If the pair of lines $6x^2-pxy-3y^2-24x+3y+q=0$ intersect on x- axis then p is equal to :- $a)3/2$ $b)-5/2$ $c)-18$ $d)-7$

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## Comments

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TopNewestFor to intersect this pair of straight line on x-axis , Let They intersect at point P ($\alpha \, 0$ )

So putting y=0 in given equation

$6\alpha ^{ 2 }\quad -\quad 24\alpha \quad +\quad q\quad =\quad 0$.

Now since Intersection point is unique so value of $\alpha$ is also unique. So Discriminant of this quadratic equation is zero !!

$q\quad =\quad 24\\ \\ \alpha \quad =\quad 2$.

Therefore Given equation representing pair of straight lines if it's Discriminant is zero !!

$\left| \begin{matrix} 6 & -p/2 & -12 \\ -p/2 & -3 & 3/2 \\ -12 & 3 & 24 \end{matrix} \right| \quad =\quad 0\\ \\ P\quad =\quad 3/2$.

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Ok! I didn't knew to use determinant after finding q and a(alpha).. Anyways ,Nice solution.

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The line(s) intersect the x-axis when y=0. The equation then becomes 6x^2 - 24x + q = 0. There are tons of values for x that will solve this equation depending on the value of q.

If y = 0 then values for p are limitless (infinite).

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Sir but can we assume that for finding value of q the equation$6x^2-24x+q=0$ should be perfect square ? If it is then the value of q is equal to 24. This is a JEE Question .

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Are there 2 lines? I see only 1 equation.

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The equation above is multiplication of those 2 lines

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What are the 2 lines?

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If this line intersects the x axis then the value for y = 0.

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