What is the perfect square form of (x^2-18x+81)?
(81=9^2) and (-18x=-2\cdot x\cdot9). In exams, identify the subtraction square from the negative middle term.
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SubjectsMathematics
TOPIC PRACTICE
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(81=9^2) and (-18x=-2\cdot x\cdot9). In exams, identify the subtraction square from the negative middle term.
View question details(16x^2=(4x)^2), (25=5^2), and (40x=2\cdot4x\cdot5). In exams, check the square roots of first and last terms.
View question details(49=7^2), so the middle term is (2\cdot x\cdot7=14x). In exams, take the positive middle coefficient for a positive perfect square.
View question detailsFor \(2x^2+3x+5\), the discriminant is \(b^2-4ac=3^2-4(2)(5)=-31\). Since it is negative, the expression has no real linear factors, hence no integer-coefficient factorisation. Exam tip: check the discriminant first.
View question details(x^2-12x+36=(x-6)^2) and (20=36-16). In exams take half of the middle coefficient and form its square.
View question details\((x-5)^2=x^2-2\cdot5x+5^2=x^2-10x+25\), so the correct factorisation is \((x-5)^2\). In contrast, \((x-5)(x+5)\) gives \(x^2-25\). Exam tip: check whether the middle term is \(\pm2ab\).
View question details\(p^2-16q^2=p^2-(4q)^2=(p-4q)(p+4q)\), so its factors are conjugate binomials. C is a repeated-binomial square. Exam tip: spot two squares separated by a minus sign.
View question detailsSince \(a\ne0\), the \(x^2\) term is present, so the highest power is 2. \(b\) or \(c\) may be zero, and real roots are not guaranteed. Exam tip: check the highest power first.
View question details(-4-8=-12) and ((-4)(-8)=32). In exams, use both negative signs when the constant is positive and middle term is negative.
View question details(25x^2=(5x)^2) and (36=6^2), so the middle term is (2\cdot5x\cdot6=60x). In exams, multiply twice the two square roots.
View question detailsIn \(5x^2-3x\), the highest power of x is 2, so it is quadratic. A constant term is not compulsory. \(5x^3-3x\) is cubic because its highest power is 3. Exam tip: identify the degree from the highest exponent.
View question detailsSince (49=7^2) and the middle term is (-14x), the form is ((x-7)^2). Exam tip: pay attention to the negative sign.
View question detailsUsing \((a-b)^2=a^2-2ab+b^2\), take \(a=2x\) and \(b=3\). Then \((2x-3)^2=(2x)^2-2(2x)(3)+3^2=4x^2-12x+9\). Hence, option B is correct. Option A has an incorrect coefficient of the middle term, while option D has the wrong sign. Exam tip: the middle term in \((a-b)^2\) is always \(-2ab\).
View question detailsThe expansion is (x^2+3x-10), so the coefficient of (x) is (3). Exam tip: the sum of constants gives the middle coefficient.
View question detailsThe constant term is ((-7)\cdot3=-21). Exam tip: multiply the constant numbers to get the constant term.
View question detailsWe get (2^2+5\cdot2-6=8). Exam tip: do powers first, then multiplication, then addition and subtraction.
View question detailsFor a perfect square, \(x^2+bx+c=(x+b/2)^2\). Thus the constant term must be \(b^2/4\), giving \(b^2=4c\). Exam tip: equivalently, check whether the discriminant \(b^2-4c\) is zero.
View question detailsHalf of (-18) is (-9), and ((-9)^2=81). Exam tip: the square becomes positive even with a negative sign.
View question details\(9x^2+12x+4=(3x)^2+2(3x)(2)+2^2=(3x+2)^2\), so it is a perfect square. In option B, the constant should be \(4\), not \(2\). Exam tip: match the middle term with \(2ab\).
View question detailsOption A is \(4x^2-12x+9=(2x-3)^2=(2x-3)(2x-3)\), so it is the product of two identical binomials. Option B is the closest distractor, but \((2x-3)^2\) has constant term \(9\), not \(8\). Exam tip: for a perfect-square trinomial \(ax^2+bx+c\), check whether \(b^2=4ac\).
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