 Arts. 75 and 76 a trinomial is a perfect square when its first and last terms are perfect squares and positive, and the middle term is twice the product of the square roots of the end terms.  A School Algebra Complete - Σελίδα 84των Fletcher Durell, Edward Rutledge Robbins - 1897 - 417 σελίδεςΠλήρης προβολή - Σχετικά με αυτό το βιβλίο
 | Horatio Nelson Robinson - 1844 - 184 σελίδες
...preceive 1st. That it consists of three terms. 2d. Two of these terms, the first and last are squares. 3d. The middle term is twice the product of the square roots of the first and last. Now let us suppose that a2 is lost, and we have only x2-\-2a&'. We know these terms... | |
 | James Thomson - 1844 - 328 σελίδες
...62 = 4«e. This is evident from § 151. ; for we saw there, that when a trinomial is a square, its middle term is twice the product of the square roots of the first and third. Hence, that ax* i-bx+c may be a square, we must have to=2o5c% ; and from this, by... | |
 | James Thomson - 1845 - 328 σελίδες
...if 62 = 4ac. This is evident from § 151. ; for we saw there, that when a trinomial is a square, its middle term is twice the product of the square roots of the first and third. Hence, that ax2 + bx + c may be a square, we must have bx=i2.a*eix ; and from this,... | |
 | Charles William Hackley - 1846 - 542 σελίδες
...expression • 9a«— 48a<62+64asi< is a perfect square ; for the two extreme terms are perfect squares, and the middle term is twice the product of the square roots of the extreme terms; hence the square root of the trinomial is Or, . Sa^—Sab*. An expression such as 4a!+... | |
 | Horatio Nelson Robinson - 1846 - 276 σελίδες
...1st. That it consists of three terms; 2d. Two of its terms, the first and the third, are squares; 3d. The middle term is twice the product of the square roots of the first and last term. Now let us suppose the third term, a3, to be lost, and we have only xa-|-2ax.... | |
 | Charles William Hackley - 1846 - 544 σελίδες
...expression 9a6— 48a'&3+64a«6< is a perfect square ; for the two extreme terms are perfect squares, and the middle term is twice the product of the square roots pf the extreme terms; hence the square root of the trinomial is Or, 3a»— 8aZ>'. An expression such... | |
 | Joseph Ray - 1848 - 252 σελίδες
...preceding section. 1st. Any trinomial can be separated into two binomial factors, when the extremes are squares and positive, and the middle term is twice the product of the square roots of the extremes. See Articles 79 and 80. Thus: a'+2ab+b*=(a+b)(a+b). a'—2ab+b'=(a—b)(a—b). 2d. Any binomial,... | |
 | Horatio Nelson Robinson - 1848 - 354 σελίδες
...1st. That it consists of three terms; 2d. Two of its terms, the first and the third, are squares; 3d. The middle term is twice the product of the square roots of the first and last term. Now let us suppose the third term, a2, to be lost, and we have only x*-\-2ax.... | |
 | Joseph Ray - 1848 - 250 σελίδες
...preceding section 1st. Any trinomial can be separated into two binomial factors. when the extremes are squares and positive, and the middle term is twice the product of the square roots of the extremes. Se» Articles 79 and 80. Thus: o*+2ab+^=(a+b)(a+b). a2— 2a6+6*=(a— b)(a— 6). 2d. Any... | |
 | Joseph Ray - 1852 - 408 σελίδες
...prime factors. 1st. Any trinomial can be separated into two binomial factors, when the extremes are squares and positive, and the middle term is twice the product of the square roots of the extreme terms. The factors will be the sum or difference of the square roots of the extreme terms,... | |
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