Question

The equation of a curve is given as y = x2 + 2-3x.
The curve intersects the x-axis at
(1) (1, 0)
(2) (2, 0)
(3) Both (1) and (2)
(4) No where



Answer = (3)
​

Answer 1

$\maltese \: \: \: \large{ \bf{ \underline{ \underline{Question \: : }}}}$

• Question says y = x² - 3x + 2 intersects x-axis at the pt.?

${ \large{ \bull }}\: \: \: \mathfrak{ \underline{ \underline{Let's \: \: understand \: \: the \: \: concept} \: \: }}_{ \bigstar \star}$

f( x ) = x² - 3x + 2 is a quadratic equation. As we know quadratic equations are always having two roots or zeros. Let's assume this graph cuts x-axis at pt

$\big(x_1,0) \: \: \& \: \: (x_2,0)$

Hence, we are seeing y or f( x ) is always 0 where it cuts the x-axis. So we can say x² - 3x + 2 would be 0 when it cuts x-axis

Now we can solve this eqn to get the pts of intersection.

Let's Solve it!!

$\longrightarrow \sf \: \: f(x) = 0$

$\longrightarrow \sf \: \: {x}^{2} - 3x + 2 = 0$

$\longrightarrow \: \: \sf {x}^{2} - 2x - x + 2 = 0$

$\longrightarrow \: \: \sf x \bigg(x - 2 \bigg) - 1 \bigg(x - 2 \bigg) = 0$

$\longrightarrow \: \sf \: \bigg(x - 1 \bigg) \bigg(x - 2 \bigg) = 0$

$\longrightarrow \: \: \underline{ \boxed{\bf x = 1 \: \: \: \& \: \: \: \: x = 2} \: \: }_{ \bigstar \star}$

$\therefore \sf \: The \: \: graph \: \: cuts \: \: x-axis \: \: at \: \: pt. \big(1,0 \big) \: \& \: \big(2,0 \big)$