The quadratic equation in the image is:<\/p>\n\n\n\n
$$y^2 – 6y – 8 = 0$$<\/p>\n\n\n\n
To factorise a quadratic of the form $y^2 + by + c = 0$ into $(y + p)(y + q) = 0$:<\/p>\n\n\n\n
The integer factor pairs of $-8$ are:<\/p>\n\n\n\n
None of these pairs add up to $-6$. Therefore, this expression cannot be factorised using whole numbers (integers)<\/strong>.<\/p>\n\n\n\n To factorise over real numbers, rewrite the quadratic into vertex form first:<\/p>\n\n\n\n Set each factor equal to zero:<\/p>\n\n\n\n $$y – 3 – \\sqrt{17} = 0 \\implies y = 3 + \\sqrt{17}$$<\/p>\n\n\n\n $$y – 3 + \\sqrt{17} = 0 \\implies y = 3 – \\sqrt{17}$$<\/p>\n\n\n\n Final Solutions:<\/strong><\/p>\n\n\n\n $$y = 3 \\pm \\sqrt{17}$$<\/p>\n\n\n\n <\/p>\n","protected":false},"excerpt":{"rendered":" The quadratic equation in the image is: $$y^2 – 6y – 8 = 0$$ Step 1: Check for simple integer factors To factorise a quadratic of the form $y^2 + by + c = 0$ into $(y + p)(y + q) = 0$: The integer factor pairs of $-8$ are: None of these pairs add […]<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":""},"class_list":["post-421","page","type-page","status-publish","hentry"],"_links":{"self":[{"href":"https:\/\/sebstack.com\/index.php\/wp-json\/wp\/v2\/pages\/421","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/sebstack.com\/index.php\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/sebstack.com\/index.php\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/sebstack.com\/index.php\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/sebstack.com\/index.php\/wp-json\/wp\/v2\/comments?post=421"}],"version-history":[{"count":5,"href":"https:\/\/sebstack.com\/index.php\/wp-json\/wp\/v2\/pages\/421\/revisions"}],"predecessor-version":[{"id":429,"href":"https:\/\/sebstack.com\/index.php\/wp-json\/wp\/v2\/pages\/421\/revisions\/429"}],"wp:attachment":[{"href":"https:\/\/sebstack.com\/index.php\/wp-json\/wp\/v2\/media?parent=421"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Step 2: Factorise by Completing the Square<\/h3>\n\n\n\n
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$$\\text{Half of } -6 = -3 \\implies (-3)^2 = 9$$<\/li>\n\n\n\n
$$y^2 – 6y + 9 – 9 – 8 = 0$$<\/li>\n\n\n\n
$$(y – 3)^2 – 17 = 0$$<\/li>\n\n\n\n
$$(y – 3)^2 – (\\sqrt{17})^2 = 0$$$$\\bigl(y – 3 – \\sqrt{17}\\bigr)\\bigl(y – 3 + \\sqrt{17}\\bigr) = 0$$<\/li>\n<\/ol>\n\n\n\nStep 3: Solve for $y$<\/h3>\n\n\n
Multivariable Function \/ System<\/strong><\/td> Yielded Quadratic Equation(s)<\/strong><\/td> Source Context<\/strong><\/td><\/tr><\/thead> $f(x, y) = x^3 + 3x^2 – 3xy^2 + 18xy – 24x + 1$
<\/sup><\/td>$y^2 – 6y + 8 = 0$
$x^2 + 2x + 1 = 0$<\/td>MST224 2014 Exam, Q7<\/sup><\/td><\/tr> $f(x, y) = 2x^2 + xy^2 – 6xy + 5x$
<\/sup><\/td>$y^2 – 6y + 5 = 0$<\/td> MST224 2015 Exam, Q21<\/sup><\/td><\/tr> $f(x, y) = 2x^2 + xy^2 + 6xy + 5x + 2$
<\/sup><\/td>$y^2 + 6y + 5 = 0$<\/td> MST224 2016 Exam, Q15<\/sup><\/td><\/tr> $f(x, y) = x^2 + xy^2 – 6xy + 5x$
<\/sup><\/td>$y^2 – 6y + 5 = 0$<\/td> MST224 2018 Exam, Q21<\/sup><\/td><\/tr> $f(x, y) = 2x^3 – 9x^2 – 3xy^2 + 12xy$
<\/sup><\/td>$y^2 – 4y = 0$
$x^2 – 3x + 2 = 0$<\/td>MST224 2019 Exam, Q7<\/sup><\/td><\/tr> $\\begin{cases} f_x = 4x + y^2 – 6y + 5 = 0 \\\\ f_y = 2xy – 6x = 0 \\end{cases}$
<\/sup><\/td>$y^2 – 6y + 5 = 0$<\/td> MST224 2022 Exam, Q30<\/sup><\/td><\/tr> $\\begin{cases} f_x = 3x^2 – 12x – 3y^2 + 12 = 0 \\\\ f_y = -6xy + 6y = 0 \\end{cases}$
<\/sup><\/td>$x^2 – 4x + 4 = 0$
$y^2 – 1 = 0$<\/td>MST224 2023 Exam, Q12<\/sup><\/td><\/tr> $\\begin{cases} f_x = 36x + y^2 – 4y – 32 = 0 \\\\ f_y = 2xy – 4x = 0 \\end{cases}$
<\/sup><\/td>$y^2 – 4y – 32 = 0$<\/td> MST224 2023 Exam, Q30<\/sup><\/td><\/tr> $\\begin{cases} \\dot{x} = -4y + 2xy – 8 = 0 \\\\ \\dot{y} = 4y^2 – x^2 = 0 \\end{cases}$
<\/sup><\/td>$y^2 – y – 2 = 0$
$y^2 + y + 2 = 0$
<\/sup><\/td>Book 4 (Unit 13), Example 6<\/sup><\/td><\/tr> $\\begin{cases} \\dot{x} = (1 + x – 2y)x = 0 \\\\ \\dot{y} = (x – 1)y = 0 \\end{cases}$
<\/sup><\/td>$x^2 + x = 0$
<\/sup><\/td>Book 4 (Unit 13), Exercise 25<\/sup><\/td><\/tr> $\\begin{cases} \\dot{x} = x(20 – y) = 0 \\\\ \\dot{y} = y(10 – y)(10 – x) = 0 \\end{cases}$
<\/sup><\/td>$y^2 – 10y = 0$
<\/sup><\/td>Book 4 (Unit 13), Exercise 9<\/sup><\/td><\/tr> $\\begin{cases} \\dot{x} = 0.5x – 0.00005x^2 = 0 \\\\ \\dot{y} = -0.1y + 0.0004xy – 0.01y^2 = 0 \\end{cases}$
<\/sup><\/td>$0.01y^2 + 0.1y = 0$
$0.01y^2 – 3.9y = 0$
<\/sup><\/td>Book 4 (Unit 13), Exercise 14<\/sup><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n