What is the value of (x^2-5x+6) when (x=2)?
Substitution gives (2^2-5\cdot2+6=0). In exams do powers first, then multiplication, then addition or subtraction.
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SubjectsMathematics
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Substitution gives (2^2-5\cdot2+6=0). In exams do powers first, then multiplication, then addition or subtraction.
View question detailsIn option A, \((x+1)(x-1)+1=x^2-1+1=x^2\), whose highest power is 2, so it is quadratic. In option B, all variable terms cancel and only 1 remains. Exam tip: check the highest power after simplifying.
View question details(16=4^2), and in a perfect square the middle term is (2\cdot x\cdot4=8x). In exams double the root of the constant term.
View question details(-14x=-2\cdot x\cdot7), so (c=7^2=49). In exams take half of the middle coefficient and square it.
View question details((x+3)(x+7)=x^2+10x+21), so (p=10). In exams the middle term comes from the sum of the constants.
View question detailsIn 3x² + 5x + 8, the coefficients of x² and x are 3 and 5. Their sum is 3 + 5 = 8, which equals the constant term. In option B, 2 + 7 = 9, not 10. Exam tip: include signs while adding coefficients.
View question detailsUse the distributive property: \((x-4)(x+2)=x(x+2)-4(x+2)=x^2+2x-4x-8=x^2-2x-8\). Therefore, the correct expression is \(x^2-2x-8\). In \(x^2+6x-8\), the negative sign of \(-4x\) has been handled incorrectly. Exam tip: write all four products first, then combine like terms.
View question details(2x\cdot5=10x) and (-3\cdot x=-3x), so the middle term is (7x). In exams write all four products.
View question detailsTaking common (5) gives (5(x^2-4x+4)). In exams first take the common factor and then apply identities.
View question detailsFirst take out (3), and (x^2-9) is a difference of squares. In exams factorise the inside expression too.
View question details(4x^2=(2x)^2) and (12x=2\cdot2x\cdot3), so (k=9). In exams identify the second part using (2ab).
View question details(9x^2=(3x)^2), (25=5^2), so the middle term is (-2\cdot3x\cdot5=-30x). In exams follow the given sign.
View question detailsHere, \(9a^2=(3a)^2\) and \(16b^2=(4b)^2\). The middle term is \(-2\times3a\times4b=-24ab\), so the expression is \((3a-4b)^2\). Exam tip: always verify the middle coefficient.
View question detailsA perfect-square trinomial has the form \(a^2\pm2ab+b^2\). Here, \(9p^2=(3p)^2\) and \(16q^2=(4q)^2\), while \(-2(3p)(4q)=-24pq\). Therefore, \(9p^2-24pq+16q^2=(3p-4q)^2\), so option A is correct. In option B, the last term is \(15q^2\) instead of \(16q^2\), so it is not a perfect square. Exam tip: take the square roots of the first and last terms, then check whether the middle term equals \(\pm2ab\).
View question detailsActually ((3p-q)(2p+q)) gives (6p^2+pq-q^2). Check the cross terms carefully in exams.
View question details\(9x^2+12x+4=(3x)^2+2(3x)(2)+2^2=(3x+2)^2\), so A is a perfect-square trinomial. D is also a square form, but it has a negative middle term: \((3x-2)^2\). Exam tip: compare the middle term with \(2ab\).
View question detailsFirst (x^2+4x+4=(x+2)^2), then apply difference of squares. In exams do not stop at the partial form.
View question detailsFirst (x^2-6x+9=(x-3)^2), then use the (a^2-b^2) identity. In exams identify the perfect-square group.
View question details(y-p)(y-q)=y^2-(p+q)y+pq
. Comparing it with
y^2-7y+12
gives p+q=7 and pq=12. A sum of -7 would produce +7y. Exam tip: expand the signs once to verify the middle term.
((x-3)(x-6)=x^2-9x+18), so (b=9). In exams (b) remains positive because the expression has (-bx).
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