Question

in the given equation, q is an integer constant. the given equation has no real solutions. what is the largest possible value of q?

Explanation:

Step1: Assume the quadratic equation

Assume the equation is \(x^2 + qx - 3x - 15 = 0\) (maybe a typo, likely quadratic \(ax^2+bx+c = 0\) form, let's correct to standard quadratic: suppose the equation is \(x^2+qx - 15=0\) or maybe \(x^2+(q - 3)x-15 = 0\). Wait, for a quadratic equation \(ax^2+bx + c=0\), discriminant \(D=b^2 - 4ac\). If no real solutions, \(D<0\). Let's assume the equation is \(x^2+qx - 15 = 0\) (maybe the original was \(x^2+qx - 3x - 15\) which simplifies to \(x^2+(q - 3)x-15 = 0\)). Wait, maybe the original equation is \(x^2 + qx - 15 = 0\) (quadratic in \(x\)). For quadratic \(ax^2+bx + c = 0\), \(a = 1\), \(b = q\), \(c=-15\). Discriminant \(D = b^2-4ac=q^2-4(1)(-15)=q^2 + 60\). Wait, that can't be, because \(q^2+60\) is always positive. So maybe the equation is \(x^2+qx + 15 = 0\)? Then \(D=q^2-4(1)(15)=q^2 - 60\). For no real solutions, \(q^2 - 60<0\) \(\Rightarrow q^2<60\). Since \(q\) is integer, \(q\) can be from \(-7\) to \(7\) (since \(7^2 = 49\), \(8^2=64>60\)). Wait, maybe the original equation was \(x^2+qx - 3x - 15 = 0\) which is \(x^2+(q - 3)x-15 = 0\). Then \(a = 1\), \(b=(q - 3)\), \(c=-15\). Discriminant \(D=(q - 3)^2-4(1)(-15)=(q - 3)^2 + 60\). Still positive. Hmm, maybe the equation is \(x^2+qx + 15 = 0\), then \(D = q^2 - 60\). Wait, no, if it's \(x^2+qx - 15 = 0\), \(D=q^2 + 60>0\) always. So maybe the equation is \(x^2+qx + 15 = 0\), then \(D=q^2 - 60\). For no real solutions, \(q^2 - 60<0\) \(\Rightarrow q^2<60\). So \(q\) is integer, maximum \(q\) such that \(q^2<60\). \(7^2 = 49\), \(8^2=64\). So maximum integer \(q\) is \(7\)? Wait, no, \(q^2<60\), so \(q\) can be up to \(7\) (since \(7^2=49<60\), \(8^2=64>60\)). Wait, maybe the equation is \(x^2+qx - 3x - 15 = 0\) with a different sign. Wait, maybe the original equation is \(x^2 + qx - 3x - 15 = 0\) and it's a quadratic, but maybe I misread. Alternatively, maybe the equation is \(x^2 + qx + 15 = 0\), then \(D = q^2 - 60\). So \(q^2<60\), so \(q\) max is \(7\). Wait, but let's check again. If the equation is \(x^2 + qx - 15 = 0\), discriminant is \(q^2 + 60\), which is always positive, so it always has real solutions. So maybe the equation is \(x^2 + qx + 15 = 0\), then discriminant \(q^2 - 60 < 0\) \(\Rightarrow q^2 < 60\). So integer \(q\) can be from \(-7\) to \(7\), so largest \(q\) is \(7\). Alternatively, maybe the equation is \(x^2 + qx - 3x + 15 = 0\), so \(x^2+(q - 3)x + 15 = 0\), then discriminant \((q - 3)^2 - 60 < 0\) \(\Rightarrow (q - 3)^2 < 60\) \(\Rightarrow - \sqrt{60}<q - 3<\sqrt{60}\) \(\Rightarrow 3 - \sqrt{60}<q<3+\sqrt{60}\). \(\sqrt{60}\approx7.746\), so \(3 - 7.746\approx - 4.746\), \(3 + 7.746\approx10.746\). So integer \(q\) from \(-4\) to \(10\), but discriminant \((q - 3)^2 - 60 < 0\). Wait, no, \((q - 3)^2 < 60\) \(\Rightarrow q - 3\) is between \(-7\) and \(7\) (since \(7^2=49<60\), \(8^2=64>60\)). So \(q - 3\leq7\) \(\Rightarrow q\leq10\), but \((10 - 3)^2=49<60\), \((11 - 3)^2=64>60\). Wait, \(7^2=49\), \(8^2=64\). So \(q - 3\) can be up to \(7\), so \(q=10\)? Wait, no, \((q - 3)^2<60\), so \(q - 3\) is integer, maximum \(q - 3 = 7\) (since \(7^2=49<60\), \(8^2=64>60\)), so \(q=10\). Wait, this is confusing. Maybe the original equation was \(x^2 + qx - 3x - 15 = 0\) and it's a quadratic, but maybe the user made a typo. Alternatively, maybe the equation is \(x^2 + qx + 15 = 0\), then discriminant \(q^2 - 60 < 0\), so \(q^2 < 60\), so maximum integer \(q\) is \(7\) (since \(7^2=49<60\), \(8^2=64>60\)). I think the intended equation is \(x^2 + qx + 15 = 0\), so discriminant \(D = q^2 - 60\). For no real solutions, \(D < 0\) \(\Rightarrow q^2 < 60\). The largest integer \(q\) with \(q^2 < 60\) is \(7\) (since \(7^2 = 49\), \(8^2 = 64 > 60\)).

Step2: Confirm the discriminant condition

For quadratic equation \(ax^2 + bx + c = 0\), discriminant \(D = b^2 - 4ac\). If \(D < 0\), no real solutions. Assuming the equation is \(x^2 + qx + 15 = 0\) (correcting possible sign error), \(a = 1\), \(b = q\), \(c = 15\). Then \(D = q^2 - 4(1)(15) = q^2 - 60\). Set \(D < 0\): \(q^2 - 60 < 0\) \(\Rightarrow q^2 < 60\).

Step3: Find the largest integer \(q\)

We need the largest integer \(q\) such that \(q^2 < 60\). Testing integers: \(7^2 = 49 < 60\), \(8^2 = 64 > 60\). So the largest integer \(q\) is \(7\).

Answer:

\(7\)