Question

1. accurate to the nearest hundredth, the larger root of $2x^2 + 5x - 1 = 0$ is

(1) 1.85
(3) 2.47
(2) 0.19
(4) $-3.56$

Explanation:

Step1: Recall Quadratic Formula

For a quadratic equation \(ax^{2}+bx + c = 0\), the roots are given by \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\). Here, \(a = 2\), \(b = 5\), \(c=- 1\).

Step2: Calculate Discriminant

First, find the discriminant \(D=b^{2}-4ac\). Substitute the values: \(D = 5^{2}-4\times2\times(-1)=25 + 8=33\).

Step3: Find the Roots

Now, find the roots using the quadratic formula. The roots are \(x=\frac{-5\pm\sqrt{33}}{4}\). Calculate the two roots:
- For the plus sign: \(x_1=\frac{-5+\sqrt{33}}{4}\). Since \(\sqrt{33}\approx5.7446\), then \(x_1=\frac{-5 + 5.7446}{4}=\frac{0.7446}{4}\approx0.18615\).
- For the minus sign: \(x_2=\frac{-5-\sqrt{33}}{4}\). Since \(\sqrt{33}\approx5.7446\), then \(x_2=\frac{-5-5.7446}{4}=\frac{-10.7446}{4}\approx - 2.68615\).
Wait, no, wait, I made a mistake. Wait, the quadratic formula is \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\), so \(2a = 4\), \( - b=-5\). Wait, no, \(a = 2\), \(b = 5\), \(c=-1\). So \(x=\frac{-5\pm\sqrt{25+8}}{4}=\frac{-5\pm\sqrt{33}}{4}\). Wait, but \(\sqrt{33}\approx5.7446\), so \(-5+\sqrt{33}\approx - 5 + 5.7446 = 0.7446\), divided by 4 is approximately 0.186, and \(-5-\sqrt{33}\approx - 5-5.7446=-10.7446\), divided by 4 is approximately - 2.686. But that can't be right, because the options have 0.19, 1.85, etc. Wait, I think I messed up the sign of \(b\). Wait, the quadratic equation is \(2x^{2}+5x - 1=0\), so \(a = 2\), \(b = 5\), \(c=-1\). The quadratic formula is \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\), so \(-b=-5\), so \(x=\frac{-5\pm\sqrt{25 + 8}}{4}=\frac{-5\pm\sqrt{33}}{4}\). Wait, but maybe I mixed up the coefficients. Wait, no, let's re - calculate. Wait, \(\sqrt{33}\approx5.7446\), so \(-5+\sqrt{33}\approx0.7446\), divided by 4 is approximately 0.186, which is approximately 0.19 (option 2), and \(-5-\sqrt{33}\approx - 10.7446\), divided by 4 is approximately - 2.686. But the options have 1.85, 2.47, etc. Wait, I must have made a mistake in the quadratic formula. Wait, no, wait, the quadratic formula is \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\), so \(2a = 4\), but maybe I wrote the equation wrong. Wait, the equation is \(2x^{2}+5x - 1=0\), so \(a = 2\), \(b = 5\), \(c=-1\). Wait, maybe I should use the quadratic formula correctly. Wait, let's check with another method. Let's use the quadratic formula again. \(x=\frac{-5\pm\sqrt{25-4\times2\times(-1)}}{4}=\frac{-5\pm\sqrt{25 + 8}}{4}=\frac{-5\pm\sqrt{33}}{4}\). Wait, \(\sqrt{33}\approx5.7446\), so \(-5+\sqrt{33}\approx0.7446\), \(0.7446\div4\approx0.186\approx0.19\), and \(-5-\sqrt{33}\approx - 10.7446\), \(-10.7446\div4\approx - 2.686\). But the options have 0.19 (option 2), 1.85 (option 1), etc. Wait, maybe I made a mistake in the sign of \(c\). Wait, the equation is \(2x^{2}+5x - 1=0\), so \(c=-1\), so \(4ac=4\times2\times(-1)=-8\), so \(b^{2}-4ac = 25+8 = 33\), that's correct. Wait, maybe the equation is \(2x^{2}-5x - 1=0\)? No, the problem says \(2x^{2}+5x - 1=0\). Wait, let's check the options. Option (2) is 0.19, which is close to our calculated 0.186. But the other option (1) is 1.85. Wait, maybe I messed up the quadratic formula. Wait, no, the quadratic formula is \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\). Wait, if the equation was \(2x^{2}-5x - 1=0\), then \(a = 2\), \(b=-5\), \(c=-1\), then \(x=\frac{5\pm\sqrt{25 + 8}}{4}=\frac{5\pm\sqrt{33}}{4}\), then \(\frac{5+\sqrt{33}}{4}\approx\frac{5 + 5.7446}{4}=\frac{10.7446}{4}\approx2.686\), no. Wait, maybe the original equation is \(2x^{2}-5x - 1=0\)? No, the problem says \(2x^{2}+5x - 1=0\). Wait, let's recalculate \(\frac{-5+\sqrt{33}}{4}\): \(\sqrt{33}\approx5.7446\), so \(-5 + 5.7446 = 0.7446\), divided by 4 is 0.186, which is approximately 0.19 (option 2). But the option (1) is 1.85. Wait, maybe I made a mistake in the coefficient of \(x\). Wait, if the equation was \(2x^{2}-5x - 1=0\), then \(a = 2\), \(b=-5\), \(c=-1\), then \(x=\frac{5\pm\sqrt{25 + 8}}{4}=\frac{5\pm\sqrt{33}}{4}\), \(\frac{5+\sqrt{33}}{4}\approx\frac{5 + 5.7446}{4}=\frac{10.7446}{4}\approx2.686\), no. Wait, maybe the equation is \(x^{2}+5x - 2=0\)? No, the problem says \(2x^{2}+5x - 1=0\). Wait, let's check the discriminant again: \(b^{2}-4ac=25 - 4\times2\times(-1)=25 + 8 = 33\), correct. \(\sqrt{33}\approx5.7446\). Then \(\frac{-5+\sqrt{33}}{4}\approx\frac{-5 + 5.7446}{4}=\frac{0.7446}{4}\approx0.186\approx0.19\) (option 2), and \(\frac{-5-\sqrt{33}}{4}\approx - 2.686\). But the option (1) is 1.85. Wait, maybe there's a typo in my calculation. Wait, no, let's use a calculator. \(\sqrt{33}\approx5.7446\), so \(-5 + 5.7446 = 0.7446\), divided by 4: 0.7446÷4 = 0.18615, which is approximately 0.19. So the larger root is approximately 0.19, which is option (2). But wait, the options also have 1.85. Wait, maybe I misread the equation. Let me check the problem again: "the larger root of \(2x^{2}+5x - 1=0\)". Wait, maybe I made a mistake in the sign of \(b\). Wait, the quadratic formula is \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\), so if \(b = 5\), then \(-b=-5\). So the two roots are \(\frac{-5+\sqrt{33}}{4}\) and \(\frac{-5-\sqrt{33}}{4}\). The first one is approximately 0.19, the second is negative. So the larger root is approximately 0.19, which is option (2). But wait, maybe the equation is \(2x^{2}-5x + 1=0\)? Let's check: \(a = 2\), \(b=-5\), \(c = 1\), discriminant \(25-8 = 17\), \(\sqrt{17}\approx4.123\), then \(x=\frac{5\pm4.123}{4}\), so \(x_1=\frac{5 + 4.123}{4}=\frac{9.123}{4}\approx2.28\), \(x_2=\frac{5 - 4.123}{4}=\frac{0.877}{4}\approx0.219\). No, not matching. Wait, maybe the original equation is \(2x^{2}-5x - 1=0\). Let's try: \(a = 2\), \(b=-5\), \(c=-1\), discriminant \(25 + 8=33\), roots \(x=\frac{5\pm\sqrt{33}}{4}\), \(\frac{5+\sqrt{33}}{4}\approx\frac{5 + 5.7446}{4}=\frac{10.7446}{4}\approx2.686\), \(\frac{5-\sqrt{33}}{4}\approx\frac{5 - 5.7446}{4}=\frac{-0.7446}{4}\approx - 0.186\). No. Wait, the options have 0.19 (option 2), which is close to our first calculation. So maybe the answer is option (2) 0.19. But wait, the user's options: (1) 1.85, (2) 0.19, (3) 2.47, (4) - 3.56. Wait, maybe I made a mistake in the quadratic formula. Wait, let's re - derive the quadratic formula. For \(ax^{2}+bx + c=0\), \(ax^{2}+bx=-c\), \(x^{2}+\frac{b}{a}x=-\frac{c}{a}\), \(x^{2}+\frac{b}{a}x+\frac{b^{2}}{4a^{2}}=\frac{b^{2}}{4a^{2}}-\frac{c}{a}\), \((x + \frac{b}{2a})^{2}=\frac{b^{2}-4ac}{4a^{2}}\), so \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\). So that's correct. So for \(2x^{2}+5x - 1=0\), \(a = 2\), \(b = 5\), \(c=-1\), so \(x=\frac{-5\pm\sqrt{25 + 8}}{4}=\frac{-5\pm\sqrt{33}}{4}\). \(\sqrt{33}\approx5.7446\), so \(-5+\sqrt{33}\approx0.7446\), divided by 4 is approximately 0.186, which is 0.19 when rounded to the nearest hundredth. So the larger root is approximately 0.19, which is option (2).

Answer:

(2) 0.19