Some impossible-looking questions about quadratic equations which can be solved with a bit of thinking.
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", "definition": "random(0,1,2)"}, "d": {"templateType": "anything", "group": "Ungrouped variables", "name": "d", "description": "", "definition": "random(1000..20000)*random(-1,1)"}}, "statement": "These questions might look impossible, but they're not: you can find the answers without doing any long calculations.
", "variable_groups": [], "parts": [{"marks": 1, "variableReplacements": [], "minValue": "if(d>0,0,2)", "correctAnswerFraction": false, "correctAnswerStyle": "plain", "notationStyles": ["plain", "en", "si-en"], "showCorrectAnswer": true, "allowFractions": false, "mustBeReduced": false, "maxValue": "if(d>0,0,2)", "type": "numberentry", "prompt": "How many real solutions does the following equation have?
\n$\\simplify{x^2+{d}} = 0$
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\n$\\simplify{ {a}x^2+{b}x+{c} } = 0$
", "scripts": {}, "mustBeReducedPC": 0, "showFeedbackIcon": true, "variableReplacementStrategy": "originalfirst"}, {"showFeedbackIcon": true, "matrix": ["1", "1", 0, 0, 0, 0], "marks": 0, "variableReplacements": [], "displayColumns": 0, "minAnswers": 0, "showCorrectAnswer": true, "warningType": "none", "maxAnswers": 0, "prompt": "Select the numbers below which are solutions to the equation
\n$\\simplify{ x^2+{-c1-c2}x + {c1*c2}} = 0$
", "shuffleChoices": true, "displayType": "checkbox", "choices": ["$\\var{c1}$
", "$\\var{c2}$
", "$\\var{mod(c1,100)}$
", "$\\var{c1-mod(c1,10)}$
", "$\\var{c2+mod(c1,10)}$
", "$\\var{c1+c2-mod(c1,100)}$
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\n\\[ \\simplify{x^2 = {-d}} \\]
\nAny number squared is positive, so this equation has {if(d>0,'no','two')} solutions.
\nThe discriminant of a quadratic equation show how many roots the equation has.
\nLook at the quadratic formula:
\n\\[ x = \\frac{-b \\pm \\sqrt{b^2-4ac}}{2a} \\]
\nThe value of $\\sqrt{b^2-4ac}$ is called the discriminant, and the number of solutions to the equation depends on whether it is positive, negative, or zero:
\nIn this case, $a = 1$, $b = \\var{b}$ and $c = \\var{c}$.
\nSince $a = 1$, the discriminant can be written $\\sqrt{b^2-4c}$.
\nWe can tell that $b^2$ will have $\\var{2*floor(log(b,10))+if(b/10^floor(log(b,10))>=sqrt(10),2,1)}$ digits. $c = \\var{c}$ has $\\var{ceil(log(c,10))}$ digits.
\nSo, $b^2-4c$ is clearly {switch(num_roots=2,'positive',num_roots=0,'negative','zero')}, so the equation has {num_roots} real {pluralise(num_roots,'root','roots')}.
\nIn a quadratic equation $x^2+bx+c$, $-b$ is the sum of the two roots of the equation and $c$ is their product.
\nSo, we first need to find a pair of numbers which sum to $\\var{c1+c2}$. In fact, there are three pairs:
\nThe pair whose product is equal to $\\var{c1*c2}$ are the solutions to the equation. We can work out which pair it is without doing the (very long!) multiplication.
\n