratic has two real and equal roots (a repeated root). The distractor 400 is incorrect because it is just \(b^2\) without subtracting \(4ac\). Option 100 arises from a common arithmetic mistake in computing \(4ac\), and \(-20\) comes from sign or substitution errors. Exam tip: check for a perfect square trinomial — here \(25x^2-20x+4=(5x-2)^2\), which immediately shows \(D=0\).

Compute the discriminant \(\Delta=b^2-4ac\). Here \(a=1,\; b=14,\; c=49\) so \(\Delta=14^2-4\cdot1\cdot49=196-196=0\). When \(\Delta=0\) a quadratic has two equal (repeated) real roots; using \(x=-\dfrac{b}{2a}\) gives \(x=-7\) as the repeated root. Why other options are wrong: A would require \(\Delta>0\), C would require \(\Delta<0\), and D is impossible for a quadratic (at most two roots). Exam tip: Always evaluate \(\Delta\) first — it directly tells you whether roots are distinct, equal, or complex.

The discriminant is given by \(D=b^2-4ac\). Here \(a=4,\; b=1,\; c=6\). So \(D=1^2-4\cdot4\cdot6=1-96=-95\). Since \(D<0\), the discriminant is negative; the equation has no real roots (roots are complex). The closest distractor 'zero' is wrong because that would require \(b^2=4ac\), which is not true here. Exam tip: compute \(4ac\) carefully and watch signs to avoid arithmetic errors.

Monic means the coefficient of \(x^2\) is 1. For roots \(-5\) and \(4\) the factors are \((x+5)\) and \((x-4)\). Multiplying gives \((x+5)(x-4)=x^2+x-20\), so the monic quadratic is \(x^2+x-20=0\). The closest distractor is B, which has the wrong sign on the middle term; since the sum of the roots is \(-1\), the middle coefficient must be \(-\text{(sum) }=1\), not \(-1\). Exam tip: For roots \(\alpha,\beta\) use \(x^2-(\alpha+\beta)x+\alpha\beta=0\) to write the monic equation quickly.

The sentence means "subtract 9x from x^2 and the result is 36", so it gives \(x^2-9x=36\). Moving all terms to one side yields \(x^2-9x-36=0\), which is option A. Option B is wrong because it has the wrong sign on the linear term (it treats subtraction as addition). Option C wrongly changes the coefficient of \(x^2\), and option D swaps the numeric roles of 9 and 36; neither matches the given statement. Exam tip: translate the sentence into an equation step by step, then bring all terms to one side to form a standard quadratic equal to zero.

Take the smaller integer as \(x\); the two consecutive integers are \(x\) and \(x+1\). Their squares sum to \(x^2+(x+1)^2=145\). Expanding gives \(x^2+x^2+2x+1=145\) or \(2x^2+2x+1=145\). Subtracting 145 yields \(2x^2+2x-144=0\), which is option A. The closest distractor, B, is wrong because it ignores the \(+1\) from \((x+1)^2\) and thus miscalculates the constant term. Exam tip: divide the equation by 2 to get \(x^2+x-72=0\), factor to \((x+9)(x-8)=0\) and choose the positive root \(x=8\).

The breadth is \(x\) units, so the length is \(x+7\) units. Using area, \(x(x+7)=120\). On expanding, \(x^2+7x=120\), so \(x^2+7x-120=0\). Option B incorrectly subtracts 7, whereas the length is 7 more than the breadth. Exam tip: In rectangle word problems, first express both dimensions and set their product equal to the area.

Factor the polynomial: \(x^2-10x = x(x-10)\). By the zero-product property, \(x(x-10)=0\) implies \(x=0\) or \(x-10=0\), giving roots \(0\) and \(10\). Thus (0, 10) is correct. Option B wrongly takes the second root as −10 (sign error); options C and D reflect incorrect factorisation or arithmetic. Exam tip: factor first; then set each factor equal to zero to find the roots quickly and avoid sign mistakes.

(49x^2=64), so (x^2=\frac{64}{49}) and (x=\pm\frac{8}{7}). Take both signs while taking square roots.

For the quadratic, the constant term is 30 and the coefficient of x is 11. Find two numbers whose product is 30 and whose sum is 11 — these are 5 and 6 because \(5\cdot6=30\) and \(5+6=11\). Hence the correct factorization is \((x+5)(x+6)=0\). The closest distractor \((x+3)(x+10)=0\) has the correct product but a sum of \(3+10=13\), so it does not match the middle coefficient. Exam tip: to factor quickly, match the product (constant) and the sum (middle coefficient) before expanding.

For a quadratic equation \(ax^2+bx+c=0\), \(ac\) is the product of coefficients \(a\) and \(c\). Here \(a=4\) and \(c=3\), so \(ac=4\times3=12\). Option B (13) is incorrect because 13 is the value of \(b\), not \(ac\); option C (16) would be wrong if someone mistakenly took both coefficients as 4. Exam tip: always identify \(a, b, c\) explicitly before computing any derived quantity.

Multiplying the whole equation by (5) gives (x^2+10x-35=0). To remove fractions, multiply the whole equation.

Denominators 4 and 2 have LCM = 4, so multiply the entire equation by 4: \(\frac{3}{4}x^2\times4=3x^2,\; -\frac{1}{2}x\times4=-2x,\; 2\times4=8\). Thus the integer-coefficient form is \(3x^2-2x+8=0\). Option B is incorrect because the constant term was not multiplied correctly (it should become 8). Exam tip: to remove fractional coefficients, multiply the whole equation by the LCM of all denominators — do not change individual terms differently.

(\frac{1}{x^2}=x^{-2}), which is not polynomial form. A usual quadratic equation does not have a negative power of the variable.

In the standard form \(ax^2+bx+c=0\), \(b\) is the coefficient of \(x\). In the given equation the coefficient of \(x\) is \(6\sqrt{5}\), so \(b=6\sqrt{5}\). Option B is incorrect because it has the wrong sign; option C is the constant term \(c\); option D equals the coefficient \(a\). Exam tip: always write the quadratic in standard form and read off the coefficients for \(a, b, c\).

The product of the roots is simply their multiplication: \\((-4)(-9)=36\\). The product of two negative numbers is positive, so 36 is correct. The closest distractor \\(-36\\) is wrong due to an incorrect sign; \\(13\\) and \\(-13\\) confuse product with the sum (the sum is \\(-13\\)). Exam tip: check signs first — if both roots are negative, the product is positive.